https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Sweetmango77&feedformat=atom AoPS Wiki - User contributions [en] 2021-01-22T18:56:48Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_14&diff=142745 2020 AMC 10A Problems/Problem 14 2021-01-19T15:34:46Z <p>Sweetmango77: Video solution goes at the back</p> <hr /> <div>== Problem ==<br /> Real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; satisfy &lt;math&gt;x + y = 4&lt;/math&gt; and &lt;math&gt;x \cdot y = -2&lt;/math&gt;. What is the value of <br /> <br /> &lt;cmath&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 360\qquad\textbf{(B)}\ 400\qquad\textbf{(C)}\ 420\qquad\textbf{(D)}\ 440\qquad\textbf{(E)}\ 480&lt;/math&gt;<br /> <br /> == Solutions ==<br /> <br /> === Solution 1 ===<br /> &lt;cmath&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y=x+\frac{x^3}{y^2}+y+\frac{y^3}{x^2}=\frac{x^3}{x^2}+\frac{y^3}{x^2}+\frac{y^3}{y^2}+\frac{x^3}{y^2}&lt;/cmath&gt;<br /> <br /> Continuing to combine &lt;cmath&gt;\frac{x^3+y^3}{x^2}+\frac{x^3+y^3}{y^2}=\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}=\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}&lt;/cmath&gt;<br /> From the givens, it can be concluded that &lt;math&gt;x^2y^2=4&lt;/math&gt;. Also, &lt;cmath&gt;(x+y)^2=x^2+2xy+y^2=16&lt;/cmath&gt; This means that &lt;math&gt;x^2+y^2=20&lt;/math&gt;. Substituting this information into &lt;math&gt;\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}&lt;/math&gt;, we have &lt;math&gt;\frac{(20)(4)(22)}{4}=20\cdot 22=\boxed{\textbf{(D)}\ 440}&lt;/math&gt;. ~PCChess<br /> <br /> === Solution 2 ===<br /> As above, we need to calculate &lt;math&gt;\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}&lt;/math&gt;. Note that &lt;math&gt;x,y,&lt;/math&gt; are the roots of &lt;math&gt;x^2-4x-2&lt;/math&gt; and so &lt;math&gt;x^3=4x^2+2x&lt;/math&gt; and &lt;math&gt;y^3=4y^2+2y&lt;/math&gt;. Thus &lt;math&gt;x^3+y^3=4(x^2+y^2)+2(x+y)=4(20)+2(4)=88&lt;/math&gt; where &lt;math&gt;x^2+y^2=20&lt;/math&gt; and &lt;math&gt;x^2y^2=4&lt;/math&gt; as in the previous solution. Thus the answer is &lt;math&gt;\frac{(20)(88)}{4}=\boxed{\textbf{(D)}\ 440}&lt;/math&gt;. Note(&lt;math&gt;x^2+y^2=(x+y)^2-2xy=20&lt;/math&gt;, and &lt;math&gt;x^2y^2 = (xy)^2 = 4&lt;/math&gt;)<br /> <br /> === Solution 3 ===<br /> Note that &lt;math&gt;( x^3 + y^3 ) ( \frac{1}{y^2} + \frac{1}{x^2} ) = x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y.&lt;/math&gt; Now, we only need to find the values of &lt;math&gt;x^3 + y^3&lt;/math&gt; and &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2}.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Recall that &lt;math&gt;x^3 + y^3 = (x + y) (x^2 - xy + y^2),&lt;/math&gt; and that &lt;math&gt;x^2 - xy + y^2 = (x + y)^2 - 3xy.&lt;/math&gt; We are able to solve the second equation, and doing so gets us &lt;math&gt;4^2 - 3(-2) = 22.&lt;/math&gt; Plugging this into the first equation, we get &lt;math&gt;x^3 + y^3 = 4(22) = 88.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> In order to find the value of &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2},&lt;/math&gt; we find a common denominator so that we can add them together. This gets us &lt;math&gt;\frac{x^2}{x^2 y^2} + \frac{y^2}{x^2 y^2} = \frac{x^2 + y^2}{(xy)^2}.&lt;/math&gt; Recalling that &lt;math&gt;x^2 + y^2 = (x+y)^2 - 2xy&lt;/math&gt; and solving this equation, we get &lt;math&gt;4^2 - 2(-2) = 20.&lt;/math&gt; Plugging this into the first equation, we get &lt;math&gt;\frac{1}{y^2} + \frac{1}{x^2} = \frac{20}{(-2)^2} = 5.&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Solving the original equation, we get &lt;math&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y = (88)(5) = \boxed{\textbf{(D)}\ 440}.&lt;/math&gt; ~[[User:emerald_block|emerald_block]]<br /> <br /> === Solution 4 (Bashing) ===<br /> This is basically bashing using Vieta's formulas to find &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; (which I highly do not recommend, I only wrote this solution for fun).<br /> <br /> We use Vieta's to find a quadratic relating &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt;. We set &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; to be the roots of the quadratic &lt;math&gt; Q ( n ) = n^2 - 4n - 2 &lt;/math&gt; (because &lt;math&gt; x + y = 4 &lt;/math&gt;, and &lt;math&gt; xy = -2 &lt;/math&gt;). We can solve the quadratic to get the roots &lt;math&gt; 2 + \sqrt{6} &lt;/math&gt; and &lt;math&gt; 2 - \sqrt{6} &lt;/math&gt;. &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; y &lt;/math&gt; are &quot;interchangeable&quot;, meaning that it doesn't matter which solution &lt;math&gt; x &lt;/math&gt; or &lt;math&gt; y &lt;/math&gt; is, because it'll return the same result when plugged in. So we plug in &lt;math&gt; 2 + \sqrt{6} &lt;/math&gt; for &lt;math&gt; x &lt;/math&gt; and &lt;math&gt; 2 - \sqrt{6} &lt;/math&gt; and get &lt;math&gt; \boxed{\textbf{(D)}\ 440} &lt;/math&gt; as our answer.<br /> <br /> ~Baolan<br /> <br /> === Solution 5 (Bashing Part 2) ===<br /> This usually wouldn't work for most problems like this, but we're lucky that we can quickly expand and factor this expression in this question.<br /> <br /> We first change the original expression to &lt;math&gt;4 + \frac{x^5 + y^5}{x^2 y^2}&lt;/math&gt;, because &lt;math&gt;x + y = 4&lt;/math&gt;. This is equal to &lt;math&gt;4 + \frac{(x + y)(x^4 - x^3 y + x^2 y^2 - x y^3 + y^4)}{4} = x^4 + y^4 - x^3 y - x y^3 + 8&lt;/math&gt;. We can factor and reduce &lt;math&gt;x^4 + y^4&lt;/math&gt; to &lt;math&gt;(x^2 + y^2)^2 - 2 x^2 y^2 = ((x + y)^2 - 2xy)^2 - 8 = 400 - 8 = 392&lt;/math&gt;. Now our expression is just &lt;math&gt;400 - (x^3 y + x y^3)&lt;/math&gt;. We factor &lt;math&gt;x^3 y + x y^3&lt;/math&gt; to get &lt;math&gt;(xy)(x^2 + y^2) = -40&lt;/math&gt;. So the answer would be &lt;math&gt;400 - (-40)<br /> = \boxed{\textbf{(D)} 440} &lt;/math&gt;.<br /> <br /> === Solution 6 (Complete Binomial Theorem) ===<br /> <br /> We first simplify the expression to &lt;cmath&gt;x + y + \frac{x^5 + y^5}{x^2y^2}.&lt;/cmath&gt;<br /> Then, we can solve for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; given the system of equations in the problem.<br /> Since &lt;math&gt;xy = -2,&lt;/math&gt; we can substitute &lt;math&gt;\frac{-2}{x}&lt;/math&gt; for &lt;math&gt;y&lt;/math&gt;. <br /> Thus, this becomes the equation &lt;cmath&gt;x - \frac{2}{x} = 4.&lt;/cmath&gt;<br /> Multiplying both sides by &lt;math&gt;x&lt;/math&gt;, we obtain &lt;math&gt;x^2 - 2 = 4x,&lt;/math&gt; or <br /> &lt;cmath&gt;x^2 - 4x - 2 = 0.&lt;/cmath&gt;<br /> By the quadratic formula we obtain &lt;math&gt;x = 2 \pm \sqrt{6}&lt;/math&gt;. <br /> We also easily find that given &lt;math&gt;x = 2 \pm \sqrt{6}&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; equals the conjugate of &lt;math&gt;x&lt;/math&gt;. <br /> Thus, plugging our values in for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, our expression equals<br /> &lt;cmath&gt;4 + \frac{(2 - \sqrt{6})^5 + (2 + \sqrt{6})^5}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}&lt;/cmath&gt;<br /> By the binomial theorem, we observe that every second terms of the expansions &lt;math&gt;x^5&lt;/math&gt; and &lt;math&gt;y^5&lt;/math&gt; will cancel out (since a positive plus a negative of the same absolute value equals zero). We also observe that the other terms not canceling out are doubled when summing the expansions of &lt;math&gt;x^5 + y^5&lt;/math&gt;.<br /> Thus, our expression equals<br /> &lt;cmath&gt;4 + \frac{2(2^5 + \tbinom{5}{2}2^3 \times 6 + \tbinom{5}{4}2 \times 36)}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}.&lt;/cmath&gt;<br /> which equals<br /> &lt;cmath&gt;4 + \frac{2(872)}{4}&lt;/cmath&gt;<br /> which equals &lt;math&gt;\boxed{\textbf{(D)} 440}&lt;/math&gt;.<br /> <br /> ~ fidgetboss_4000<br /> <br /> === Solution 7 ===<br /> As before, simplify the expression to &lt;cmath&gt;x + y + \frac{x^5 + y^5}{x^2y^2}.&lt;/cmath&gt;<br /> Since &lt;math&gt;x + y = 4&lt;/math&gt; and &lt;math&gt;x^2y^2 = 4&lt;/math&gt;, we substitute that in to obtain &lt;cmath&gt; 4 + \frac{x^5 + y^5}{4}.&lt;/cmath&gt;<br /> Now, we must solve for &lt;math&gt;x^5 + y^5&lt;/math&gt;. Start by squaring &lt;math&gt;x + y&lt;/math&gt;, to obtain &lt;cmath&gt;x^2 + 2xy + y^2 = 16&lt;/cmath&gt;<br /> Simplifying, &lt;math&gt;x^2 + y^2 = 20&lt;/math&gt;. Squaring once more, we obtain &lt;cmath&gt;x^4 + y^4 + 2x^2y^2 = 400&lt;/cmath&gt;<br /> Once again simplifying, &lt;math&gt;x^4 + y^4 = 392&lt;/math&gt;. Now, to obtain the fifth powers of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, we multiply both sides by &lt;math&gt;x + y&lt;/math&gt;.<br /> We now have <br /> &lt;cmath&gt;x^5 + x^4y + xy^4 + y^5 = 1568&lt;/cmath&gt;, or <br /> &lt;cmath&gt;x^5 + y^5 + xy(x^3 + y^3) = 1568&lt;/cmath&gt;<br /> We now solve for &lt;math&gt;x^3 + y^3&lt;/math&gt;. &lt;math&gt;(x + y)^3=x^3 + y^3 + 3xy(x + y) = 64&lt;/math&gt;, so &lt;math&gt;x^3 + y^3 = 88&lt;/math&gt;. <br /> Plugging this back into&lt;math&gt;x^5 + x^4y + xy^4 + y^5 = 1568&lt;/math&gt;, we find that &lt;math&gt;x^5 + y^5 = 1744&lt;/math&gt;, so we have &lt;cmath&gt; 4 + \frac{1744}{4}.&lt;/cmath&gt;. This equals 440, so our answer is &lt;math&gt;\boxed{\textbf{(D)} 440}&lt;/math&gt;.<br /> <br /> ~Binderclips1<br /> <br /> === Solution 8 ===<br /> We can use Newton Sums to solve this problem.<br /> We start by noticing that we can rewrite the equation as &lt;math&gt;\frac{x^3}{y^2} + \frac{y^3}{x^2} + x + y.&lt;/math&gt;<br /> Then, we know that &lt;math&gt;x + y = 4,&lt;/math&gt; so we have &lt;math&gt;\frac{x^3}{y^2} + \frac{y^3}{x^2} + 4.&lt;/math&gt;<br /> We can use the equation &lt;math&gt;x \cdot y = -2&lt;/math&gt; to write &lt;math&gt;x = \frac{-2}{y}&lt;/math&gt; and &lt;math&gt;y = \frac{-2}{x}.&lt;/math&gt;<br /> Next, we can plug in these values of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; to get &lt;math&gt;\frac{x^3}{y^2} + \frac{y^3}{x^2} = \frac{x^5}{4} + \frac{y^5}{4},&lt;/math&gt; which is the same as &lt;cmath&gt;\frac{x^3}{y^2} + \frac{y^3}{x^2} = \frac{x^5 + y^5}{4}.&lt;/cmath&gt;<br /> Then, we use Newton sums where &lt;math&gt;S_n&lt;/math&gt; is the elementary symmetric sum of the sequence and &lt;math&gt;P_n&lt;/math&gt; is the power sum (&lt;math&gt;x^n + y^n&lt;/math&gt;). Using this, we can make the following Newton sums:<br /> &lt;cmath&gt;P_1 = S_1&lt;/cmath&gt;<br /> &lt;cmath&gt;P_2 = P_1 S_1 - 2S_2&lt;/cmath&gt;<br /> &lt;cmath&gt;P_3 = P_2 S_1 - P_1 S_2&lt;/cmath&gt;<br /> &lt;cmath&gt;P_4 = P_3 S_1 - P_2 S_2&lt;/cmath&gt;<br /> &lt;cmath&gt;P_5 = P_4 S_1 - P_3 S_2.&lt;/cmath&gt;<br /> We also know that &lt;math&gt;S_1&lt;/math&gt; is 4 because &lt;math&gt;x + y&lt;/math&gt; is four, and we know that &lt;math&gt;S_2&lt;/math&gt; is &lt;math&gt;-2&lt;/math&gt; because &lt;math&gt;x \cdot y&lt;/math&gt; is &lt;math&gt;-2&lt;/math&gt; as well.<br /> Then, we can plug in values! We have<br /> &lt;cmath&gt;P_1 = S_1 = 4&lt;/cmath&gt;<br /> &lt;cmath&gt;P_2 = P_1 S_1 - 2S_2 = 16 - (-4) = 20&lt;/cmath&gt;<br /> &lt;cmath&gt;P_3 = P_2 S_1 - P_1 S_2 = 80 - (-8) = 88&lt;/cmath&gt;<br /> &lt;cmath&gt;P_4 = P_3 S_1 - P_2 S_2 = 88 \cdot 4 - (-40) = 392&lt;/cmath&gt;<br /> &lt;cmath&gt;P_5 = P_4 S_1 - P_3 S_2 = 392 \cdot 4 - (-2) \cdot 88 = 1744.&lt;/cmath&gt;<br /> We earlier noted that &lt;math&gt;\frac{x^3}{y^2} + \frac{y^3}{x^2} = \frac{x^5 + y^5}{4},&lt;/math&gt; so we have that this equals &lt;math&gt;\frac{1744}{4},&lt;/math&gt; or &lt;math&gt;436.&lt;/math&gt; Then, plugging this back into the original equation, this is &lt;math&gt;436 + 4&lt;/math&gt; or &lt;math&gt;440,&lt;/math&gt; so our answer is &lt;math&gt;\boxed{\textbf{(D)}\ 440}.&lt;/math&gt;<br /> <br /> ~Coolpeep<br /> <br /> === Solution 9 ===<br /> As in the first solution, we get the expression to be &lt;math&gt;\frac{x^3+y^3}{x^2} + \frac{x^3+y^3}{y^2}.&lt;/math&gt;<br /> <br /> Then, since the numerators are the same, we can put the two fractions as a common denominator and multiply the numerator by &lt;math&gt;x^2y^2.&lt;/math&gt; This gets us &lt;math&gt;\frac{(x^2 + y^2)(x^3 + y^3)}{x^2y^2}.&lt;/math&gt;<br /> <br /> Now, since we know &lt;math&gt;x+y=4&lt;/math&gt; and &lt;math&gt;xy=-2,&lt;/math&gt; instead of solving for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y,&lt;/math&gt; we will try to manipulate the above expression them into a manner that we can substitute the sum and product that we know. Also, another form of &lt;math&gt;x^3+y^3&lt;/math&gt; is &lt;math&gt;(x+y)(x^2-xy+y^2).&lt;/math&gt;<br /> <br /> Thus, we can convert the current expression to &lt;math&gt;\frac{(x^2 + y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}.&lt;/math&gt;<br /> <br /> Doing some algebraic multiplications, we get &lt;math&gt;\frac{((x+y)^2 - 2xy)(x+y)((x+y)^2 - 3xy)}{(xy)^2}.&lt;/math&gt;<br /> <br /> Since we know &lt;math&gt;x+y=4&lt;/math&gt; and &lt;math&gt;xy=-2,&lt;/math&gt; we have &lt;math&gt;\frac{(16-(-4))(4)(16-(-6))}{4} = \frac{20 \cdot 4 \cdot 22}{4} = 20 \cdot 22 = 440.&lt;/math&gt;<br /> <br /> Therefore the answer is &lt;math&gt;\boxed{\textbf{(D)} 440}.&lt;/math&gt;<br /> <br /> ~mathboy282<br /> <br /> <br /> <br /> === Video Solution ===<br /> <br /> Education, The Study of Everything<br /> <br /> https://youtu.be/PNkRlUKWCzg<br /> <br /> https://youtu.be/ZGwAasE32Y4<br /> <br /> ~IceMatrix<br /> <br /> https://youtu.be/XEtzvxfFEJk<br /> <br /> ~savannahsolver<br /> <br /> https://youtu.be/ba6w1OhXqOQ?t=3551<br /> <br /> ~ pi_is_3.14<br /> <br /> <br /> == See Also ==<br /> {{AMC10 box|year=2020|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_15&diff=142739 2009 AMC 10B Problems/Problem 15 2021-01-19T15:31:36Z <p>Sweetmango77: /* Solution 4 */</p> <hr /> <div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #15]] and [[2009 AMC 12B Problems|2009 AMC 12B #8]]}}<br /> <br /> == Problem ==<br /> When a bucket is two-thirds full of water, the bucket and water weigh &lt;math&gt;a&lt;/math&gt; kilograms. When the bucket is one-half full of water the total weight is &lt;math&gt;b&lt;/math&gt; kilograms. In terms of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, what is the total weight in kilograms when the bucket is full of water?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac23a + \frac13b\qquad<br /> \mathrm{(B)}\ \frac32a - \frac12b\qquad<br /> \mathrm{(C)}\ \frac32a + b\qquad<br /> \mathrm{(D)}\ \frac32a + 2b\qquad<br /> \mathrm{(E)}\ 3a - 2b&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Let &lt;math&gt;x&lt;/math&gt; be the weight of the bucket and let &lt;math&gt;y&lt;/math&gt; be the weight of the water in a full bucket. Then we are given that &lt;math&gt;x + \frac 23y = a&lt;/math&gt; and &lt;math&gt;x + \frac 12y = b&lt;/math&gt;. Hence &lt;math&gt;\frac 16y = a-b&lt;/math&gt;, so &lt;math&gt;y = 6a-6b&lt;/math&gt;. Thus &lt;math&gt;x = b - \frac 12 (6a-6b) = -3a + 4b&lt;/math&gt;. Finally &lt;math&gt;x + y = \boxed {3a-2b}&lt;/math&gt;. The answer is &lt;math&gt;\mathrm{(E)}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Imagine that we take three buckets of the first type, to get rid of the fraction. We will have three buckets and two buckets' worth of water.<br /> <br /> On the other hand, if we take two buckets of the second type, we will have two buckets and enough water to fill one bucket.<br /> <br /> The difference between these is exactly one bucket full of water, hence the answer is &lt;math&gt;3a-2b&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> We are looking for an expression of the form &lt;math&gt;xa + yb&lt;/math&gt;.<br /> <br /> We must have &lt;math&gt;x+y=1&lt;/math&gt;, as the desired result contains exactly one bucket. Also, we must have &lt;math&gt;\frac 23 x + \frac 12 y = 1&lt;/math&gt;, as the desired result contains exactly one bucket of water.<br /> <br /> At this moment, it is easiest to check that only the options (A), (B), and (E) satisfy &lt;math&gt;x+y=1&lt;/math&gt;, and out of these only (E) satisfies the second equation.