https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mutinykids&feedformat=atom AoPS Wiki - User contributions [en] 2021-06-19T23:01:46Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=Anthony_Yang&diff=153488 Anthony Yang 2021-05-10T18:37:30Z <p>Mutinykids: </p> <hr /> <div>Pro future janitor that lives in NC. The only person that doesn't silly that I know of. AKA mutinykids<br /> <br /> aYO why you gotta expose me</div> Mutinykids https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_5&diff=104825 2019 AIME II Problems/Problem 5 2019-03-23T00:47:26Z <p>Mutinykids: /* Solution */</p> <hr /> <div>==Problem==<br /> Four ambassadors and one advisor for each of them are to be seated at a round table with &lt;math&gt;12&lt;/math&gt; chairs numbered in order &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;12&lt;/math&gt;. Each ambassador must sit in an even-numbered chair. Each advisor must sit in a chair adjacent to his or her ambassador. There are &lt;math&gt;N&lt;/math&gt; ways for the &lt;math&gt;8&lt;/math&gt; people to be seated at the table under these conditions. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> ==Solution==<br /> There are &lt;math&gt;4&lt;/math&gt; ambassadors and there are &lt;math&gt;6&lt;/math&gt; seats for them.<br /> So we consider the position of the blank seats.<br /> There are &lt;math&gt;15&lt;/math&gt; kinds of versions:<br /> If the two seats are adjacent to each other, there are &lt;math&gt;6&lt;/math&gt; options, and the ambassadors are sitting in four adjacent seats, and there are five seats that their advisors can sit. Choose any of them and the advisors’ seats are fixed, so there are &lt;math&gt;5&lt;/math&gt; kinds of solutions for the advisors to sit. And that’s a &lt;math&gt;6\cdot5&lt;/math&gt; if we don’t consider the order of the ambassadors.<br /> We can also get that if the blank seats are opposite, it will be &lt;math&gt;3\cdot9&lt;/math&gt;, if they are not adjacent and not opposite, it will be &lt;math&gt;6\cdot8&