<br /> <br /> Alternatively, we can directly solve the system, getting &lt;math&gt;x=3&lt;/math&gt; and &lt;math&gt;y=-2&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; is one bucket plus two-thirds of the total amount of water, and &lt;math&gt;b&lt;/math&gt; is one bucket plus one-half of the total amount of water, &lt;math&gt;a-b&lt;/math&gt; would equal &lt;math&gt;\cancel{\text{bucket}}+\frac23\cdot\text{water}-(\cancel{\text{bucket}}+\frac12\cdot\text{water})=\frac16\cdot\text{water}&lt;/math&gt;. Therefore, &lt;math&gt;a-b&lt;/math&gt; is one-sixth of the total mass of the water. Starting from &lt;math&gt;a&lt;/math&gt;, we add two one-sixths of the total amount of the water to become full. Therefore, the full bucket of water is &lt;math&gt;a+2(a-b)=a+2a-2b=3a-2b\Rightarrow\textbf{(E)}&lt;/math&gt;<br /> <br /> -[[User:Sweetmango77|SweetMango77]]<br /> <br /> == Video Solution ==<br /> https://youtu.be/fXXSk8zUXgo<br /> <br /> == See also ==<br /> {{AMC10 box|year=2009|ab=B|num-b=14|num-a=16}}<br /> {{AMC12 box|year=2009|ab=B|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=User:Sweetmango77&diff=142738 User:Sweetmango77 2021-01-19T15:30:47Z <p>Sweetmango77: /* SweetMango77: */</p> <hr /> <div>==&lt;font color=&quot;black&quot; style=&quot;font-family: Times New Roman&quot;&gt;SweetMango77:&lt;/font&gt;==<br /> <br /> &lt;font color=&quot;black&quot; style=&quot;font-family: Times New Roman&quot;&gt;<br /> A person who made a 25x100 multiplication chart: https://docs.google.com/spreadsheets/d/1Tdy1sz8y7LuO9yqADUnAXZ583IqnFlmTM88XsaX1W6g/edit#gid=0<br /> <br /> Please friend me in Hypixel, my username is &lt;i&gt;Dat_Is_Stupid&lt;/i&gt;.<br /> <br /> Also good at coding with HTML, CSS, and JavaScript.<br /> &lt;hr&gt;<br /> PM me if you would like to edit this, otherwise I will undo all of your changes.<br /> <br /> &lt;hr&gt;<br /> <br /> &lt;math&gt;\Huge{\textbf{BYE}}&lt;/math&gt;<br /> &lt;/font&gt;</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=2020_CIME_II_Problems/Problem_12&diff=142737 2020 CIME II Problems/Problem 12 2021-01-19T15:27:36Z <p>Sweetmango77: </p> <hr /> <div>Note that the &lt;math&gt;\text{lcm}(\text{gcd}(a,b),c)&lt;/math&gt; and its iterations are all divisible by 180. This implies that 2 of &lt;math&gt;a,b,c&lt;/math&gt; are divisible by 4, 2 are divisible by 9 and 2 are divisible by 5. <br /> &lt;cmath&gt;a,b,c = d 20, e 36, f 45&lt;/cmath&gt;<br /> <br /> Next we note that iterations are 180,&lt;math&gt;2\times 180&lt;/math&gt;,&lt;math&gt;3 \times 180&lt;/math&gt;. This implies that &lt;math&gt;d&lt;/math&gt; or &lt;math&gt;e&lt;/math&gt; must have an additional factor of 2 and &lt;math&gt;e&lt;/math&gt; or &lt;math&gt;f&lt;/math&gt; must have an additional factor of 3. The sum is minimized if d=2 and e=3 and f=1.<br /> <br /> &lt;cmath&gt;a,b,c = 40,45,108&lt;/cmath&gt;<br /> &lt;cmath&gt;a+b+c = 193&lt;/cmath&gt;</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_15&diff=142705 2009 AMC 10B Problems/Problem 15 2021-01-19T02:40:34Z <p>Sweetmango77: /* Video Solution */</p> <hr /> <div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #15]] and [[2009 AMC 12B Problems|2009 AMC 12B #8]]}}<br /> <br /> == Problem ==<br /> When a bucket is two-thirds full of water, the bucket and water weigh &lt;math&gt;a&lt;/math&gt; kilograms. When the bucket is one-half full of water the total weight is &lt;math&gt;b&lt;/math&gt; kilograms. In terms of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, what is the total weight in kilograms when the bucket is full of water?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac23a + \frac13b\qquad<br /> \mathrm{(B)}\ \frac32a - \frac12b\qquad<br /> \mathrm{(C)}\ \frac32a + b\qquad<br /> \mathrm{(D)}\ \frac32a + 2b\qquad<br /> \mathrm{(E)}\ 3a - 2b&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Let &lt;math&gt;x&lt;/math&gt; be the weight of the bucket and let &lt;math&gt;y&lt;/math&gt; be the weight of the water in a full bucket. Then we are given that &lt;math&gt;x + \frac 23y = a&lt;/math&gt; and &lt;math&gt;x + \frac 12y = b&lt;/math&gt;. Hence &lt;math&gt;\frac 16y = a-b&lt;/math&gt;, so &lt;math&gt;y = 6a-6b&lt;/math&gt;. Thus &lt;math&gt;x = b - \frac 12 (6a-6b) = -3a + 4b&lt;/math&gt;. Finally &lt;math&gt;x + y = \boxed {3a-2b}&lt;/math&gt;. The answer is &lt;math&gt;\mathrm{(E)}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Imagine that we take three buckets of the first type, to get rid of the fraction. We will have three buckets and two buckets' worth of water.<br /> <br /> On the other hand, if we take two buckets of the second type, we will have two buckets and enough water to fill one bucket.<br /> <br /> The difference between these is exactly one bucket full of water, hence the answer is &lt;math&gt;3a-2b&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> We are looking for an expression of the form &lt;math&gt;xa + yb&lt;/math&gt;.<br /> <br /> We must have &lt;math&gt;x+y=1&lt;/math&gt;, as the desired result contains exactly one bucket. Also, we must have &lt;math&gt;\frac 23 x + \frac 12 y = 1&lt;/math&gt;, as the desired result contains exactly one bucket of water.<br /> <br /> At this moment, it is easiest to check that only the options (A), (B), and (E) satisfy &lt;math&gt;x+y=1&lt;/math&gt;, and out of these only (E) satisfies the second equation.<br /> <br /> Alternatively, we can directly solve the system, getting &lt;math&gt;x=3&lt;/math&gt; and &lt;math&gt;y=-2&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; is one bucket plus two-thirds of the total amount of water, and &lt;math&gt;b&lt;/math&gt; is one bucket plus one-half of the total amount of water, &lt;math&gt;a-b&lt;/math&gt; would equal &lt;math&gt;\cancel{\text{bucket}}+\frac23\cdot\text{water}-(\cancel{\text{bucket}}+\frac12\cdot\text{water})=\frac16\cdot\text{water}&lt;/math&gt;. Therefore, &lt;math&gt;a-b&lt;/math&gt; is one-sixth of the total mass of the water. Starting from &lt;math&gt;a&lt;/math&gt;, we add two one-sixths of the total amount of the water to become full. Therefore, the full bucket of water is &lt;math&gt;a+2(a-b)=a+2a-2b=3a-2b\Rightarrow\textbf{(E)}&lt;/math&gt;<br /> <br /> == Video Solution ==<br /> https://youtu.be/fXXSk8zUXgo<br /> <br /> == See also ==<br /> {{AMC10 box|year=2009|ab=B|num-b=14|num-a=16}}<br /> {{AMC12 box|year=2009|ab=B|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_15&diff=142704 2009 AMC 10B Problems/Problem 15 2021-01-19T02:39:57Z <p>Sweetmango77: Solution 3</p> <hr /> <div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #15]] and [[2009 AMC 12B Problems|2009 AMC 12B #8]]}}<br /> <br /> == Problem ==<br /> When a bucket is two-thirds full of water, the bucket and water weigh &lt;math&gt;a&lt;/math&gt; kilograms. When the bucket is one-half full of water the total weight is &lt;math&gt;b&lt;/math&gt; kilograms. In terms of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, what is the total weight in kilograms when the bucket is full of water?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ \frac23a + \frac13b\qquad<br /> \mathrm{(B)}\ \frac32a - \frac12b\qquad<br /> \mathrm{(C)}\ \frac32a + b\qquad<br /> \mathrm{(D)}\ \frac32a + 2b\qquad<br /> \mathrm{(E)}\ 3a - 2b&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Let &lt;math&gt;x&lt;/math&gt; be the weight of the bucket and let &lt;math&gt;y&lt;/math&gt; be the weight of the water in a full bucket. Then we are given that &lt;math&gt;x + \frac 23y = a&lt;/math&gt; and &lt;math&gt;x + \frac 12y = b&lt;/math&gt;. Hence &lt;math&gt;\frac 16y = a-b&lt;/math&gt;, so &lt;math&gt;y = 6a-6b&lt;/math&gt;. Thus &lt;math&gt;x = b - \frac 12 (6a-6b) = -3a + 4b&lt;/math&gt;. Finally &lt;math&gt;x + y = \boxed {3a-2b}&lt;/math&gt;. The answer is &lt;math&gt;\mathrm{(E)}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Imagine that we take three buckets of the first type, to get rid of the fraction. We will have three buckets and two buckets' worth of water.<br /> <br /> On the other hand, if we take two buckets of the second type, we will have two buckets and enough water to fill one bucket.<br /> <br /> The difference between these is exactly one bucket full of water, hence the answer is &lt;math&gt;3a-2b&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> We are looking for an expression of the form &lt;math&gt;xa + yb&lt;/math&gt;.<br /> <br /> We must have &lt;math&gt;x+y=1&lt;/math&gt;, as the desired result contains exactly one bucket. Also, we must have &lt;math&gt;\frac 23 x + \frac 12 y = 1&lt;/math&gt;, as the desired result contains exactly one bucket of water.<br /> <br /> At this moment, it is easiest to check that only the options (A), (B), and (E) satisfy &lt;math&gt;x+y=1&lt;/math&gt;, and out of these only (E) satisfies the second equation.<br /> <br /> Alternatively, we can directly solve the system, getting &lt;math&gt;x=3&lt;/math&gt; and &lt;math&gt;y=-2&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; is one bucket plus two-thirds of the total amount of water, and &lt;math&gt;b&lt;/math&gt; is one bucket plus one-half of the total amount of water, &lt;math&gt;a-b&lt;/math&gt; would equal &lt;math&gt;\cancel{\text{bucket}}+\frac23\cdot\text{water}-(\cancel{\text{bucket}}+\frac12\cdot\text{water})=\frac16\cdot\text{water}&lt;/math&gt;. Therefore, &lt;math&gt;a-b&lt;/math&gt; is one-sixth of the total mass of the water. Starting from &lt;math&gt;a&lt;/math&gt;, we add two one-sixths of the total amount of the water to become full. Therefore, the full bucket of water is &lt;math&gt;a+2(a-b)=a+2a-2b=3a-2b\Rightarrow\textbf{(E)}&lt;/math&gt;<br /> <br /> == Video Solution ==<br /> https://youtu.be/fXXSk8zUXgo<br /> <br /> ~savannahsolver<br /> <br /> == See also ==<br /> {{AMC10 box|year=2009|ab=B|num-b=14|num-a=16}}<br /> {{AMC12 box|year=2009|ab=B|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=Georgeooga-Harryooga_Theorem&diff=142032 Georgeooga-Harryooga Theorem 2021-01-14T02:13:16Z <p>Sweetmango77: Undo revision 142028 by Sugar rush (talk)</p> <hr /> <div>=Definition=<br /> The Georgeooga-Harryooga Theorem states that if you have &lt;math&gt;a&lt;/math&gt; distinguishable objects and &lt;math&gt;b&lt;/math&gt; are kept away from each other, then there are &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects.<br /> <br /> <br /> Created by George and Harry of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]<br /> <br /> =Proofs=<br /> ==Proof 1==<br /> Let our group of &lt;math&gt;a&lt;/math&gt; objects be represented like so &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, ..., &lt;math&gt;a-1&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;. Let the last &lt;math&gt;b&lt;/math&gt; objects be the ones we can't have together.<br /> <br /> Then we can organize our objects like so &lt;math&gt;\square1\square2\square3\square...\square a-b-1\square a-b\square&lt;/math&gt;.<br /> <br /> We have &lt;math&gt;(a-b)!&lt;/math&gt; ways to arrange the objects in that list.<br /> <br /> Now we have &lt;math&gt;a-b+1&lt;/math&gt; blanks and &lt;math&gt;b&lt;/math&gt; other objects so we have &lt;math&gt;_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects we can't put together.<br /> <br /> By fundamental counting principal our answer is &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt;.<br /> <br /> <br /> Proof by [[User:RedFireTruck|RedFireTruck]]<br /> ==Proof 2==<br /> Let us call the &lt;math&gt;b&lt;/math&gt; people &lt;math&gt;1, 2, ... b&lt;/math&gt;<br /> <br /> Let the number of people before &lt;math&gt;1&lt;/math&gt; in line be &lt;math&gt;y_1&lt;/math&gt;, between &lt;math&gt;1, 2&lt;/math&gt; be &lt;math&gt;y_2&lt;/math&gt;, ... after &lt;math&gt;b&lt;/math&gt; be &lt;math&gt;y_{b+1}&lt;/math&gt;.<br /> We have &lt;cmath&gt;y_1 + y_2 + y_3 + \dots y_{b+1} = a-b&lt;/cmath&gt;<br /> <br /> The number of ways to determine &lt;math&gt;y_1, y_2, \dots&lt;/math&gt; is equivalent to the number of positive integer solutions to:<br /> &lt;cmath&gt;x_1 + x_2 + .. + x_{b+1} = a-b + 2&lt;/cmath&gt; where &lt;math&gt;(x_2, ... x_b) = (y_2, ..., y_b) &lt;/math&gt; and &lt;math&gt;(x_1, x_{b+1}) = (y_1 +1, y_{b+1} + 1)&lt;/math&gt;.<br /> <br /> So, by stars and bars, the number of ways to determine &lt;math&gt;(y_2, ..., y_b) &lt;/math&gt; is &lt;cmath&gt;F(a,b) = \dbinom{a-b+1}{b} = \frac {(a-b+1)!}{b!(a-2b+1)!}&lt;/cmath&gt;<br /> <br /> Furthermore, after picking positions for the people, we have &lt;math&gt;(a-b)!&lt;/math&gt; ways to order the &lt;math&gt;(a-b)&lt;/math&gt; people who can be together, and &lt;math&gt;b!&lt;/math&gt; ways to order the &lt;math&gt;b&lt;/math&gt; people who cannot be together. So for each &lt;math&gt;(y_1, y_2, ... y_{b+1}&lt;/math&gt;, we have &lt;math&gt;b! (a-b)!&lt;/math&gt; orderings.<br /> <br /> Therefore, the final answer is &lt;cmath&gt;b! (a-b)! F(a,b) = \frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/cmath&gt;<br /> <br /> <br /> Proof by Aryabhata000<br /> <br /> =Applications=<br /> ==Application 1==<br /> ===Problem===<br /> Alice, Bob, Carl, David, Eric, Fred, George, and Harry want to stand in a line to buy ice cream.<br /> Fred and George are identical twins, so they are indistinguishable.<br /> Alice, Bob, and Carl had a serious disagreement in 6th grade, so none of them can be together in the line.<br /> <br /> With these conditions, how many different ways can you arrange these kids in a line?<br /> <br /> <br /> Problem by [https://artofproblemsolving.com/community/c4h2342517p18900597 Math4Life2020]<br /> <br /> ===Solutions===<br /> ====Solution 1====<br /> If Eric and Fred were distinguishable we would have &lt;math&gt;\frac{(8-3)!(8-3+1)!}{(8-2\cdot3+1)!}=14400&lt;/math&gt; ways to arrange them by the [[Georgeooga-Harryooga Theorem]]. However, Eric and Fred are indistinguishable so we have to divide by &lt;math&gt;2!=2&lt;/math&gt;. Therefore, our answer is &lt;math&gt;\frac{14400}2=\boxed{7200}&lt;/math&gt;.<br /> <br /> <br /> Solution by [[User:RedFireTruck|RedFireTruck]]<br /> <br /> ==Application 2==<br /> ===Problem===<br /> Zara has a collection of &lt;math&gt;4&lt;/math&gt; marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> <br /> Problem by [[2020 AMC 8 Problems/Problem 10|The Mathematical Association of America's American Mathematics Competitions]]<br /> <br /> ===Solutions===<br /> ====Solution 1====<br /> By the [[Georgeooga-Harryooga Theorem]] there are &lt;math&gt;\frac{(4-2)!(4-2+1)!}{(4-2\cdot2+1)!}=\boxed{\textbf{(C) }12}&lt;/math&gt; way to arrange the marbles.<br /> <br /> <br /> Solution by [[User:RedFireTruck|RedFireTruck]]<br /> <br /> ====Solution 2====<br /> We can arrange our marbles like so &lt;math&gt;\square A\square B\square&lt;/math&gt;.<br /> <br /> To arrange the &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; we have &lt;math&gt;2!=2&lt;/math&gt; ways.<br /> <br /> To place the &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; in the blanks we have &lt;math&gt;_3P_2=6&lt;/math&gt; ways.<br /> <br /> By fundamental counting principle our final answer is &lt;math&gt;2\cdot6=\boxed{\textbf{(C) }12}&lt;/math&gt;<br /> <br /> <br /> Solution by [[User:Redfiretruck|RedFireTruck]]<br /> <br /> ====Solution 3====<br /> Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by &lt;math&gt;A,B,S,&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt;, respectively. If we ignore the constraint that &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; cannot be next to each other, we get a total of &lt;math&gt;4!=24&lt;/math&gt; ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; can be next to each other. If we place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; next to each other in that order, then there are three places that we can place them, namely in the first two slots, in the second two slots, or in the last two slots (i.e. &lt;math&gt;ST\square\square, \square ST\square, \square\square ST&lt;/math&gt;). However, we could also have placed &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; in the opposite order (i.e. &lt;math&gt;TS\square\square, \square TS\square, \square\square TS&lt;/math&gt;). Thus there are 6 ways of placing &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; directly next to each other. Next, notice that for each of these placements, we have two open slots for placing &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. Specifically, we can place &lt;math&gt;A&lt;/math&gt; in the first open slot and &lt;math&gt;B&lt;/math&gt; in the second open slot or switch their order and place &lt;math&gt;B&lt;/math&gt; in the first open slot and &lt;math&gt;A&lt;/math&gt; in the second open slot. This gives us a total of &lt;math&gt;6\times 2=12&lt;/math&gt; ways to place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; next to each other. Subtracting this from the total number of arrangements gives us &lt;math&gt;24-12=12&lt;/math&gt; total arrangements &lt;math&gt;\implies\boxed{\textbf{(C) }12}&lt;/math&gt;.&lt;br&gt;<br /> <br /> We can also solve this problem directly by looking at the number of ways that we can place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; such that they are not directly next to each other. Observe that there are three ways to place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; (in that order) into the four slots so they are not next to each other (i.e. &lt;math&gt;S\square T\square, \square S\square T, S\square\square T&lt;/math&gt;). However, we could also have placed &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; in the opposite order (i.e. &lt;math&gt;T\square S\square, \square T\square S, T\square\square S&lt;/math&gt;). Thus there are 6 ways of placing &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; so that they are not next to each other. Next, notice that for each of these placements, we have two open slots for placing &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. Specifically, we can place &lt;math&gt;A&lt;/math&gt; in the first open slot and &lt;math&gt;B&lt;/math&gt; in the second open slot or switch their order and place &lt;math&gt;B&lt;/math&gt; in the first open slot and &lt;math&gt;A&lt;/math&gt; in the second open slot. This gives us a total of &lt;math&gt;6\times 2=12&lt;/math&gt; ways to place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; such that they are not next to each other &lt;math&gt;\implies\boxed{\textbf{(C) }12}&lt;/math&gt;.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ====Solution 4====<br /> Let's try complementary counting. There &lt;math&gt;4!&lt;/math&gt; ways to arrange the 4 marbles. However, there are &lt;math&gt;2\cdot3!&lt;/math&gt; arrangements where Steelie and Tiger are next to each other. (Think about permutations of the element ST, A, and B or TS, A, and B). Thus, &lt;cmath&gt;4!-2\cdot3!=\boxed{12 \textbf{(C)}}&lt;/cmath&gt;<br /> <br /> ====Solution 5====<br /> <br /> We use complementary counting: we will count the numbers of ways where Steelie and Tiger are together and subtract that from the total count. Treat the Steelie and the Tiger as a &quot;super marble.&quot; There are &lt;math&gt;2!&lt;/math&gt; ways to arrange Steelie and Tiger within this &quot;super marble.&quot; Then there are &lt;math&gt;3!&lt;/math&gt; ways to arrange the &quot;super marble&quot; and Zara's two other marbles in a row. Since there are &lt;math&gt;4!&lt;/math&gt; ways to arrange the marbles without any restrictions, the answer is given by &lt;math&gt;4!-2!\cdot 3!=\textbf{(C) }12&lt;/math&gt;<br /> <br /> -franzliszt<br /> <br /> ====Solution 6====<br /> <br /> We will use the following<br /> <br /> &lt;math&gt;\textbf{Georgeooga-Harryooga Theorem:}&lt;/math&gt; The [[Georgeooga-Harryooga Theorem]] states that if you have &lt;math&gt;a&lt;/math&gt; distinguishable objects and &lt;math&gt;b&lt;/math&gt; of them cannot be together, then there are &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects.<br /> <br /> &lt;math&gt;\textit{Proof. (Created by AoPS user RedFireTruck)}&lt;/math&gt;<br /> <br /> Let our group of &lt;math&gt;a&lt;/math&gt; objects be represented like so &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, ..., &lt;math&gt;a-1&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;. Let the last &lt;math&gt;b&lt;/math&gt; objects be the ones we can't have together.<br /> <br /> Then we can organize our objects like so &lt;math&gt;\square1\square2\square3\square...\square a-b-1\square a-b\square&lt;/math&gt;.<br /> <br /> We have &lt;math&gt;(a-b)!&lt;/math&gt; ways to arrange the objects in that list.<br /> <br /> Now we have &lt;math&gt;a-b+1&lt;/math&gt; blanks and &lt;math&gt;b&lt;/math&gt; other objects so we have &lt;math&gt;_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects we can't put together.<br /> <br /> By fundamental counting principal our answer is &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt;.<br /> <br /> <br /> Proof by [[User:RedFireTruck|RedFireTruck]]<br /> <br /> <br /> Back to the problem. By the [[Georgeooga-Harryooga Theorem]], our answer is &lt;math&gt;\frac{(4-2)!(4-2+1)!}{(4-2\cdot2+1)!}=\textbf{(C) }12&lt;/math&gt;.<br /> <br /> -franzliszt<br /> <br /> ====Solution 7====<br /> https://youtu.be/pB46JzBNM6g<br /> <br /> ~savannahsolver<br /> <br /> =Testimonials=<br /> &quot;Thanks for rediscovering our theorem [[User:Redfiretruck|RedFireTruck]]&quot; - George and Harry of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]<br /> <br /> &quot;Wow! George and Harry are alive???&quot; ~ samrocksnature<br /> <br /> &quot;Hi&quot; ~ jasperE3<br /> <br /> &quot;I used this theorem on the AMC 8 and got a 25. Very useful!&quot; - [[User:Redfiretruck|RedFireTruck]]<br /> <br /> &quot;This is very complicated, but great.&quot; - Jiseop55406<br /> <br /> I used this theorem on the AMC 8 too! ~ ilp<br /> <br /> &quot;Very nice theorem&quot; ~ IdkHowToAddNumbers<br /> <br /> &quot;This theorem is really worth of being used in the AMC 8! Very useful!&quot; ~ [[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] ([[User talk:Aops-g5-gethsemanea2|talk]]) 20:48, 21 December 2020 (EST)<br /> <br /> &quot;I have realized the way&quot; ~ hi..<br /> <br /> &quot;This theorem even works on AMC10 and 12&quot; - TryhardMathlete</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=CrystalFlower&diff=141049 CrystalFlower 2020-12-30T18:34:52Z <p>Sweetmango77: </p> <hr /> <div>hi<br /> <br /> {{delete|no use}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_19&diff=140834 2020 AMC 10B Problems/Problem 19 2020-12-28T16:34:25Z <p>Sweetmango77: /* Solution 5 (Very Factor Bashy CRT) */</p> <hr /> <div>==Problem==<br /> <br /> In a certain card game, a player is dealt a hand of &lt;math&gt;10&lt;/math&gt; cards from a deck of &lt;math&gt;52&lt;/math&gt; distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as &lt;math&gt;158A00A4AA0&lt;/math&gt;. What is the digit &lt;math&gt;A&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;math&gt;158A00A4AA0 \equiv 1+5+8+A+0+0+A+4+A+A+0 \equiv 4A \pmod{9}&lt;/math&gt;<br /> <br /> We're looking for the amount of ways we can get &lt;math&gt;10&lt;/math&gt; cards from a deck of &lt;math&gt;52&lt;/math&gt;, which is represented by &lt;math&gt;\binom{52}{10}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}&lt;/math&gt;<br /> <br /> We need to get rid of the multiples of &lt;math&gt;3&lt;/math&gt;, which will subsequently get rid of the multiples of &lt;math&gt;9&lt;/math&gt; (if we didn't, the zeroes would mess with the equation since you can't divide by 0)<br /> <br /> &lt;math&gt;9\cdot5=45&lt;/math&gt;, &lt;math&gt;8\cdot6=48&lt;/math&gt;, &lt;math&gt;\frac{51}{3}&lt;/math&gt; leaves us with 17.<br /> <br /> &lt;math&gt;\frac{52\cdot\cancel{51}^{17}\cdot50\cdot49\cdot\cancel{48}\cdot47\cdot46\cdot\cancel{45}\cdot44\cdot43}{10\cdot\cancel{9}\cdot\cancel{8}\cdot7\cdot\cancel{6}\cdot\cancel{5}\cdot4\cdot\cancel{3}\cdot2\cdot1}&lt;/math&gt;<br /> <br /> Converting these into&lt;math&gt;\pmod{9}&lt;/math&gt;, we have<br /> <br /> &lt;math&gt;\binom{52}{10}\equiv \frac{(-2)\cdot(-1)\cdot(-4)\cdot4\cdot2\cdot1\cdot(-1)\cdot(-2)}{1\cdot(-2)\cdot4\cdot2\cdot1} \equiv (-1)\cdot(-4)\cdot(-1)\cdot(-2) \equiv 8 \pmod{9} &lt;/math&gt;<br /> <br /> &lt;math&gt;4A\equiv8\pmod{9} \implies A=\boxed{\textbf{(A) }2}&lt;/math&gt; ~quacker88<br /> <br /> ==Solution 2==<br /> &lt;math&gt;\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=26\cdot17\cdot5\cdot7\cdot47\cdot46\cdot11\cdot43&lt;/math&gt;<br /> <br /> Since this number is divisible by &lt;math&gt;4&lt;/math&gt; but not &lt;math&gt;8&lt;/math&gt;, the last &lt;math&gt;2&lt;/math&gt; digits must be divisible by &lt;math&gt;4&lt;/math&gt; but the last &lt;math&gt;3&lt;/math&gt; digits cannot be divisible by &lt;math&gt;8&lt;/math&gt;. This narrows the options down to &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;. <br /> <br /> Also, the number cannot be divisible by &lt;math&gt;3&lt;/math&gt;. Adding up the digits, we get &lt;math&gt;18+4A&lt;/math&gt;. If &lt;math&gt;A=6&lt;/math&gt;, then the expression equals &lt;math&gt;42&lt;/math&gt;, a multiple of &lt;math&gt;3&lt;/math&gt;. This would mean that the entire number would be divisible by &lt;math&gt;3&lt;/math&gt;, which is not what we want. Therefore, the only option is &lt;math&gt;\boxed{\textbf{(A) }2}&lt;/math&gt;-PCChess<br /> <br /> ==Solution 3==<br /> It is not hard to check that &lt;math&gt;13&lt;/math&gt; divides the number,<br /> &lt;cmath&gt;\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=26\cdot17\cdot5\cdot7\cdot47\cdot46\cdot11\cdot43.&lt;/cmath&gt; As &lt;math&gt;10^3\equiv-1\pmod{13}&lt;/math&gt;, using &lt;math&gt;\pmod{13}&lt;/math&gt; we have &lt;math&gt;13|\overline{AA0}-\overline{0A4}+\overline{8A0}-\overline{15}=110A+781&lt;/math&gt;. Thus &lt;math&gt;6A+1\equiv0\pmod{13}&lt;/math&gt;, implying &lt;math&gt;A\equiv2\pmod{13}&lt;/math&gt; so the answer is &lt;math&gt;\boxed{\textbf{(A) }2}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{- Emathmaster}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> As mentioned above, &lt;br&gt;<br /> &lt;cmath&gt;\binom{52}{10}=\frac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47 \cdot 46 \cdot 45 \cdot 44 \cdot 43}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = {10 \cdot 17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43} = 158A00A4AA0.&lt;/cmath&gt;<br /> We can divide both sides of &lt;math&gt;10 \cdot 17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43 = 158A00A4AA0&lt;/math&gt; by 10 to obtain<br /> &lt;cmath&gt;17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43 = 158A00A4AA,&lt;/cmath&gt;<br /> which means &lt;math&gt;A&lt;/math&gt; is simply the units digit of the left-hand side. This value is<br /> &lt;cmath&gt;7 \cdot 3 \cdot 7 \cdot 7 \cdot 6 \cdot 1 \cdot 3 \equiv \boxed{\textbf{(A) }2} \pmod{10}.&lt;/cmath&gt;<br /> ~[[User:i_equal_tan_90|i_equal_tan_90]], revised by [[User:emerald_block|emerald_block]]<br /> <br /> ==Solution 5 (Very Factor Bashy CRT)==<br /> We note that:<br /> &lt;cmath&gt; \frac{(52)(51)(50)(49)(48)(47)(46)(45)(44)(43)}{(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)} = (13)(17)(7)(47)(46)(5)(22)(43). &lt;/cmath&gt;<br /> Let &lt;math&gt;K=(13)(17)(7)(47)(46)(5)(22)(43)&lt;/math&gt;. This will help us find the last two digits modulo &lt;math&gt;4&lt;/math&gt; and modulo &lt;math&gt;25&lt;/math&gt;.<br /> It is obvious that &lt;math&gt;K \equiv 0 (mod 4)&lt;/math&gt;. Also (although this not so obvious), &lt;math&gt;K \equiv (13)(17)(7)(47)(46)(5)(22)(43) \equiv (13)(-8)(7)(-3)(-4)(5)(-3)(-7) \equiv (13)(-96)(21)(35) \equiv (13)(4)(-4)(10) \equiv (13)(-16)(10) \equiv (13)(9)(10) \equiv (117)(10) \equiv (-8)(10) \equiv 20 (mod 25)&lt;/math&gt;. Therefore, &lt;math&gt;K \equiv 20 \mod 100&lt;/math&gt;. Thus &lt;math&gt;K=20&lt;/math&gt;, implying that &lt;math&gt;A=2&lt;/math&gt;. &lt;math&gt;\textbf{A}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/3BvJeZU3T-M<br /> <br /> ~IceMatrix<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=ApqZFuuQJ18&amp;list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&amp;index=6 ~ MathEx<br /> <br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_19&diff=140775 2020 AMC 10B Problems/Problem 19 2020-12-27T22:33:11Z <p>Sweetmango77: /* Solution 5 (Very Factor Bashy CRT) */</p> <hr /> <div>==Problem==<br /> <br /> In a certain card game, a player is dealt a hand of &lt;math&gt;10&lt;/math&gt; cards from a deck of &lt;math&gt;52&lt;/math&gt; distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as &lt;math&gt;158A00A4AA0&lt;/math&gt;. What is the digit &lt;math&gt;A&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;math&gt;158A00A4AA0 \equiv 1+5+8+A+0+0+A+4+A+A+0 \equiv 4A \pmod{9}&lt;/math&gt;<br /> <br /> We're looking for the amount of ways we can get &lt;math&gt;10&lt;/math&gt; cards from a deck of &lt;math&gt;52&lt;/math&gt;, which is represented by &lt;math&gt;\binom{52}{10}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}&lt;/math&gt;<br /> <br /> We need to get rid of the multiples of &lt;math&gt;3&lt;/math&gt;, which will subsequently get rid of the multiples of &lt;math&gt;9&lt;/math&gt; (if we didn't, the zeroes would mess with the equation since you can't divide by 0)<br /> <br /> &lt;math&gt;9\cdot5=45&lt;/math&gt;, &lt;math&gt;8\cdot6=48&lt;/math&gt;, &lt;math&gt;\frac{51}{3}&lt;/math&gt; leaves us with 17.<br /> <br /> &lt;math&gt;\frac{52\cdot\cancel{51}^{17}\cdot50\cdot49\cdot\cancel{48}\cdot47\cdot46\cdot\cancel{45}\cdot44\cdot43}{10\cdot\cancel{9}\cdot\cancel{8}\cdot7\cdot\cancel{6}\cdot\cancel{5}\cdot4\cdot\cancel{3}\cdot2\cdot1}&lt;/math&gt;<br /> <br /> Converting these into&lt;math&gt;\pmod{9}&lt;/math&gt;, we have<br /> <br /> &lt;math&gt;\binom{52}{10}\equiv \frac{(-2)\cdot(-1)\cdot(-4)\cdot4\cdot2\cdot1\cdot(-1)\cdot(-2)}{1\cdot(-2)\cdot4\cdot2\cdot1} \equiv (-1)\cdot(-4)\cdot(-1)\cdot(-2) \equiv 8 \pmod{9} &lt;/math&gt;<br /> <br /> &lt;math&gt;4A\equiv8\pmod{9} \implies A=\boxed{\textbf{(A) }2}&lt;/math&gt; ~quacker88<br /> <br /> ==Solution 2==<br /> &lt;math&gt;\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=26\cdot17\cdot5\cdot7\cdot47\cdot46\cdot11\cdot43&lt;/math&gt;<br /> <br /> Since this number is divisible by &lt;math&gt;4&lt;/math&gt; but not &lt;math&gt;8&lt;/math&gt;, the last &lt;math&gt;2&lt;/math&gt; digits must be divisible by &lt;math&gt;4&lt;/math&gt; but the last &lt;math&gt;3&lt;/math&gt; digits cannot be divisible by &lt;math&gt;8&lt;/math&gt;. This narrows the options down to &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;. <br /> <br /> Also, the number cannot be divisible by &lt;math&gt;3&lt;/math&gt;. Adding up the digits, we get &lt;math&gt;18+4A&lt;/math&gt;. If &lt;math&gt;A=6&lt;/math&gt;, then the expression equals &lt;math&gt;42&lt;/math&gt;, a multiple of &lt;math&gt;3&lt;/math&gt;. This would mean that the entire number would be divisible by &lt;math&gt;3&lt;/math&gt;, which is not what we want. Therefore, the only option is &lt;math&gt;\boxed{\textbf{(A) }2}&lt;/math&gt;-PCChess<br /> <br /> ==Solution 3==<br /> It is not hard to check that &lt;math&gt;13&lt;/math&gt; divides the number,<br /> &lt;cmath&gt;\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=26\cdot17\cdot5\cdot7\cdot47\cdot46\cdot11\cdot43.&lt;/cmath&gt; As &lt;math&gt;10^3\equiv-1\pmod{13}&lt;/math&gt;, using &lt;math&gt;\pmod{13}&lt;/math&gt; we have &lt;math&gt;13|\overline{AA0}-\overline{0A4}+\overline{8A0}-\overline{15}=110A+781&lt;/math&gt;. Thus &lt;math&gt;6A+1\equiv0\pmod{13}&lt;/math&gt;, implying &lt;math&gt;A\equiv2\pmod{13}&lt;/math&gt; so the answer is &lt;math&gt;\boxed{\textbf{(A) }2}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{- Emathmaster}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> As mentioned above, &lt;br&gt;<br /> &lt;cmath&gt;\binom{52}{10}=\frac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47 \cdot 46 \cdot 45 \cdot 44 \cdot 43}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = {10 \cdot 17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43} = 158A00A4AA0.&lt;/cmath&gt;<br /> We can divide both sides of &lt;math&gt;10 \cdot 17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43 = 158A00A4AA0&lt;/math&gt; by 10 to obtain<br /> &lt;cmath&gt;17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43 = 158A00A4AA,&lt;/cmath&gt;<br /> which means &lt;math&gt;A&lt;/math&gt; is simply the units digit of the left-hand side. This value is<br /> &lt;cmath&gt;7 \cdot 3 \cdot 7 \cdot 7 \cdot 6 \cdot 1 \cdot 3 \equiv \boxed{\textbf{(A) }2} \pmod{10}.&lt;/cmath&gt;<br /> ~[[User:i_equal_tan_90|i_equal_tan_90]], revised by [[User:emerald_block|emerald_block]]<br /> <br /> ==Solution 5 (Very Factor Bashy CRT)==<br /> We note that:<br /> &lt;cmath&gt; \frac{(52)(51)(50)(49)(48)(47)(46)(45)(44)(43)}{(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)} = (13)(17)(7)(47)(46)(5)(22)(43). &lt;/cmath&gt;<br /> Let &lt;math&gt;K=(13)(17)(7)(47)(46)(5)(22)(43)&lt;/math&gt;. This will help us find the last two digits modulo &lt;math&gt;4&lt;/math&gt; and modulo &lt;math&gt;25&lt;/math&gt;.<br /> It is obvious that &lt;math&gt;K \equiv 0 (mod 4)&lt;/math&gt;. Also (although this not so obvious), &lt;math&gt;K \equiv (13)(17)(7)(47)(46)(5)(22)(43) \equiv (13)(-8)(7)(-3)(-4)(5)(-3)(-7) \equiv (13)(-96)(21)(35) \equiv (13)(4)(-4)(10) \equiv (13)(-16)(10) \equiv (13)(9)(10) \equiv (117)(10) \equiv (-8)(10) \equiv 20 (mod 25)&lt;/math&gt;. Therefore, &lt;math&gt;K \equiv 20 (mod 100)&lt;/math&gt;. Thus &lt;math&gt;K=20&lt;/math&gt;, implying that &lt;math&gt;A=2&lt;/math&gt;. &lt;math&gt;\textbf{A}&lt;/math&gt;<br /> <br /> (I need help with finding the modulo congruence symbol. Thanks!)<br /> <br /> It is \equiv<br /> <br /> ==Video Solution==<br /> https://youtu.be/3BvJeZU3T-M<br /> <br /> ~IceMatrix<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=ApqZFuuQJ18&amp;list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&amp;index=6 ~ MathEx<br /> <br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10A_Problems/Problem_15&diff=139503 2007 AMC 10A Problems/Problem 15 2020-12-12T22:26:31Z <p>Sweetmango77: why is that other problem on the top</p> <hr /> <div>== Problem ==<br /> Four circles of radius &lt;math&gt;1&lt;/math&gt; are each tangent to two sides of a square and externally tangent to a circle of radius &lt;math&gt;2&lt;/math&gt;, as shown. What is the area of the square?<br /> &lt;center&gt;<br /> &lt;asy&gt;<br /> unitsize(5mm);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> real h=3*sqrt(2)/2;<br /> pair O0=(0,0), O1=(h,h), O2=(-h,h), O3=(-h,-h), O4=(h,-h);<br /> pair X=O0+2*dir(30), Y=O2+dir(45);<br /> draw((-h-1,-h-1)--(-h-1,h+1)--(h+1,h+1)--(h+1,-h-1)--cycle);<br /> draw(Circle(O0,2));<br /> draw(Circle(O1,1));<br /> draw(Circle(O2,1));<br /> draw(Circle(O3,1));<br /> draw(Circle(O4,1));<br /> draw(O0--X);<br /> draw(O2--Y);<br /> label(&quot;$2$&quot;,midpoint(O0--X),NW);<br /> label(&quot;$1$&quot;,midpoint(O2--Y),SE);<br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> &lt;math&gt;\text{(A)}\ 32 \qquad \text{(B)}\ 22 + 12\sqrt {2}\qquad \text{(C)}\ 16 + 16\sqrt {3}\qquad \text{(D)}\ 48 \qquad \text{(E)}\ 36 + 16\sqrt {2}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Draw a square connecting the centers of the four small circles of radius &lt;math&gt;1&lt;/math&gt;. This square has a diagonal of length &lt;math&gt;6&lt;/math&gt;, as it includes the diameter of the big circle of radius &lt;math&gt;2&lt;/math&gt; and two radii of the small circles of radius &lt;math&gt;1&lt;/math&gt;. Therefore, the side length of this square is &lt;cmath&gt;\frac{6}{\sqrt{2}} = 3\sqrt{2}.&lt;/cmath&gt; The radius of the large square has a side length &lt;math&gt;2&lt;/math&gt; units larger than the one found by connecting the midpoints, so its side length is &lt;cmath&gt;2 + 3\sqrt{2}.&lt;/cmath&gt; The area of this square is &lt;math&gt;(2+3\sqrt{2})^2 = 22 + 12\sqrt{2}&lt;/math&gt; &lt;math&gt;(B).&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> <br /> We draw the long diagonal of the square. This diagonal yields &lt;math&gt;2\sqrt{2}+1+1+2+2=2\sqrt{2}+6&lt;/math&gt;. We know that the side length &lt;math&gt;s&lt;/math&gt; in terms of the diagonal &lt;math&gt;d&lt;/math&gt; is &lt;math&gt;s=\frac{d}{\sqrt{2}}&lt;/math&gt;, so our side length is &lt;math&gt;\frac{2\sqrt{2}+6}{\sqrt{2}}&lt;/math&gt;. However, we are trying to look for the area, so squaring &lt;math&gt;\frac{2\sqrt{2}+6}{\sqrt{2}}&lt;/math&gt; yields &lt;math&gt;\frac{44+24\sqrt{2}}{2}=\boxed{\text{(B)}22+12\sqrt{2}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2007|ab=A|num-b=14|num-a=16}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=Arkansas_MathCounts&diff=138907 Arkansas MathCounts 2020-12-02T01:27:51Z <p>Sweetmango77: </p> <hr /> <div>{{delete|Blank page}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=3Blue1Brown&diff=138906 3Blue1Brown 2020-12-02T01:27:20Z <p>Sweetmango77: </p> <hr /> <div>I think that this might be a good page to have. I am not sure how to set up a wiki page, so please help make a nice page out of this. There is a wikipedia page on this that could be somewhat modeled after.<br /> <br /> {{delete|WHY DOES THIS EXIST, JUST SEARCH UP 3BLUE1BROWN ON YOUTUBE}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=Northeastern_WOOTers_Mock_AIME_I_Problems/Problem_1&diff=138905 Northeastern WOOTers Mock AIME I Problems/Problem 1 2020-12-02T01:24:56Z <p>Sweetmango77: </p> <hr /> <div>{{delete|This has nothing to do with WOOT or AIME, probably spam.}}<br /> <br /> I am an environmentalist so I like bicycles</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=User:Sweetmango77&diff=138904 User:Sweetmango77 2020-12-02T01:22:50Z <p>Sweetmango77: /* SweetMango77: */</p> <hr /> <div>==&lt;font color=&quot;black&quot; style=&quot;font-family: Roman&quot;&gt;SweetMango77:&lt;/font&gt;==<br /> <br /> &lt;font color=&quot;black&quot; style=&quot;font-family: Roman&quot;&gt;<br /> A person who made a 25x100 multiplication chart: https://docs.google.com/spreadsheets/d/1Tdy1sz8y7LuO9yqADUnAXZ583IqnFlmTM88XsaX1W6g/edit#gid=0<br /> <br /> Please friend me in Hypixel, my username is &lt;i&gt;Dat_Is_Stupid&lt;/i&gt;.<br /> <br /> Also good at coding with HTML, CSS, and JavaScript.<br /> &lt;hr&gt;<br /> PM me if you would like to edit this, otherwise I will undo all of your changes.<br /> <br /> &lt;hr&gt;<br /> <br /> &lt;math&gt;\Huge{\text{BYE}}&lt;/math&gt;<br /> &lt;/font&gt;</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=Euclid&diff=138903 Euclid 2020-12-02T01:21:43Z <p>Sweetmango77: /* Results Attributed to Euclid */</p> <hr /> <div>'''Euclid''' (also referred to as '''Euclid of Alexandria''') (Greek: Εὐκλείδης) (c. 325–c. 265 BC) was a Greek mathematician who lived in Alexandria in Hellenistic Egypt, and almost certainly during the reign of Ptolemy I (323 BC–283 BC), is often considered to be the &quot;father of [[geometry]]&quot;. Very little biographical information is known about Euclid.<br /> <br /> ==Euclid's Elements==<br /> {{main|Euclid's Elements}}<br /> Euclid's most popular work, [[Euclid's Elements|Elements]], is thought to be one of the most successful textbooks in the history of [[mathematics]]. It is divided into thirteen volumes, each consisting of &quot;common notions&quot; (common [[arithmetic]]al [[axiom]]s), [[postulate]]s ([[geometry|geometrical]] axioms), and &quot;propositions&quot;, or theorems. Several propositions in fact should have been either common notions or postulates, as some of Euclid's methods of proof were faulty.<br /> <br /> <br /> <br /> == Results Attributed to Euclid ==<br /> There are many results still attributed to or named after Euclid in use today. They include:<br /> <br /> *The most known type of geometry is known as Euclidean Geometry.<br /> *[[Euclid's lemma]]<br /> *[[Euclidean Division Algorithm]]<br /> *[[Euclid's proof on the infinitude of primes]]<br /> <br /> [[Category:Famous mathematicians]]<br /> <br /> {{stub}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=Skew&diff=138902 Skew 2020-12-02T01:19:38Z <p>Sweetmango77: </p> <hr /> <div>Skew lines in 3-D do not intersect and are not parallel. They also point in different directions.<br /> <br /> {{stub}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=Geometric_mean_theorem&diff=138901 Geometric mean theorem 2020-12-02T01:16:29Z <p>Sweetmango77: </p> <hr /> <div> The geometric mean theorems relate the legs and altitude of a right triangle to the hypotenuse of the right triangle.<br /> <br /> {{delete|this does not tell us what the geometric mean theorem is, it only tells us part of it}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12B&diff=138900 2022 AMC 12B 2020-12-02T01:14:33Z <p>Sweetmango77: </p> <hr /> <div>See this page next year<br /> <br /> {{delete|way into the future, no use to put it here}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=/wiki/i&diff=138898 /wiki/i 2020-12-02T01:09:04Z <p>Sweetmango77: </p> <hr /> <div>{{delete|spam and possibly ad}}Keala Kanae Has Lot To Offer In Quick Time<br /> <br /> <br /> <br /> Keala Kanae affiliate promoting has become quickly turning into a favourite way of increasing site visitors or product sales to acquire a organization. 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The event of social-networking celeb has authorized businesses to simply take a brand-new method. With tens of countless numbers and tens of countless numbers, at times even tens of countless numbers and tens of countless numbers of follower’s advertising marketing campaign by accessing social websites influencers to enroll and talk about your connections.</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=Gmaas_Language&diff=138897 Gmaas Language 2020-12-02T01:02:16Z <p>Sweetmango77: </p> <hr /> <div>All who worship GMAAS must learn the GMAAS language. AR1US is the messenger delivering this mystic tongue and GMAAS is the one who has chosen him to do it. <br /> <br /> THE GMAAS ALPHABET: <br /> <br /> A a <br /> Aa aa <br /> Ar ar <br /> Aar aar <br /> B b <br /> C c <br /> D d <br /> E e <br /> Ee ee <br /> Er er <br /> Eer eer <br /> F f <br /> G g <br /> H h <br /> I i <br /> Ii ii <br /> Ir ir <br /> Iir iir <br /> J j <br /> K k <br /> L l <br /> Ll ll <br /> M m <br /> N n <br /> Ng ng <br /> Ngu ngu <br /> O o <br /> Oo oo <br /> Or or <br /> Oor oor <br /> P p <br /> Q q <br /> R r <br /> Rr rr <br /> Rw rw <br /> S s <br /> Sy sy <br /> T t <br /> U u <br /> Uu uu <br /> Ur ur <br /> Uur uur <br /> V v <br /> W w <br /> X x <br /> Y y <br /> Z z <br /> Zy zy <br /> <br /> '''This page will be edited.'''<br /> <br /> {{delete|Really, stop with the Gmaas thing}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=User:Sweetmango77&diff=138895 User:Sweetmango77 2020-12-02T00:53:01Z <p>Sweetmango77: </p> <hr /> <div>==&lt;font color=&quot;black&quot; style=&quot;font-family: Roman&quot;&gt;SweetMango77:&lt;/font&gt;==<br /> <br /> &lt;font color=&quot;black&quot; style=&quot;font-family: Roman&quot;&gt;<br /> A person who made a 25x100 multiplication chart: https://docs.google.com/spreadsheets/d/1Tdy1sz8y7LuO9yqADUnAXZ583IqnFlmTM88XsaX1W6g/edit#gid=0<br /> <br /> Please friend me in Hypixel, my username is &lt;i&gt;Dat_Is_Stupid&lt;/i&gt;.<br /> <br /> Also good at coding with HTML, CSS, and JavaScript.<br /> &lt;hr&gt;<br /> PM me if you would like to edit this, otherwise I will undo all of your changes.<br /> <br /> &lt;hr&gt;<br /> <br /> &lt;math&gt;\Huge{\text{BYE}}&lt;/math&gt;<br /> &lt;/font&gt;<br /> <br /> &lt;br&gt;<br /> &lt;b&gt;Currently inactive, will be back after I do a lot of Mathcounts Trainer&lt;/b&gt;.</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=User:Sweetmango77&diff=138894 User:Sweetmango77 2020-12-02T00:52:44Z <p>Sweetmango77: </p> <hr /> <div>==&lt;font color=&quot;black&quot; style=&quot;font-family: Roman&quot;&gt;SweetMango77:&lt;/font&gt;==<br /> <br /> &lt;font color=&quot;black&quot; style=&quot;font-family: Roman&quot;&gt;<br /> A person who made a 25x100 multiplication chart: https://docs.google.com/spreadsheets/d/1Tdy1sz8y7LuO9yqADUnAXZ583IqnFlmTM88XsaX1W6g/edit#gid=0<br /> <br /> Please friend me in Hypixel, my username is &lt;i&gt;Dat_Is_Stupid&lt;/i&gt;.<br /> <br /> Also good at coding with HTML, CSS, and JavaScript.<br /> &lt;hr&gt;<br /> PM me if you would like to edit this, otherwise I will undo all of your changes.<br /> <br /> &lt;hr&gt;<br /> <br /> &lt;math&gt;\Huge{\text{BYE}}&lt;/math&gt;<br /> &lt;/font&gt;<br /> <br /> &lt;b&gt;Currently inactive, will be back after I do a lot of Mathcounts Trainer&lt;/b&gt;.</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=User:Sweetmango77&diff=138893 User:Sweetmango77 2020-12-02T00:50:10Z <p>Sweetmango77: </p> <hr /> <div>==&lt;font color=&quot;black&quot; style=&quot;font-family: Roman&quot;&gt;SweetMango77:&lt;/font&gt;==<br /> <br /> &lt;font color=&quot;black&quot; style=&quot;font-family: Roman&quot;&gt;<br /> A person who made a 25x100 multiplication chart: https://docs.google.com/spreadsheets/d/1Tdy1sz8y7LuO9yqADUnAXZ583IqnFlmTM88XsaX1W6g/edit#gid=0<br /> <br /> Please friend me in Hypixel, my username is &lt;i&gt;Dat_Is_Stupid&lt;/i&gt;.<br /> <br /> Also good at coding with HTML, CSS, and JavaScript.<br /> &lt;hr&gt;<br /> PM me if you would like to edit this, otherwise I will undo all of your changes.<br /> <br /> &lt;hr&gt;<br /> <br /> &lt;math&gt;\Huge{\text{BYE}}&lt;/math&gt;<br /> &lt;/font&gt;</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=User:Sweetmango77&diff=138892 User:Sweetmango77 2020-12-02T00:14:35Z <p>Sweetmango77: </p> <hr /> <div>&lt;h1&gt;404&lt;/h1&gt;<br /> <br /> There is currently no text in this page. You can search for this page title in other pages, search the related logs, or create this page.</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_14&diff=138749 2018 AMC 10A Problems/Problem 14 2020-11-29T16:15:36Z <p>Sweetmango77: </p> <hr /> <div>==Problem==<br /> <br /> What is the greatest integer less than or equal to &lt;cmath&gt;\frac{3^{100}+2^{100}}{3^{96}+2^{96}}?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }80\qquad<br /> \textbf{(B) }81 \qquad<br /> \textbf{(C) }96 \qquad<br /> \textbf{(D) }97 \qquad<br /> \textbf{(E) }625\qquad<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We write<br /> &lt;cmath&gt;\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=\frac{3^{96}}{3^{96}+2^{96}}\cdot\frac{3^{100}}{3^{96}}+\frac{2^{96}}{3^{96}+2^{96}}\cdot\frac{2^{100}}{2^{96}}=\frac{3^{96}}{3^{96}+2^{96}}\cdot 81+\frac{2^{96}}{3^{96}+2^{96}}\cdot 16.&lt;/cmath&gt;<br /> Hence we see that our number is a weighted average of 81 and 16, extremely heavily weighted toward 81. Hence the number is ever so slightly less than 81, so the answer is &lt;math&gt;\boxed{\textbf{(A) }80}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let's set this value equal to &lt;math&gt;x&lt;/math&gt;. We can write<br /> &lt;cmath&gt;\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=x.&lt;/cmath&gt;<br /> Multiplying by &lt;math&gt;3^{96}+2^{96}&lt;/math&gt; on both sides, we get <br /> &lt;cmath&gt;3^{100}+2^{100}=x(3^{96}+2^{96}).&lt;/cmath&gt;<br /> Now let's take a look at the answer choices. We notice that &lt;math&gt;81&lt;/math&gt;, choice &lt;math&gt;B&lt;/math&gt;, can be written as &lt;math&gt;3^4&lt;/math&gt;. Plugging this into our equation above, we get<br /> &lt;cmath&gt;3^{100}+2^{100} \stackrel{?}{=} 3^4(3^{96}+2^{96}) \Rightarrow 3^{100}+2^{100} \stackrel{?}{=} 3^{100}+3^4\cdot 2^{96}.&lt;/cmath&gt;<br /> The right side is larger than the left side because <br /> &lt;cmath&gt;2^{100} \leq 2^{96}\cdot 3^4.&lt;/cmath&gt;<br /> This means that our original value, &lt;math&gt;x&lt;/math&gt;, must be less than &lt;math&gt;81&lt;/math&gt;. The only answer that is less than &lt;math&gt;81&lt;/math&gt; is &lt;math&gt;80&lt;/math&gt; so our answer is &lt;math&gt;\boxed{A}&lt;/math&gt;.<br /> <br /> ~Nivek<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=\frac{2^{96}(\frac{3^{100}}{2^{96}})+2^{96}(2^{4})}{2^{96}(\frac{3}{2})^{96}+2^{96}(1)}=\frac{\frac{3^{100}}{2^{96}}+2^{4}}{(\frac{3}{2})^{96}+1}=\frac{\frac{3^{100}}{2^{100}}*2^{4}+2^{4}}{(\frac{3}{2})^{96}+1}=\frac{2^{4}(\frac{3^{100}}{2^{100}}+1)}{(\frac{3}{2})^{96}+1}&lt;/math&gt;.<br /> <br /> We can ignore the 1's on the end because they won't really affect the fraction. So, the answer is very very very close but less than the new fraction.<br /> <br /> &lt;math&gt;\frac{2^{4}(\frac{3^{100}}{2^{100}}+1)}{(\frac{3}{2})^{96}+1}&lt;\frac{2^{4}(\frac{3^{100}}{2^{100}})}{(\frac{3}{2})^{96}}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2^{4}(\frac{3^{100}}{2^{100}})}{(\frac{3}{2})^{96}}=\frac{3^{4}}{2^{4}}*2^{4}=3^{4}=81&lt;/math&gt;<br /> <br /> So, our final answer is very close but not quite 81, and therefore the greatest integer less than the number is &lt;math&gt;\boxed{(A) 80}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> <br /> Let &lt;math&gt;x=3^{96}&lt;/math&gt; and &lt;math&gt;y=2^{96}&lt;/math&gt;. Then our fraction can be written as<br /> &lt;math&gt;\frac{81x+16y}{x+y}=\frac{16x+16y}{x+y}+\frac{65x}{x+y}=16+\frac{65x}{x+y}&lt;/math&gt;.<br /> Notice that <br /> &lt;math&gt;\frac{65x}{x+y}&lt;\frac{65x}{x}=65&lt;/math&gt;.<br /> So ,<br /> &lt;math&gt;16+\frac{65x}{x+y}&lt;16+65=81&lt;/math&gt;.<br /> And our only answer choice less than 81 is &lt;math&gt;\boxed{(A) 80}&lt;/math&gt; (RegularHexagon)<br /> <br /> ==Solution 5==<br /> Let &lt;math&gt;x=\frac{3^{100}+2^{100}}{3^{96}+2^{96}}&lt;/math&gt;. Multiply both sides by &lt;math&gt;(3^{96}+2^{96})&lt;/math&gt;, and expand. Rearranging the terms, we get &lt;math&gt;3^{96}(3^4-x)+2^{96}(2^4-x)=0&lt;/math&gt;. The left side is decreasing, and it is negative when &lt;math&gt;x=81&lt;/math&gt;. This means that the answer must be less than &lt;math&gt;81&lt;/math&gt;; therefore the answer is &lt;math&gt;\boxed{(A)}&lt;/math&gt;.<br /> <br /> ==Solution 6 (eyeball it)==<br /> A faster solution. Recognize that for exponents of this size &lt;math&gt;3^{n}&lt;/math&gt; will be enormously greater than &lt;math&gt;2^{n}&lt;/math&gt;, so the terms involving &lt;math&gt;2&lt;/math&gt; will actually have very little effect on the quotient. Now we know the answer will be very close to &lt;math&gt;81&lt;/math&gt;.<br /> <br /> Notice that the terms being added on to the top and bottom are in the ratio &lt;math&gt;\frac{1}{16}&lt;/math&gt; with each other, so they must pull the ratio down from 81 very slightly. (In the same way that a new test score lower than your current cumulative grade always must pull that grade downward.) Answer: &lt;math&gt;\boxed{(A)}&lt;/math&gt;.<br /> <br /> ==Solution 7==<br /> Notice how &lt;math&gt;\frac{3^{100}+2^{100}}{3^{96}+2^{96}}&lt;/math&gt; can be rewritten as &lt;math&gt;\frac{81(3^{96})+16(2^{96})}{3^{96}+2^{96}}=\frac{81(3^{96})+81(2^{96})}{3^{96}+2^{96}}-\frac{65(2^{96})}{3^{96}+2^{96}}=81-\frac{65(2^{96})}{3^{96}+2^{96}}&lt;/math&gt;. Note that &lt;math&gt;\frac{65(2^{96})}{3^{96}+2^{96}}&lt;1&lt;/math&gt;, so the greatest integer less than or equal to &lt;math&gt;\frac{3^{100}+2^{100}}{3^{96}+2^{96}}&lt;/math&gt; is &lt;math&gt;80&lt;/math&gt; or &lt;math&gt;\boxed{\textbf{(A)}}&lt;/math&gt;<br /> ~blitzkrieg21<br /> <br /> ==Solution 8==<br /> For positive &lt;math&gt;a, b, c, d&lt;/math&gt;, if &lt;math&gt;\frac{a}{b}&lt;\frac{c}{d}&lt;/math&gt; then &lt;math&gt;\frac{c+a}{d+b}&lt;\frac{c}{d}&lt;/math&gt;. Let &lt;math&gt;a=2^{100}, b=2^{96}, c=3^{100}, d=3^{96}&lt;/math&gt;. Then &lt;math&gt;\frac{c}{d}=3^4&lt;/math&gt;. So answer is less than 81, which leaves only one choice, 80.<br /> * Note that the algebra here is synonymous to the explanation given in Solution 6. This is the algebraic reason to the logic of if you get a test score with a lower percentage than your average (no matter how many points/percentage of your total grade it was worth), it will pull your overall grade down.<br /> <br /> ~ ccx09<br /> <br /> ==Solution 9==<br /> Try long division, and notice putting &lt;math&gt;3^4=81&lt;/math&gt; as the denominator is too big and putting &lt;math&gt;3^4-1=80&lt;/math&gt; is too small. So we know that the answer is between &lt;math&gt;80&lt;/math&gt; and &lt;math&gt;81&lt;/math&gt;, yielding &lt;math&gt;80&lt;/math&gt; as our answer.<br /> <br /> ==Solution 10 (Using the answer choices)==<br /> ===Solution 10.1===<br /> <br /> We can compare the given value to each of our answer choices. We already know that it is greater than &lt;math&gt;80&lt;/math&gt; because otherwise there would have been a smaller answer, so we move onto &lt;math&gt;81&lt;/math&gt;. We get:<br /> <br /> &lt;math&gt;\frac{3^{100}+2^{100}}{3^{96}+2^{96}} \text{ ? } 3^4&lt;/math&gt;<br /> <br /> Cross multiply to get:<br /> <br /> &lt;math&gt;3^{100}+2^{100} \text{ ? }3^{100}+(2^{96})(3^4)&lt;/math&gt;<br /> <br /> Cancel out &lt;math&gt;3^{100}&lt;/math&gt; and divide by &lt;math&gt;2^{96}&lt;/math&gt; to get &lt;math&gt;2^{4} \text{ ? }3^4&lt;/math&gt;. We know that &lt;math&gt;2^4 &lt; 3^4&lt;/math&gt;, which means the expression is less than &lt;math&gt;81&lt;/math&gt; so the answer is &lt;math&gt;\boxed{(A)}&lt;/math&gt;.<br /> <br /> ===Solution 10.2===<br /> <br /> We know this will be between 16 and 81 because &lt;math&gt;\frac{3^{100}}{3^{96}} = 3^4 = 81&lt;/math&gt; and &lt;math&gt;\frac{2^{100}}{2^{96}} = 2^4 = 16&lt;/math&gt;. &lt;math&gt;80=\boxed{(A)}&lt;/math&gt; is the only option choice in this range.<br /> <br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_14&diff=138747 2018 AMC 10A Problems/Problem 14 2020-11-29T16:12:03Z <p>Sweetmango77: /* Solution 8 (The Slick Solution) */</p> <hr /> <div>==Problem==<br /> <br /> What is the greatest integer less than or equal to &lt;cmath&gt;\frac{3^{100}+2^{100}}{3^{96}+2^{96}}?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }80\qquad<br /> \textbf{(B) }81 \qquad<br /> \textbf{(C) }96 \qquad<br /> \textbf{(D) }97 \qquad<br /> \textbf{(E) }625\qquad<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We write<br /> &lt;cmath&gt;\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=\frac{3^{96}}{3^{96}+2^{96}}\cdot\frac{3^{100}}{3^{96}}+\frac{2^{96}}{3^{96}+2^{96}}\cdot\frac{2^{100}}{2^{96}}=\frac{3^{96}}{3^{96}+2^{96}}\cdot 81+\frac{2^{96}}{3^{96}+2^{96}}\cdot 16.&lt;/cmath&gt;<br /> Hence we see that our number is a weighted average of 81 and 16, extremely heavily weighted toward 81. Hence the number is ever so slightly less than 81, so the answer is &lt;math&gt;\boxed{\textbf{(A) }80}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let's set this value equal to &lt;math&gt;x&lt;/math&gt;. We can write<br /> &lt;cmath&gt;\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=x.&lt;/cmath&gt;<br /> Multiplying by &lt;math&gt;3^{96}+2^{96}&lt;/math&gt; on both sides, we get <br /> &lt;cmath&gt;3^{100}+2^{100}=x(3^{96}+2^{96}).&lt;/cmath&gt;<br /> Now let's take a look at the answer choices. We notice that &lt;math&gt;81&lt;/math&gt;, choice &lt;math&gt;B&lt;/math&gt;, can be written as &lt;math&gt;3^4&lt;/math&gt;. Plugging this into our equation above, we get<br /> &lt;cmath&gt;3^{100}+2^{100} \stackrel{?}{=} 3^4(3^{96}+2^{96}) \Rightarrow 3^{100}+2^{100} \stackrel{?}{=} 3^{100}+3^4\cdot 2^{96}.&lt;/cmath&gt;<br /> The right side is larger than the left side because <br /> &lt;cmath&gt;2^{100} \leq 2^{96}\cdot 3^4.&lt;/cmath&gt;<br /> This means that our original value, &lt;math&gt;x&lt;/math&gt;, must be less than &lt;math&gt;81&lt;/math&gt;. The only answer that is less than &lt;math&gt;81&lt;/math&gt; is &lt;math&gt;80&lt;/math&gt; so our answer is &lt;math&gt;\boxed{A}&lt;/math&gt;.<br /> <br /> ~Nivek<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=\frac{2^{96}(\frac{3^{100}}{2^{96}})+2^{96}(2^{4})}{2^{96}(\frac{3}{2})^{96}+2^{96}(1)}=\frac{\frac{3^{100}}{2^{96}}+2^{4}}{(\frac{3}{2})^{96}+1}=\frac{\frac{3^{100}}{2^{100}}*2^{4}+2^{4}}{(\frac{3}{2})^{96}+1}=\frac{2^{4}(\frac{3^{100}}{2^{100}}+1)}{(\frac{3}{2})^{96}+1}&lt;/math&gt;.<br /> <br /> We can ignore the 1's on the end because they won't really affect the fraction. So, the answer is very very very close but less than the new fraction.<br /> <br /> &lt;math&gt;\frac{2^{4}(\frac{3^{100}}{2^{100}}+1)}{(\frac{3}{2})^{96}+1}&lt;\frac{2^{4}(\frac{3^{100}}{2^{100}})}{(\frac{3}{2})^{96}}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2^{4}(\frac{3^{100}}{2^{100}})}{(\frac{3}{2})^{96}}=\frac{3^{4}}{2^{4}}*2^{4}=3^{4}=81&lt;/math&gt;<br /> <br /> So, our final answer is very close but not quite 81, and therefore the greatest integer less than the number is &lt;math&gt;\boxed{(A) 80}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> <br /> Let &lt;math&gt;x=3^{96}&lt;/math&gt; and &lt;math&gt;y=2^{96}&lt;/math&gt;. Then our fraction can be written as<br /> &lt;math&gt;\frac{81x+16y}{x+y}=\frac{16x+16y}{x+y}+\frac{65x}{x+y}=16+\frac{65x}{x+y}&lt;/math&gt;.<br /> Notice that <br /> &lt;math&gt;\frac{65x}{x+y}&lt;\frac{65x}{x}=65&lt;/math&gt;.<br /> So ,<br /> &lt;math&gt;16+\frac{65x}{x+y}&lt;16+65=81&lt;/math&gt;.<br /> And our only answer choice less than 81 is &lt;math&gt;\boxed{(A) 80}&lt;/math&gt; (RegularHexagon)<br /> <br /> ==Solution 5==<br /> Let &lt;math&gt;x=\frac{3^{100}+2^{100}}{3^{96}+2^{96}}&lt;/math&gt;. Multiply both sides by &lt;math&gt;(3^{96}+2^{96})&lt;/math&gt;, and expand. Rearranging the terms, we get &lt;math&gt;3^{96}(3^4-x)+2^{96}(2^4-x)=0&lt;/math&gt;. The left side is decreasing, and it is negative when &lt;math&gt;x=81&lt;/math&gt;. This means that the answer must be less than &lt;math&gt;81&lt;/math&gt;; therefore the answer is &lt;math&gt;\boxed{(A)}&lt;/math&gt;.<br /> <br /> ==Solution 6 (eyeball it)==<br /> A faster solution. Recognize that for exponents of this size &lt;math&gt;3^{n}&lt;/math&gt; will be enormously greater than &lt;math&gt;2^{n}&lt;/math&gt;, so the terms involving &lt;math&gt;2&lt;/math&gt; will actually have very little effect on the quotient. Now we know the answer will be very close to &lt;math&gt;81&lt;/math&gt;.<br /> <br /> Notice that the terms being added on to the top and bottom are in the ratio &lt;math&gt;\frac{1}{16}&lt;/math&gt; with each other, so they must pull the ratio down from 81 very slightly. (In the same way that a new test score lower than your current cumulative grade always must pull that grade downward.) Answer: &lt;math&gt;\boxed{(A)}&lt;/math&gt;.<br /> <br /> ==Solution 7 (Using the answer choices)==<br /> We can compare the given value to each of our answer choices. We already know that it is greater than &lt;math&gt;80&lt;/math&gt; because otherwise there would have been a smaller answer, so we move onto &lt;math&gt;81&lt;/math&gt;. We get:<br /> <br /> &lt;math&gt;\frac{3^{100}+2^{100}}{3^{96}+2^{96}} \text{ ? } 3^4&lt;/math&gt;<br /> <br /> Cross multiply to get:<br /> <br /> &lt;math&gt;3^{100}+2^{100} \text{ ? }3^{100}+(2^{96})(3^4)&lt;/math&gt;<br /> <br /> Cancel out &lt;math&gt;3^{100}&lt;/math&gt; and divide by &lt;math&gt;2^{96}&lt;/math&gt; to get &lt;math&gt;2^{4} \text{ ? }3^4&lt;/math&gt;. We know that &lt;math&gt;2^4 &lt; 3^4&lt;/math&gt;, which means the expression is less than &lt;math&gt;81&lt;/math&gt; so the answer is &lt;math&gt;\boxed{(A)}&lt;/math&gt;.<br /> <br /> ==Solution 8==<br /> Notice how &lt;math&gt;\frac{3^{100}+2^{100}}{3^{96}+2^{96}}&lt;/math&gt; can be rewritten as &lt;math&gt;\frac{81(3^{96})+16(2^{96})}{3^{96}+2^{96}}=\frac{81(3^{96})+81(2^{96})}{3^{96}+2^{96}}-\frac{65(2^{96})}{3^{96}+2^{96}}=81-\frac{65(2^{96})}{3^{96}+2^{96}}&lt;/math&gt;. Note that &lt;math&gt;\frac{65(2^{96})}{3^{96}+2^{96}}&lt;1&lt;/math&gt;, so the greatest integer less than or equal to &lt;math&gt;\frac{3^{100}+2^{100}}{3^{96}+2^{96}}&lt;/math&gt; is &lt;math&gt;80&lt;/math&gt; or &lt;math&gt;\boxed{\textbf{(A)}}&lt;/math&gt;<br /> ~blitzkrieg21<br /> <br /> ==Solution 9==<br /> For positive &lt;math&gt;a, b, c, d&lt;/math&gt;, if &lt;math&gt;\frac{a}{b}&lt;\frac{c}{d}&lt;/math&gt; then &lt;math&gt;\frac{c+a}{d+b}&lt;\frac{c}{d}&lt;/math&gt;. Let &lt;math&gt;a=2^{100}, b=2^{96}, c=3^{100}, d=3^{96}&lt;/math&gt;. Then &lt;math&gt;\frac{c}{d}=3^4&lt;/math&gt;. So answer is less than 81, which leaves only one choice, 80.<br /> * Note that the algebra here is synonymous to the explanation given in Solution 6. This is the algebraic reason to the logic of if you get a test score with a lower percentage than your average (no matter how many points/percentage of your total grade it was worth), it will pull your overall grade down.<br /> <br /> ~ ccx09<br /> <br /> ==Solution 10==<br /> Try long division, and notice putting &lt;math&gt;3^4=81&lt;/math&gt; as the denominator is too big and putting &lt;math&gt;3^4-1=80&lt;/math&gt; is too small. So we know that the answer is between &lt;math&gt;80&lt;/math&gt; and &lt;math&gt;81&lt;/math&gt;, yielding &lt;math&gt;80&lt;/math&gt; as our answer.<br /> <br /> ==Solution 11(Using the Answer Choices)==<br /> We know this will be between 16 and 81 because &lt;math&gt;\frac{3^{100}}{3^{96}} = 3^4 = 81&lt;/math&gt; and &lt;math&gt;\frac{2^{100}}{2^{96}} = 2^4 = 16&lt;/math&gt;. &lt;math&gt;80=\boxed{(A)}&lt;/math&gt; is the only option choice in this range.<br /> ~Latex edits by Argonauts16<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_3&diff=138742 2018 AMC 10A Problems/Problem 3 2020-11-29T16:01:34Z <p>Sweetmango77: These two solutions are basically the same, so we only need one of them. The second solution explains it better, so I removed the first one.</p> <hr /> <div>== Problem ==<br /> <br /> A unit of blood expires after &lt;math&gt;10!=10\cdot 9 \cdot 8 \cdots 1&lt;/math&gt; seconds. Yasin donates a unit of blood at noon of January 1. On what day does his unit of blood expire?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{January 2}\qquad\textbf{(B) }\text{January 12}\qquad\textbf{(C) }\text{January 22}\qquad\textbf{(D) }\text{February 11}\qquad\textbf{(E) }\text{February 12}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/bPfLeXu9kx0<br /> <br /> Education, the Study of Everything<br /> <br /> == Solution ==<br /> <br /> The problem says there are &lt;math&gt;10! = 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1&lt;/math&gt; seconds.<br /> Convert &lt;math&gt;10!&lt;/math&gt; seconds to minutes by dividing by &lt;math&gt;60&lt;/math&gt;: &lt;math&gt;9\cdot 8\cdot 7\cdot 5\cdot 4\cdot 3\cdot 2&lt;/math&gt; minutes.<br /> Convert minutes to hours by again, dividing by &lt;math&gt;60&lt;/math&gt;: &lt;math&gt;9\cdot 8\cdot 7\cdot 2&lt;/math&gt; hours.<br /> Convert hours to days by dividing by &lt;math&gt;24&lt;/math&gt;: &lt;math&gt;3\cdot 7\cdot 2 = 42&lt;/math&gt; days.<br /> <br /> Now we need to count &lt;math&gt;42&lt;/math&gt; days after January 1. Since we start on Jan. 1, then we can't count that as a day itself. When we reach Jan. 31(The end of the month), we only have counted 30 days. &lt;math&gt;42 - 30 = 12&lt;/math&gt;. Count &lt;math&gt;12&lt;/math&gt; more days, resulting &lt;math&gt;\fbox{\textbf{(E) }\text{February 12}}&lt;/math&gt;<br /> ~nosysnow and Max0815<br /> <br /> ==Video Solution==<br /> https://youtu.be/vO-ELYmgRI8<br /> <br /> https://youtu.be/FbSYTL8tPwo<br /> <br /> ~savannahsolver<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_18&diff=138741 2020 AMC 8 Problems/Problem 18 2020-11-29T15:59:01Z <p>Sweetmango77: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; is inscribed in a semicircle with diameter &lt;math&gt;\overline{FE},&lt;/math&gt; as shown in the figure. Let &lt;math&gt;DA=16,&lt;/math&gt; and let &lt;math&gt;FD=AE=9.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt; <br /> draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(&quot;$A$&quot;,(8,0), 1.25*S); dot(&quot;$B$&quot;,(8,15), 1.25*N); dot(&quot;$C$&quot;,(-8,15), 1.25*N); dot(&quot;$D$&quot;,(-8,0), 1.25*S); dot(&quot;$E$&quot;,(17,0), 1.25*S); dot(&quot;$F$&quot;,(-17,0), 1.25*S); label(&quot;$16$&quot;,(0,0),N); label(&quot;$9$&quot;,(12.5,0),N); label(&quot;$9$&quot;,(-12.5,0),N); <br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;asy&gt; draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(&quot;$A$&quot;,(8,0), 1.25*S); dot(&quot;$B$&quot;,(8,15), 1.25*N); dot(&quot;$C$&quot;,(-8,15), 1.25*N); dot(&quot;$D$&quot;,(-8,0), 1.25*S); dot(&quot;$E$&quot;,(17,0), 1.25*S); dot(&quot;$F$&quot;,(-17,0), 1.25*S); label(&quot;$16$&quot;,(0,0),N); label(&quot;$9$&quot;,(12.5,0),N); label(&quot;$9$&quot;,(-12.5,0),N); dot(&quot;$O$&quot;, (0,0), 1.25*S); draw((0,0)--(-8,15));&lt;/asy&gt;<br /> <br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the semicircle. The diameter of the semicircle is &lt;math&gt;9+16+9=34&lt;/math&gt;, so &lt;math&gt;OC = 17&lt;/math&gt;. By symmetry, &lt;math&gt;O&lt;/math&gt; is in fact the midpoint of &lt;math&gt;DA&lt;/math&gt;, so &lt;math&gt;OD=OA=\frac{16}{2}= 8&lt;/math&gt;. By the Pythagorean theorem in right-angled triangle &lt;math&gt;ODC&lt;/math&gt; (or &lt;math&gt;OBA&lt;/math&gt;), we have that &lt;math&gt;CD&lt;/math&gt; (or &lt;math&gt;AB&lt;/math&gt;) is &lt;math&gt;\sqrt{17^2-8^2}=15&lt;/math&gt;. Accordingly, the area of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;16\cdot 15=\boxed{\textbf{(A) }240}&lt;/math&gt;.<br /> <br /> ==Solution 2 (coordinate geometry)==<br /> Let the midpoint of segment &lt;math&gt;FE&lt;/math&gt; be the origin. Evidently, point &lt;math&gt;D=(-8,0)&lt;/math&gt; and &lt;math&gt;A=(8,0)&lt;/math&gt;. Since points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; share &lt;math&gt;x&lt;/math&gt;-coordinates with &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt; respectively, it suffices to find the &lt;math&gt;y&lt;/math&gt;-coordinate of &lt;math&gt;B&lt;/math&gt; (which will be the height of the rectangle) and multiply this by &lt;math&gt;DA&lt;/math&gt; (which we know is &lt;math&gt;16&lt;/math&gt;). The radius of the semicircle is &lt;math&gt;\frac{9+16+9}{2} = 17&lt;/math&gt;, so the whole circle has equation &lt;math&gt;x^2+y^2=289&lt;/math&gt;; as already stated, &lt;math&gt;B&lt;/math&gt; has the same &lt;math&gt;x&lt;/math&gt;-coordinate as &lt;math&gt;A&lt;/math&gt;, i.e. &lt;math&gt;8&lt;/math&gt;, so substituting this into the equation shows that &lt;math&gt;y=\pm15&lt;/math&gt;. Since &lt;math&gt;y&gt;0&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt;, the y-coordinate of &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;15&lt;/math&gt;. Therefore, the answer is &lt;math&gt;16\cdot 15 = \boxed{\textbf{(A) }240}&lt;/math&gt;.<br /> <br /> (Note that the synthetic solution (Solution 1) is definitely faster and more elegant. However, this is the solution that you should use if you can't see any other easier strategy.)<br /> <br /> ==Solution 3==<br /> We can use a result from the Art of Problem Solving &lt;i&gt;Introduction to Algebra&lt;/i&gt; book Sidenote: for a semicircle with diameter &lt;math&gt;(1+n)&lt;/math&gt;, such that the &lt;math&gt;1&lt;/math&gt; part is on one side and the &lt;math&gt;n&lt;/math&gt; part is on the other side, the height from the end of the &lt;math&gt;1&lt;/math&gt; side (or the start of the &lt;math&gt;n&lt;/math&gt; side) is &lt;math&gt;\sqrt{n}&lt;/math&gt;. To use this formula, we scale the figure down by &lt;math&gt;9&lt;/math&gt;; this will give the height a length of &lt;math&gt;\sqrt{\frac{16+9}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}&lt;/math&gt;. Now, scaling back up by &lt;math&gt;9&lt;/math&gt;, the height &lt;math&gt;DC&lt;/math&gt; is &lt;math&gt;9 \cdot \frac{5}{3} = 15&lt;/math&gt;. The answer is then &lt;math&gt;15 \cdot 16 = \boxed{\textbf{(A) }240}&lt;/math&gt;.<br /> -[[User:Sweetmango77|SweetMango77]]<br /> <br /> ==Video Solution==<br /> https://youtu.be/VnOecUiP-SA<br /> <br /> {{AMC8 box|year=2020|num-b=17|num-a=19}}<br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_18&diff=138740 2020 AMC 8 Problems/Problem 18 2020-11-29T15:57:38Z <p>Sweetmango77: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; is inscribed in a semicircle with diameter &lt;math&gt;\overline{FE},&lt;/math&gt; as shown in the figure. Let &lt;math&gt;DA=16,&lt;/math&gt; and let &lt;math&gt;FD=AE=9.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt; <br /> draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(&quot;$A$&quot;,(8,0), 1.25*S); dot(&quot;$B$&quot;,(8,15), 1.25*N); dot(&quot;$C$&quot;,(-8,15), 1.25*N); dot(&quot;$D$&quot;,(-8,0), 1.25*S); dot(&quot;$E$&quot;,(17,0), 1.25*S); dot(&quot;$F$&quot;,(-17,0), 1.25*S); label(&quot;$16$&quot;,(0,0),N); label(&quot;$9$&quot;,(12.5,0),N); label(&quot;$9$&quot;,(-12.5,0),N); <br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;asy&gt; draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(&quot;$A$&quot;,(8,0), 1.25*S); dot(&quot;$B$&quot;,(8,15), 1.25*N); dot(&quot;$C$&quot;,(-8,15), 1.25*N); dot(&quot;$D$&quot;,(-8,0), 1.25*S); dot(&quot;$E$&quot;,(17,0), 1.25*S); dot(&quot;$F$&quot;,(-17,0), 1.25*S); label(&quot;$16$&quot;,(0,0),N); label(&quot;$9$&quot;,(12.5,0),N); label(&quot;$9$&quot;,(-12.5,0),N); dot(&quot;$O$&quot;, (0,0), 1.25*S); draw((0,0)--(-8,15));&lt;/asy&gt;<br /> <br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the semicircle. The diameter of the semicircle is &lt;math&gt;9+16+9=34&lt;/math&gt;, so &lt;math&gt;OC = 17&lt;/math&gt;. By symmetry, &lt;math&gt;O&lt;/math&gt; is in fact the midpoint of &lt;math&gt;DA&lt;/math&gt;, so &lt;math&gt;OD=OA=\frac{16}{2}= 8&lt;/math&gt;. By the Pythagorean theorem in right-angled triangle &lt;math&gt;ODC&lt;/math&gt; (or &lt;math&gt;OBA&lt;/math&gt;), we have that &lt;math&gt;CD&lt;/math&gt; (or &lt;math&gt;AB&lt;/math&gt;) is &lt;math&gt;\sqrt{17^2-8^2}=15&lt;/math&gt;. Accordingly, the area of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;16\cdot 15=\boxed{\textbf{(A) }240}&lt;/math&gt;.<br /> <br /> ==Solution 2 (coordinate geometry)==<br /> Let the midpoint of segment &lt;math&gt;FE&lt;/math&gt; be the origin. Evidently, point &lt;math&gt;D=(-8,0)&lt;/math&gt; and &lt;math&gt;A=(8,0)&lt;/math&gt;. Since points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; share &lt;math&gt;x&lt;/math&gt;-coordinates with &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt; respectively, it suffices to find the &lt;math&gt;y&lt;/math&gt;-coordinate of &lt;math&gt;B&lt;/math&gt; (which will be the height of the rectangle) and multiply this by &lt;math&gt;DA&lt;/math&gt; (which we know is &lt;math&gt;16&lt;/math&gt;). The radius of the semicircle is &lt;math&gt;\frac{9+16+9}{2} = 17&lt;/math&gt;, so the whole circle has equation &lt;math&gt;x^2+y^2=289&lt;/math&gt;; as already stated, &lt;math&gt;B&lt;/math&gt; has the same &lt;math&gt;x&lt;/math&gt;-coordinate as &lt;math&gt;A&lt;/math&gt;, i.e. &lt;math&gt;8&lt;/math&gt;, so substituting this into the equation shows that &lt;math&gt;y=\pm15&lt;/math&gt;. Since &lt;math&gt;y&gt;0&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt;, the y-coordinate of &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;15&lt;/math&gt;. Therefore, the answer is &lt;math&gt;16\cdot 15 = \boxed{\textbf{(A) }240}&lt;/math&gt;.<br /> <br /> (Note that the synthetic solution (Solution 1) is definitely faster and more elegant. However, this is the solution that you should use if you can't see any other easier strategy.)<br /> <br /> ==Solution 3==<br /> We can use a result from the Art of Problem Solving &lt;i&gt;Introduction to Algebra&lt;/i&gt; book Sidenote: for a semicircle with diameter &lt;math&gt;(1+n)&lt;/math&gt;, such that the &lt;math&gt;1&lt;/math&gt; part is on one side and the &lt;math&gt;n&lt;/math&gt; part is on the other side, the height from the end of the &lt;math&gt;1&lt;/math&gt; side (or the start of the &lt;math&gt;n&lt;/math&gt; side) is &lt;math&gt;\sqrt{n}&lt;/math&gt;. To use this formula, we scale the figure down by &lt;math&gt;9&lt;/math&gt;; this will give the height a length of &lt;math&gt;\sqrt{\frac{16+9}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}&lt;/math&gt;. Now, scaling back up by &lt;math&gt;9&lt;/math&gt;, the height &lt;math&gt;DC&lt;/math&gt; is &lt;math&gt;9 \cdot \frac{5}{3} = 15&lt;/math&gt;. The answer is then &lt;math&gt;15 \cdot 16 = \boxed{\textbf{(A) }240}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/VnOecUiP-SA<br /> <br /> {{AMC8 box|year=2020|num-b=17|num-a=19}}<br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_14&diff=138656 2005 AMC 10A Problems/Problem 14 2020-11-28T19:55:56Z <p>Sweetmango77: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 41\qquad \mathrm{(B) \ } 42\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 44\qquad \mathrm{(E) \ } 45 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits. <br /> <br /> Doing some [[casework]]: <br /> <br /> If the middle digit is &lt;math&gt;1&lt;/math&gt;, possible numbers range from &lt;math&gt;111&lt;/math&gt; to &lt;math&gt;210&lt;/math&gt;. So there are &lt;math&gt;2&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;2&lt;/math&gt;, possible numbers range from &lt;math&gt;123&lt;/math&gt; to &lt;math&gt;420&lt;/math&gt;. So there are &lt;math&gt;4&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;3&lt;/math&gt;, possible numbers range from &lt;math&gt;135&lt;/math&gt; to &lt;math&gt;630&lt;/math&gt;. So there are &lt;math&gt;6&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;4&lt;/math&gt;, possible numbers range from &lt;math&gt;147&lt;/math&gt; to &lt;math&gt;840&lt;/math&gt;. So there are &lt;math&gt;8&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;5&lt;/math&gt;, possible numbers range from &lt;math&gt;159&lt;/math&gt; to &lt;math&gt;951&lt;/math&gt;. So there are &lt;math&gt;9&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;6&lt;/math&gt;, possible numbers range from &lt;math&gt;369&lt;/math&gt; to &lt;math&gt;963&lt;/math&gt;. So there are &lt;math&gt;7&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;7&lt;/math&gt;, possible numbers range from &lt;math&gt;579&lt;/math&gt; to &lt;math&gt;975&lt;/math&gt;. So there are &lt;math&gt;5&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;8&lt;/math&gt;, possible numbers range from &lt;math&gt;789&lt;/math&gt; to &lt;math&gt;987&lt;/math&gt;. So there are &lt;math&gt;3&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;9&lt;/math&gt;, the only possible number is &lt;math&gt;999&lt;/math&gt;. So there is &lt;math&gt;1&lt;/math&gt; number in this case. <br /> <br /> So the total number of three-digit numbers that satisfy the property is &lt;math&gt;2+4+6+8+9+7+5+3+1=45\Rightarrow E&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> Alternatively, we could note that the middle digit is uniquely defined by the first and third digits since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into 2 cases:<br /> <br /> If both the first digit and the last digit are odd, then we have 1, 3, 5, 7, or 9 as choices for each of these digits, and there are &lt;math&gt;5\cdot5=25&lt;/math&gt; numbers in this case.<br /> <br /> If both the first and last digits are even, then we have 2, 4, 6, 8 as our choices for the first digit and 0, 2, 4, 6, 8 for the third digit. There are &lt;math&gt;4\cdot5=20&lt;/math&gt; numbers here.<br /> <br /> The total number, then, is &lt;math&gt;20+25=45\Rightarrow E&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> As we noted in Solution 2, we note that the sum of the first and third digits has to be even. The first digit can have &lt;math&gt;9&lt;/math&gt; possibilities (&lt;math&gt;1-9&lt;/math&gt;), and the third digit can have &lt;math&gt;10&lt;/math&gt; possibilities (&lt;math&gt;0-9&lt;/math&gt;). This means there can be &lt;math&gt;9\cdot10=90&lt;/math&gt; possible two-digit numbers in which the first digit and the third digit are digits. Exactly half of these would have their sum be divisible by &lt;math&gt;2&lt;/math&gt; (since &lt;math&gt;90&lt;/math&gt; is even), so our answer is &lt;math&gt;90/2=\textbf{(E) }45&lt;/math&gt;.&lt;br&gt;<br /> - [[User:Sweetmango77|SweetMango77]]<br /> <br /> ==Video Solution==<br /> CHECK OUT Video Solution: https://youtu.be/eItd9O8cCnQ<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=13|num-a=15}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_14&diff=138655 2005 AMC 10A Problems/Problem 14 2020-11-28T19:55:44Z <p>Sweetmango77: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 41\qquad \mathrm{(B) \ } 42\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 44\qquad \mathrm{(E) \ } 45 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits. <br /> <br /> Doing some [[casework]]: <br /> <br /> If the middle digit is &lt;math&gt;1&lt;/math&gt;, possible numbers range from &lt;math&gt;111&lt;/math&gt; to &lt;math&gt;210&lt;/math&gt;. So there are &lt;math&gt;2&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;2&lt;/math&gt;, possible numbers range from &lt;math&gt;123&lt;/math&gt; to &lt;math&gt;420&lt;/math&gt;. So there are &lt;math&gt;4&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;3&lt;/math&gt;, possible numbers range from &lt;math&gt;135&lt;/math&gt; to &lt;math&gt;630&lt;/math&gt;. So there are &lt;math&gt;6&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;4&lt;/math&gt;, possible numbers range from &lt;math&gt;147&lt;/math&gt; to &lt;math&gt;840&lt;/math&gt;. So there are &lt;math&gt;8&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;5&lt;/math&gt;, possible numbers range from &lt;math&gt;159&lt;/math&gt; to &lt;math&gt;951&lt;/math&gt;. So there are &lt;math&gt;9&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;6&lt;/math&gt;, possible numbers range from &lt;math&gt;369&lt;/math&gt; to &lt;math&gt;963&lt;/math&gt;. So there are &lt;math&gt;7&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;7&lt;/math&gt;, possible numbers range from &lt;math&gt;579&lt;/math&gt; to &lt;math&gt;975&lt;/math&gt;. So there are &lt;math&gt;5&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;8&lt;/math&gt;, possible numbers range from &lt;math&gt;789&lt;/math&gt; to &lt;math&gt;987&lt;/math&gt;. So there are &lt;math&gt;3&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;9&lt;/math&gt;, the only possible number is &lt;math&gt;999&lt;/math&gt;. So there is &lt;math&gt;1&lt;/math&gt; number in this case. <br /> <br /> So the total number of three-digit numbers that satisfy the property is &lt;math&gt;2+4+6+8+9+7+5+3+1=45\Rightarrow E&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> Alternatively, we could note that the middle digit is uniquely defined by the first and third digits since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into 2 cases:<br /> <br /> If both the first digit and the last digit are odd, then we have 1, 3, 5, 7, or 9 as choices for each of these digits, and there are &lt;math&gt;5\cdot5=25&lt;/math&gt; numbers in this case.<br /> <br /> If both the first and last digits are even, then we have 2, 4, 6, 8 as our choices for the first digit and 0, 2, 4, 6, 8 for the third digit. There are &lt;math&gt;4\cdot5=20&lt;/math&gt; numbers here.<br /> <br /> The total number, then, is &lt;math&gt;20+25=45\Rightarrow E&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> As we noted in Solution 2, we note that the sum of the first and third digits has to be even. The first digit can have &lt;math&gt;9&lt;/math&gt; possibilities (&lt;math&gt;1-9&lt;/math&gt;), and the third digit can have &lt;math&gt;10&lt;/math&gt; possibilities (&lt;math&gt;0-9&lt;/math&gt;). This means there can be &lt;math&gt;9\cdot10=90&lt;/math&gt; possible two-digit numbers in which the first digit and the third digit are digits. Exactly half of these would have their sum be divisible by &lt;math&gt;2&lt;/math&gt; (since &lt;math&gt;90&lt;/math&gt; is even), so our answer is &lt;math&gt;90/2=\textbf{(E) }45&lt;/math&gt;.<br /> - [[User:Sweetmango77|SweetMango77]]<br /> <br /> ==Video Solution==<br /> CHECK OUT Video Solution: https://youtu.be/eItd9O8cCnQ<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=13|num-a=15}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=Bestzack66%27s_AMC10_Study_Plan&diff=138471 Bestzack66's AMC10 Study Plan 2020-11-25T18:44:09Z <p>Sweetmango77: </p> <hr /> <div>AMC-10 Preparation Plan<br /> <br /> Topics<br /> <br /> Checkmark - ✅<br /> <br /> Algebra<br /> Logarithmic equations<br /> Solving cubic and other exponential equations<br /> Graphing functions<br /> [[Parabola|Parabolas]]<br /> Ellipses<br /> [[Hyperbola|Hyperbolas]]<br /> Conics<br /> <br /> Geometry<br /> Trigonometry<br /> Finding exact values of trig functions <br /> Graphing functions<br /> Using them in triangle problems<br /> Law of Sines<br /> Law of Cosines<br /> Shoelace bash<br /> [[Wooga Looga Theorem]]<br /> <br /> Intermediate Algebra<br /> Advanced Functions<br /> Cauchy-Schwarz Inequality<br /> Power Mean Inequality<br /> [[Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality|RMS-AM-GM-HM]]<br /> Advanced Complex numbers<br /> [[Summation|Summations]]<br /> Products<br /> <br /> CP<br /> Problems with digits<br /> Geometry with CP<br /> Distinguishability<br /> [[Georgeooga-Harryooga Theorem]]<br /> <br /> Speed<br /> Faster Arithmetic<br /> Addition<br /> Subtraction<br /> Multiplication<br /> Division<br /> Equation solving<br /> More problems<br /> <br /> <br /> == Feedback ==<br /> <br /> Please post comments here.<br /> ----------------------------------------<br /> RedFireTruck was here. - RedFireTruck<br /> <br /> Memorize the [[Wooga Looga Theorem]] and the [[Georgeooga-Harryooga Theorem]]. - RedFireTruck<br /> <br /> Ahhhh completely forgot about [[Wooga Looga]] and [[Georgeooga-Harryooga Theroem]]. Adding them now. Also gonna add some counting and probability stuff. - bestzack66<br /> <br /> bestzach66 OP - IceSheep2764<br /> <br /> HIA2020 was here<br /> <br /> yofro was not here<br /> <br /> jiseop55406 was here, it was great!</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=Bestzack66%27s_AMC10_Study_Plan&diff=138470 Bestzack66's AMC10 Study Plan 2020-11-25T18:40:23Z <p>Sweetmango77: </p> <hr /> <div>AMC-10 Preparation Plan<br /> <br /> Topics<br /> <br /> Checkmark - ✅<br /> <br /> Algebra<br /> Logarithmic equations<br /> Solving cubic and other exponential equations<br /> Graphing functions<br /> Parabolas<br /> Ellipses<br /> Hyperbolas<br /> Conics<br /> <br /> Geometry<br /> Trigonometry<br /> Finding exact values of trig functions <br /> Graphing functions<br /> Using them in triangle problems<br /> Law of Sines<br /> Law of Cosines<br /> Shoelace bash<br /> [[Wooga Looga Theorem]]<br /> <br /> Intermediate Algebra<br /> Advanced Functions<br /> Cauchy-Schwarz Inequality<br /> Power Mean Inequality<br /> [[Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality|RMS-AM-GM-HM]]<br /> Advanced Complex numbers<br /> Summations<br /> Products<br /> <br /> CP<br /> Problems with digits<br /> Geometry with CP<br /> Distinguishability<br /> [[Georgeooga-Harryooga Theorem]]<br /> <br /> Speed<br /> Faster Arithmetic<br /> Addition<br /> Subtraction<br /> Multiplication<br /> Division<br /> Equation solving<br /> More problems<br /> <br /> <br /> == Feedback ==<br /> <br /> Please post comments here.<br /> ----------------------------------------<br /> RedFireTruck was here. - RedFireTruck<br /> <br /> Memorize the [[Wooga Looga Theorem]] and the [[Georgeooga-Harryooga Theorem]]. - RedFireTruck<br /> <br /> Ahhhh completely forgot about [[Wooga Looga]] and [[Georgeooga-Harryooga Theroem]]. Adding them now. Also gonna add some counting and probability stuff. - bestzack66<br /> <br /> bestzach66 OP - IceSheep2764<br /> <br /> HIA2020 was here<br /> <br /> yofro was not here<br /> <br /> jiseop55406 was here, it was great!</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_14&diff=138469 2005 AMC 10A Problems/Problem 14 2020-11-25T18:37:54Z <p>Sweetmango77: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 41\qquad \mathrm{(B) \ } 42\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 44\qquad \mathrm{(E) \ } 45 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits. <br /> <br /> Doing some [[casework]]: <br /> <br /> If the middle digit is &lt;math&gt;1&lt;/math&gt;, possible numbers range from &lt;math&gt;111&lt;/math&gt; to &lt;math&gt;210&lt;/math&gt;. So there are &lt;math&gt;2&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;2&lt;/math&gt;, possible numbers range from &lt;math&gt;123&lt;/math&gt; to &lt;math&gt;420&lt;/math&gt;. So there are &lt;math&gt;4&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;3&lt;/math&gt;, possible numbers range from &lt;math&gt;135&lt;/math&gt; to &lt;math&gt;630&lt;/math&gt;. So there are &lt;math&gt;6&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;4&lt;/math&gt;, possible numbers range from &lt;math&gt;147&lt;/math&gt; to &lt;math&gt;840&lt;/math&gt;. So there are &lt;math&gt;8&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;5&lt;/math&gt;, possible numbers range from &lt;math&gt;159&lt;/math&gt; to &lt;math&gt;951&lt;/math&gt;. So there are &lt;math&gt;9&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;6&lt;/math&gt;, possible numbers range from &lt;math&gt;369&lt;/math&gt; to &lt;math&gt;963&lt;/math&gt;. So there are &lt;math&gt;7&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;7&lt;/math&gt;, possible numbers range from &lt;math&gt;579&lt;/math&gt; to &lt;math&gt;975&lt;/math&gt;. So there are &lt;math&gt;5&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;8&lt;/math&gt;, possible numbers range from &lt;math&gt;789&lt;/math&gt; to &lt;math&gt;987&lt;/math&gt;. So there are &lt;math&gt;3&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;9&lt;/math&gt;, the only possible number is &lt;math&gt;999&lt;/math&gt;. So there is &lt;math&gt;1&lt;/math&gt; number in this case. <br /> <br /> So the total number of three-digit numbers that satisfy the property is &lt;math&gt;2+4+6+8+9+7+5+3+1=45\Rightarrow E&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> Alternatively, we could note that the middle digit is uniquely defined by the first and third digits since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into 2 cases:<br /> <br /> If both the first digit and the last digit are odd, then we have 1, 3, 5, 7, or 9 as choices for each of these digits, and there are &lt;math&gt;5\cdot5=25&lt;/math&gt; numbers in this case.<br /> <br /> If both the first and last digits are even, then we have 2, 4, 6, 8 as our choices for the first digit and 0, 2, 4, 6, 8 for the third digit. There are &lt;math&gt;4\cdot5=20&lt;/math&gt; numbers here.<br /> <br /> The total number, then, is &lt;math&gt;20+25=45\Rightarrow E&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> As we noted in Solution 2, we note that the sum of the first and third digits has to be even. The first digit can have &lt;math&gt;9&lt;/math&gt; possibilities (&lt;math&gt;1-9&lt;/math&gt;), and the third digit can have &lt;math&gt;10&lt;/math&gt; possibilities (&lt;math&gt;0-9&lt;/math&gt;). This means there can be &lt;math&gt;9\cdot10=90&lt;/math&gt; possible three-digit numbers in which the first digit and the third digit are digits and the second digit is a &lt;math&gt;0&lt;/math&gt; (or any other constant). Exactly half of these would be divisible by &lt;math&gt;2&lt;/math&gt;, so our answer is &lt;math&gt;90/2=\textbf{(E) }45&lt;/math&gt;.<br /> -[[User:Sweetmango77|SweetMango77]]<br /> <br /> ==Video Solution==<br /> CHECK OUT Video Solution: https://youtu.be/eItd9O8cCnQ<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=13|num-a=15}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_14&diff=138468 2005 AMC 10A Problems/Problem 14 2020-11-25T18:37:43Z <p>Sweetmango77: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 41\qquad \mathrm{(B) \ } 42\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 44\qquad \mathrm{(E) \ } 45 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits. <br /> <br /> Doing some [[casework]]: <br /> <br /> If the middle digit is &lt;math&gt;1&lt;/math&gt;, possible numbers range from &lt;math&gt;111&lt;/math&gt; to &lt;math&gt;210&lt;/math&gt;. So there are &lt;math&gt;2&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;2&lt;/math&gt;, possible numbers range from &lt;math&gt;123&lt;/math&gt; to &lt;math&gt;420&lt;/math&gt;. So there are &lt;math&gt;4&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;3&lt;/math&gt;, possible numbers range from &lt;math&gt;135&lt;/math&gt; to &lt;math&gt;630&lt;/math&gt;. So there are &lt;math&gt;6&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;4&lt;/math&gt;, possible numbers range from &lt;math&gt;147&lt;/math&gt; to &lt;math&gt;840&lt;/math&gt;. So there are &lt;math&gt;8&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;5&lt;/math&gt;, possible numbers range from &lt;math&gt;159&lt;/math&gt; to &lt;math&gt;951&lt;/math&gt;. So there are &lt;math&gt;9&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;6&lt;/math&gt;, possible numbers range from &lt;math&gt;369&lt;/math&gt; to &lt;math&gt;963&lt;/math&gt;. So there are &lt;math&gt;7&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;7&lt;/math&gt;, possible numbers range from &lt;math&gt;579&lt;/math&gt; to &lt;math&gt;975&lt;/math&gt;. So there are &lt;math&gt;5&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;8&lt;/math&gt;, possible numbers range from &lt;math&gt;789&lt;/math&gt; to &lt;math&gt;987&lt;/math&gt;. So there are &lt;math&gt;3&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;9&lt;/math&gt;, the only possible number is &lt;math&gt;999&lt;/math&gt;. So there is &lt;math&gt;1&lt;/math&gt; number in this case. <br /> <br /> So the total number of three-digit numbers that satisfy the property is &lt;math&gt;2+4+6+8+9+7+5+3+1=45\Rightarrow E&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> Alternatively, we could note that the middle digit is uniquely defined by the first and third digits since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into 2 cases:<br /> <br /> If both the first digit and the last digit are odd, then we have 1, 3, 5, 7, or 9 as choices for each of these digits, and there are &lt;math&gt;5\cdot5=25&lt;/math&gt; numbers in this case.<br /> <br /> If both the first and last digits are even, then we have 2, 4, 6, 8 as our choices for the first digit and 0, 2, 4, 6, 8 for the third digit. There are &lt;math&gt;4\cdot5=20&lt;/math&gt; numbers here.<br /> <br /> The total number, then, is &lt;math&gt;20+25=45\Rightarrow E&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> As we noted in Solution 2, we note that the sum of the first and third digits has to be even. The first digit can have &lt;math&gt;9&lt;/math&gt; possibilities (&lt;math&gt;1-9&lt;/math&gt;), and the third digit can have &lt;math&gt;10&lt;/math&gt; possibilities (&lt;math&gt;0-9&lt;/math&gt;). This means there can be &lt;math&gt;9\cdot10=90&lt;/math&gt; possible three-digit numbers in which the first digit and the third digit are digits and the second digit is a &lt;math&gt;0&lt;/math&gt; (or any other constant). Exactly half of these would be divisible by &lt;math&gt;2&lt;/math&gt;, so our answer is &lt;math&gt;90/2=\textbf{(E) }45&lt;/math&gt;.<br /> <br /> -[[User:Sweetmango77|SweetMango77]]<br /> <br /> ==Video Solution==<br /> CHECK OUT Video Solution: https://youtu.be/eItd9O8cCnQ<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=13|num-a=15}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_14&diff=138467 2005 AMC 10A Problems/Problem 14 2020-11-25T18:37:18Z <p>Sweetmango77: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 41\qquad \mathrm{(B) \ } 42\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 44\qquad \mathrm{(E) \ } 45 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits. <br /> <br /> Doing some [[casework]]: <br /> <br /> If the middle digit is &lt;math&gt;1&lt;/math&gt;, possible numbers range from &lt;math&gt;111&lt;/math&gt; to &lt;math&gt;210&lt;/math&gt;. So there are &lt;math&gt;2&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;2&lt;/math&gt;, possible numbers range from &lt;math&gt;123&lt;/math&gt; to &lt;math&gt;420&lt;/math&gt;. So there are &lt;math&gt;4&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;3&lt;/math&gt;, possible numbers range from &lt;math&gt;135&lt;/math&gt; to &lt;math&gt;630&lt;/math&gt;. So there are &lt;math&gt;6&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;4&lt;/math&gt;, possible numbers range from &lt;math&gt;147&lt;/math&gt; to &lt;math&gt;840&lt;/math&gt;. So there are &lt;math&gt;8&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;5&lt;/math&gt;, possible numbers range from &lt;math&gt;159&lt;/math&gt; to &lt;math&gt;951&lt;/math&gt;. So there are &lt;math&gt;9&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;6&lt;/math&gt;, possible numbers range from &lt;math&gt;369&lt;/math&gt; to &lt;math&gt;963&lt;/math&gt;. So there are &lt;math&gt;7&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;7&lt;/math&gt;, possible numbers range from &lt;math&gt;579&lt;/math&gt; to &lt;math&gt;975&lt;/math&gt;. So there are &lt;math&gt;5&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;8&lt;/math&gt;, possible numbers range from &lt;math&gt;789&lt;/math&gt; to &lt;math&gt;987&lt;/math&gt;. So there are &lt;math&gt;3&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;9&lt;/math&gt;, the only possible number is &lt;math&gt;999&lt;/math&gt;. So there is &lt;math&gt;1&lt;/math&gt; number in this case. <br /> <br /> So the total number of three-digit numbers that satisfy the property is &lt;math&gt;2+4+6+8+9+7+5+3+1=45\Rightarrow E&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> Alternatively, we could note that the middle digit is uniquely defined by the first and third digits since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into 2 cases:<br /> <br /> If both the first digit and the last digit are odd, then we have 1, 3, 5, 7, or 9 as choices for each of these digits, and there are &lt;math&gt;5\cdot5=25&lt;/math&gt; numbers in this case.<br /> <br /> If both the first and last digits are even, then we have 2, 4, 6, 8 as our choices for the first digit and 0, 2, 4, 6, 8 for the third digit. There are &lt;math&gt;4\cdot5=20&lt;/math&gt; numbers here.<br /> <br /> The total number, then, is &lt;math&gt;20+25=45\Rightarrow E&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> As we noted in Solution 2, we note that the sum of the first and third digits has to be even. The first digit can have &lt;math&gt;9&lt;/math&gt; possibilities (&lt;math&gt;1-9&lt;/math&gt;), and the third digit can have &lt;math&gt;10&lt;/math&gt; possibilities (&lt;math&gt;0-9&lt;/math&gt;). This means there can be &lt;math&gt;9\cdot10=90&lt;/math&gt; possible three-digit numbers in which the first digit and the third digit are digits and the second digit is a &lt;math&gt;0&lt;/math&gt; (or any other constant). Exactly half of these would be divisible by &lt;math&gt;2&lt;/math&gt;, so our answer is &lt;math&gt;90/2=\textbf{(E) }45&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> CHECK OUT Video Solution: https://youtu.be/eItd9O8cCnQ<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=13|num-a=15}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=User:Sweetmango77&diff=138466 User:Sweetmango77 2020-11-25T18:27:32Z <p>Sweetmango77: /* SweetMango77: */</p> <hr /> <div>==&lt;font color=&quot;black&quot; style=&quot;font-family: Chalkduster&quot;&gt;SweetMango77:&lt;/font&gt;==<br /> <br /> &lt;font color=&quot;black&quot; style=&quot;font-family: Chalkduster&quot;&gt;<br /> A person who made a 25x100 multiplication chart: https://docs.google.com/spreadsheets/d/1Tdy1sz8y7LuO9yqADUnAXZ583IqnFlmTM88XsaX1W6g/edit#gid=0<br /> <br /> Username in Minecraft is &lt;strong&gt;&lt;i&gt;Dat_Is_Stupid&lt;/i&gt;&lt;/strong&gt;, please friend me in Hypixel when I am active.<br /> <br /> Also good at coding with HTML, CSS, and JavaScript.<br /> &lt;hr&gt;<br /> PM me if you would like to edit this, otherwise I will undo all of your changes, unless they are good changes.<br /> &lt;/font&gt;</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_15&diff=138465 2020 AMC 8 Problems/Problem 15 2020-11-25T18:22:00Z <p>Sweetmango77: /* Solution 4 */</p> <hr /> <div>==Problem==<br /> Suppose &lt;math&gt;15\%&lt;/math&gt; of &lt;math&gt;x&lt;/math&gt; equals &lt;math&gt;20\%&lt;/math&gt; of &lt;math&gt;y.&lt;/math&gt; What percentage of &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;y?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since &lt;math&gt;20\% = \frac{1}{5}&lt;/math&gt;, multiplying the given condition by &lt;math&gt;5&lt;/math&gt; shows that &lt;math&gt;y&lt;/math&gt; is &lt;math&gt;15 \cdot 5 = \boxed{\textbf{(C) }75}&lt;/math&gt; percent of &lt;math&gt;x&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Letting &lt;math&gt;x=100&lt;/math&gt; (without loss of generality), the condition becomes &lt;math&gt;0.15\cdot 100 = 0.2\cdot y \Rightarrow 15 = \frac{y}{5} \Rightarrow y=75&lt;/math&gt;. Clearly, it follows that &lt;math&gt;y&lt;/math&gt; is &lt;math&gt;75\%&lt;/math&gt; of &lt;math&gt;x&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C) }75}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> We have &lt;math&gt;15\%=\frac{3}{20}&lt;/math&gt; and &lt;math&gt;20\%=\frac{1}{5}&lt;/math&gt;, so &lt;math&gt;\frac{3}{20}x=\frac{1}{5}y&lt;/math&gt;. Solving for &lt;math&gt;y&lt;/math&gt;, we multiply by &lt;math&gt;5&lt;/math&gt; to give &lt;math&gt;y = \frac{15}{20}x = \frac{3}{4}x&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C) }75}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> WLOG, we can assume that &lt;math&gt;x=20&lt;/math&gt; and &lt;math&gt;y=15&lt;/math&gt; to make them equal. This means &lt;math&gt;\frac{y}{x}=\frac{15}{20}=\frac{75}{100}&lt;/math&gt;, and thus answer is &lt;math&gt;\boxed{\textbf{(C) }75}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/mjS-PHTw-GE<br /> <br /> ==See also==<br /> {{AMC8 box|year=2020|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=User:Sweetmango77&diff=138464 User:Sweetmango77 2020-11-25T17:59:59Z <p>Sweetmango77: /* SweetMango77: */</p> <hr /> <div>==&lt;font color=&quot;black&quot; style=&quot;font-family: Chalkduster&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;SweetMango77:&lt;/div&gt;&lt;/font&gt;==<br /> <br /> &lt;font color=&quot;black&quot; style=&quot;font-family: Chalkduster&quot;&gt;<br /> A person who made a 25x100 multiplication chart: https://docs.google.com/spreadsheets/d/1Tdy1sz8y7LuO9yqADUnAXZ583IqnFlmTM88XsaX1W6g/edit#gid=0<br /> <br /> Username in Minecraft is &lt;strong&gt;&lt;i&gt;Dat_Is_Stupid&lt;/i&gt;&lt;/strong&gt;, please friend me in Hypixel when I am active.<br /> <br /> Also good at coding with HTML, CSS, and JavaScript.<br /> &lt;hr&gt;<br /> PM me if you would like to edit this, otherwise I will undo all of your changes, unless they are good changes.<br /> &lt;/font&gt;</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=User:Sweetmango77&diff=138463 User:Sweetmango77 2020-11-25T17:57:33Z <p>Sweetmango77: /* SweetMango77: */</p> <hr /> <div>==&lt;font color=&quot;black&quot; style=&quot;font-family: Chalkduster&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;SweetMango77:&lt;/div&gt;&lt;/font&gt;==<br /> <br /> &lt;font color=&quot;black&quot; style=&quot;font-family: Chalkduster&quot;&gt;<br /> A person who made a 25x100 multiplication chart: https://docs.google.com/spreadsheets/d/1Tdy1sz8y7LuO9yqADUnAXZ583IqnFlmTM88XsaX1W6g/edit#gid=0<br /> &lt;hr&gt;<br /> PM me if you would like to edit this, otherwise I will undo all of your changes, unless they are good changes.<br /> <br /> Also good at coding with HTML, CSS, and JavaScript.&lt;/font&gt;</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=User:Sweetmango77&diff=138433 User:Sweetmango77 2020-11-25T01:56:23Z <p>Sweetmango77: /* SweetMango77 */</p> <hr /> <div>==&lt;font color=&quot;black&quot; style=&quot;font-family: Chalkduster&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;SweetMango77:&lt;/div&gt;&lt;/font&gt;==<br /> <br /> &lt;font color=&quot;black&quot; style=&quot;font-family: Chalkduster&quot;&gt;<br /> A person who made a 25x100 multiplication chart: https://docs.google.com/spreadsheets/d/1Tdy1sz8y7LuO9yqADUnAXZ583IqnFlmTM88XsaX1W6g/edit#gid=0<br /> <br /> Also good at coding with HTML, CSS, and JavaScript.&lt;/font&gt;<br /> &lt;hr&gt;<br /> PM me if you would like to edit this, otherwise I will undo all of your changes, unless they are good changes.</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=User:Sweetmango77&diff=138432 User:Sweetmango77 2020-11-25T01:49:42Z <p>Sweetmango77: /* SweetMango77 */</p> <hr /> <div>==SweetMango77==<br /> A person who made a 25x100 multiplication chart: https://docs.google.com/spreadsheets/d/1Tdy1sz8y7LuO9yqADUnAXZ583IqnFlmTM88XsaX1W6g/edit#gid=0<br /> <br /> Also good at coding with HTML, CSS, and JavaScript.<br /> &lt;hr&gt;<br /> PM me if you would like to edit this, otherwise I will undo all of your changes, unless they are good changes.</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=User:Sweetmango77&diff=138431 User:Sweetmango77 2020-11-25T01:49:15Z <p>Sweetmango77: /* SweetMango77 */</p> <hr /> <div>==SweetMango77==<br /> A person who made a 25x100 multiplication chart: https://docs.google.com/spreadsheets/d/1Tdy1sz8y7LuO9yqADUnAXZ583IqnFlmTM88XsaX1W6g/edit#gid=0<br /> <br /> Also good at coding with &lt;strong&gt;&lt;i&gt;HTML, CSS, and JavaScript&lt;/i&gt;&lt;/strong&gt;<br /> &lt;hr&gt;<br /> PM me if you would like to edit this, otherwise I will undo all of your changes, unless they are good changes.</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=User:Sweetmango77&diff=138430 User:Sweetmango77 2020-11-25T01:48:46Z <p>Sweetmango77: /* SweetMango77 */</p> <hr /> <div>==SweetMango77==<br /> A person who made a 25x100 multiplication chart: https://docs.google.com/spreadsheets/d/1Tdy1sz8y7LuO9yqADUnAXZ583IqnFlmTM88XsaX1W6g/edit#gid=0<br /> <br /> Also good at coding with [b][i]HTML, CSS, and JavaScript[/i][/b]<br /> &lt;hr&gt;<br /> PM me if you would like to edit this, otherwise I will undo all of your changes, unless they are good changes.</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=Georgeooga-Harryooga_Theorem&diff=138426 Georgeooga-Harryooga Theorem 2020-11-25T01:45:28Z <p>Sweetmango77: </p> <hr /> <div>&lt;h1&gt;Overview&lt;/h1&gt;<br /> <br /> This is not a legit theorem<br /> <br /> &lt;i&gt;@Sugar rush&lt;/i&gt; Even though this is not a real theorem, it could be useful to use this, so I will bring parts of it back:<br /> <br /> &lt;h1&gt;Definition&lt;/h1&gt;<br /> The Georgeooga-Harryooga Theorem states that if you have &lt;math&gt;a&lt;/math&gt; distinguishable objects and &lt;math&gt;b&lt;/math&gt; are kept away from each other, then there are &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects.<br /> <br /> &lt;h1&gt;Proof&lt;/h1&gt;<br /> Let our group of &lt;math&gt;a&lt;/math&gt; objects be represented like so &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, ..., &lt;math&gt;a-1&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;. Let the last &lt;math&gt;b&lt;/math&gt; objects be the ones we can't have together.<br /> <br /> Then we can organize our objects like so &lt;math&gt;\square1\square2\square3\square...\square a-b-1\square a-b\square&lt;/math&gt;.<br /> <br /> We have &lt;math&gt;(a-b)!&lt;/math&gt; ways to arrange the objects in that list.<br /> <br /> Now we have &lt;math&gt;a-b+1&lt;/math&gt; blanks and &lt;math&gt;b&lt;/math&gt; other objects so we have &lt;math&gt;_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects we can't put together.<br /> <br /> By fundamental counting principal our answer is &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt;.<br /> <br /> <br /> Proof by [[User:Redfiretruck|RedFireTruck]]<br /> <br /> &lt;h1&gt;Application&lt;/h1&gt;<br /> <br /> Alice, Bob, Carl, David, Eric, Fred, George, and Harry want to stand in a line to buy ice cream. Fred and George are identical twins, so they are indistinguishable. Alice, Bob, and Carl had a serious disagreement in 6th grade, so none of them can be together in the line.<br /> With these conditions, how many different ways can you arrange these kids in a line?<br /> <br /> Problem by Math4Life2020<br /> <br /> &lt;h2&gt;Solution&lt;/h2&gt;<br /> <br /> If Eric and Fred were distinguishable we would have &lt;math&gt;\frac{(8-3)!(8-3+1)!}{(8-2\cdot3+1)!}=14400&lt;/math&gt; ways to arrange them by the Georgeooga-Harryooga Theorem. However, Eric and Fred are indistinguishable so we have to divide by &lt;math&gt;2!=2&lt;/math&gt;. Therefore, our answer is &lt;math&gt;\frac{14400}2=\boxed{7200}&lt;/math&gt;.<br /> <br /> <br /> Solution by [[User:Redfiretruck|RedFireTruck]]<br /> &lt;hr&gt;<br /> &lt;strong&gt;ALL THINGS ABOVE EXCEPT FOR THE OVERVIEW TAB AND THE PROBLEM IS MADE BY RedFireTruck &lt;/strong&gt;</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=Georgeooga-Harryooga_Theorem&diff=138425 Georgeooga-Harryooga Theorem 2020-11-25T01:44:41Z <p>Sweetmango77: </p> <hr /> <div>&lt;h1&gt;Overview&lt;/h1&gt;<br /> <br /> This is not a legit theorem<br /> <br /> &lt;i&gt;@Sugar rush&lt;/i&gt; Even though this is not a real theorem, it could be useful to use this, so I will bring parts of it back:<br /> <br /> &lt;h1&gt;Definition&lt;/h1&gt;<br /> The Georgeooga-Harryooga Theorem states that if you have &lt;math&gt;a&lt;/math&gt; distinguishable objects and &lt;math&gt;b&lt;/math&gt; are kept away from each other, then there are &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects.<br /> <br /> &lt;h1&gt;Proof&lt;/h1&gt;<br /> Let our group of &lt;math&gt;a&lt;/math&gt; objects be represented like so &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, ..., &lt;math&gt;a-1&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;. Let the last &lt;math&gt;b&lt;/math&gt; objects be the ones we can't have together.<br /> <br /> Then we can organize our objects like so &lt;math&gt;\square1\square2\square3\square...\square a-b-1\square a-b\square&lt;/math&gt;.<br /> <br /> We have &lt;math&gt;(a-b)!&lt;/math&gt; ways to arrange the objects in that list.<br /> <br /> Now we have &lt;math&gt;a-b+1&lt;/math&gt; blanks and &lt;math&gt;b&lt;/math&gt; other objects so we have &lt;math&gt;_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects we can't put together.<br /> <br /> By fundamental counting principal our answer is &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt;.<br /> <br /> <br /> Proof by [[User:Redfiretruck|RedFireTruck]]<br /> <br /> &lt;h1&gt;Application&lt;/h1&gt;<br /> <br /> Alice, Bob, Carl, David, Eric, Fred, George, and Harry want to stand in a line to buy ice cream. Fred and George are identical twins, so they are indistinguishable. Alice, Bob, and Carl had a serious disagreement in 6th grade, so none of them can be together in the line.<br /> With these conditions, how many different ways can you arrange these kids in a line?<br /> <br /> Problem by Math4Life2020<br /> <br /> &lt;h2&gt;Solution&lt;/h2&gt;<br /> <br /> If Eric and Fred were distinguishable we would have &lt;math&gt;\frac{(8-3)!(8-3+1)!}{(8-2\cdot3+1)!}=14400&lt;/math&gt; ways to arrange them by the Georgeooga-Harryooga Theorem. However, Eric and Fred are indistinguishable so we have to divide by &lt;math&gt;2!=2&lt;/math&gt;. Therefore, our answer is &lt;math&gt;\frac{14400}2=\boxed{7200}&lt;/math&gt;.<br /> <br /> <br /> Solution by [[User:Redfiretruck|RedFireTruck]]<br /> &lt;hr&gt;<br /> &lt;strong&gt;ALL THINGS ABOVE EXCEPT FOR THE OVERVIEW TAB IS MADE BY RedFireTruck.&lt;/strong&gt;</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=Georgeooga-Harryooga_Theorem&diff=138424 Georgeooga-Harryooga Theorem 2020-11-25T01:42:28Z <p>Sweetmango77: </p> <hr /> <div>&lt;h1&gt;Overview&lt;/h1&gt;<br /> <br /> This is not a legit theorem<br /> <br /> [@Sugar rush] Even though this is not a real theorem, it could be useful to use this, so I will bring parts of it back:<br /> <br /> &lt;h1&gt;Definition&lt;/h1&gt;<br /> The Georgeooga-Harryooga Theorem states that if you have &lt;math&gt;a&lt;/math&gt; distinguishable objects and &lt;math&gt;b&lt;/math&gt; are kept away from each other, then there are &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects.<br /> <br /> &lt;h1&gt;Proof&lt;/h1&gt;<br /> Let our group of &lt;math&gt;a&lt;/math&gt; objects be represented like so &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, ..., &lt;math&gt;a-1&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;. Let the last &lt;math&gt;b&lt;/math&gt; objects be the ones we can't have together.<br /> <br /> Then we can organize our objects like so &lt;math&gt;\square1\square2\square3\square...\square a-b-1\square a-b\square&lt;/math&gt;.<br /> <br /> We have &lt;math&gt;(a-b)!&lt;/math&gt; ways to arrange the objects in that list.<br /> <br /> Now we have &lt;math&gt;a-b+1&lt;/math&gt; blanks and &lt;math&gt;b&lt;/math&gt; other objects so we have &lt;math&gt;_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects we can't put together.<br /> <br /> By fundamental counting principal our answer is &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt;.<br /> <br /> <br /> Proof by [[User:Redfiretruck|RedFireTruck]]<br /> <br /> &lt;h1&gt;Application&lt;/h1&gt;<br /> <br /> Alice, Bob, Carl, David, Eric, Fred, George, and Harry want to stand in a line to buy ice cream. Fred and George are identical twins, so they are indistinguishable. Alice, Bob, and Carl had a serious disagreement in 6th grade, so none of them can be together in the line.<br /> With these conditions, how many different ways can you arrange these kids in a line?<br /> <br /> Problem by Math4Life2020<br /> <br /> &lt;h2&gt;Solution&lt;/h2&gt;<br /> <br /> If Eric and Fred were distinguishable we would have &lt;math&gt;\frac{(8-3)!(8-3+1)!}{(8-2\cdot3+1)!}=14400&lt;/math&gt; ways to arrange them by the Georgeooga-Harryooga Theorem. However, Eric and Fred are indistinguishable so we have to divide by &lt;math&gt;2!=2&lt;/math&gt;. Therefore, our answer is &lt;math&gt;\frac{14400}2=\boxed{7200}&lt;/math&gt;.<br /> <br /> <br /> Solution by [[User:Redfiretruck|RedFireTruck]]<br /> &lt;hr&gt;<br /> &lt;strong&gt;ALL THINGS ABOVE EXCEPT FOR THE OVERVIEW TAB IS MADE BY RedFireTruck.&lt;/strong&gt;</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=User:Sweetmango77&diff=138421 User:Sweetmango77 2020-11-25T01:11:06Z <p>Sweetmango77: /* SweetMango77 */</p> <hr /> <div>==SweetMango77==<br /> A person who made a 25x100 multiplication chart: https://docs.google.com/spreadsheets/d/1Tdy1sz8y7LuO9yqADUnAXZ583IqnFlmTM88XsaX1W6g/edit#gid=0</div> Sweetmango77 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_18&diff=138420 2020 AMC 8 Problems/Problem 18 2020-11-25T01:09:36Z <p>Sweetmango77: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; is inscribed in a semicircle with diameter &lt;math&gt;\overline{FE},&lt;/math&gt; as shown in the figure. Let &lt;math&gt;DA=16,&lt;/math&gt; and let &lt;math&gt;FD=AE=9.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt; <br /> draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(&quot;$A$&quot;,(8,0), 1.25*S); dot(&quot;$B$&quot;,(8,15), 1.25*N); dot(&quot;$C$&quot;,(-8,15), 1.25*N); dot(&quot;$D$&quot;,(-8,0), 1.25*S); dot(&quot;$E$&quot;,(17,0), 1.25*S); dot(&quot;$F$&quot;,(-17,0), 1.25*S); label(&quot;$16$&quot;,(0,0),N); label(&quot;$9$&quot;,(12.5,0),N); label(&quot;$9$&quot;,(-12.5,0),N); <br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;asy&gt; draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(&quot;$A$&quot;,(8,0), 1.25*S); dot(&quot;$B$&quot;,(8,15), 1.25*N); dot(&quot;$C$&quot;,(-8,15), 1.25*N); dot(&quot;$D$&quot;,(-8,0), 1.25*S); dot(&quot;$E$&quot;,(17,0), 1.25*S); dot(&quot;$F$&quot;,(-17,0), 1.25*S); label(&quot;$16$&quot;,(0,0),N); label(&quot;$9$&quot;,(12.5,0),N); label(&quot;$9$&quot;,(-12.5,0),N); dot(&quot;$O$&quot;, (0,0), 1.25*S); draw((0,0)--(-8,15));&lt;/asy&gt;<br /> <br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the semicircle. The diameter of the semicircle is &lt;math&gt;9+16+9=34&lt;/math&gt;, so &lt;math&gt;OC = 17&lt;/math&gt;. By symmetry, &lt;math&gt;O&lt;/math&gt; is in fact the midpoint of &lt;math&gt;DA&lt;/math&gt;, so &lt;math&gt;OD=OA=\frac{16}{2}= &lt;/math&gt;. By the Pythagorean theorem in right-angled triangle &lt;math&gt;ODC&lt;/math&gt; (or &lt;math&gt;OBA&lt;/math&gt;), we have that &lt;math&gt;CD&lt;/math&gt; (or &lt;math&gt;AB&lt;/math&gt;) is &lt;math&gt;\sqrt{17^2-8^2}=15&lt;/math&gt;. Accordingly, the area of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;16\cdot 15=\boxed{\textbf{(A) }240}&lt;/math&gt;.<br /> <br /> ==Solution 2 (coordinate geometry)==<br /> Let the midpoint of segment &lt;math&gt;FE&lt;/math&gt; be the origin. Evidently, point &lt;math&gt;D=(-8,0)&lt;/math&gt; and &lt;math&gt;A=(8,0)&lt;/math&gt;. Since points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; share &lt;math&gt;x&lt;/math&gt;-coordinates with &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt; respectively, it suffices to find the &lt;math&gt;y&lt;/math&gt;-coordinate of &lt;math&gt;B&lt;/math&gt; (which will be the height of the rectangle) and multiply this by &lt;math&gt;DA&lt;/math&gt; (which we know is &lt;math&gt;16&lt;/math&gt;). The radius of the semicircle is &lt;math&gt;\frac{9+16+9}{2} = 17&lt;/math&gt;, so the whole circle has equation &lt;math&gt;x^2+y^2=289&lt;/math&gt;; as already stated, &lt;math&gt;B&lt;/math&gt; has the same &lt;math&gt;x&lt;/math&gt;-coordinate as &lt;math&gt;A&lt;/math&gt;, i.e. &lt;math&gt;8&lt;/math&gt;, so substituting this into the equation shows that &lt;math&gt;y=\pm15&lt;/math&gt;. Since &lt;math&gt;y&gt;0&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt;, the y-coordinate of &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;15&lt;/math&gt;. Therefore, the answer is &lt;math&gt;16\cdot 15 = \boxed{\textbf{(A) }240}&lt;/math&gt;.<br /> <br /> (Note that the synthetic solution (Solution 1) is definitely faster and more elegant. However, this is the solution that you should use if you can't see any other easier strategy.)<br /> <br /> ==Solution 3==<br /> We can use a result from the Art of Problem Solving &lt;i&gt;Introduction to Algebra&lt;/i&gt; book: for a semicircle with diameter &lt;math&gt;(1+n)&lt;/math&gt;, such that the &lt;math&gt;1&lt;/math&gt; part is on one side and the &lt;math&gt;n&lt;/math&gt; part is on the other side, the height from the end of the &lt;math&gt;1&lt;/math&gt; side (or the start of the &lt;math&gt;n&lt;/math&gt; side) is &lt;math&gt;\sqrt{n}&lt;/math&gt;. To use this, we scale the figure down by &lt;math&gt;9&lt;/math&gt;; then the height is &lt;math&gt;\sqrt{1+\frac{16}{9}} = \sqrt{\frac{16+9}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}&lt;/math&gt;. Now, scaling back up by &lt;math&gt;9&lt;/math&gt;, the height &lt;math&gt;DC&lt;/math&gt; is &lt;math&gt;9 \cdot \frac{5}{3} = 15&lt;/math&gt;. The answer is &lt;math&gt;15 \cdot 16 = \boxed{\textbf{(A) }240}&lt;/math&gt;.&lt;br&gt;<br /> -[[User:Sweetmango77|SweetMango77]]; edited by Sevenoptimus, who made it easier to read<br /> <br /> ==Video Solution==<br /> https://youtu.be/l9wZS3qGSCg<br /> <br /> {{AMC8 box|year=2020|num-b=17|num-a=19}}<br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Sweetmango77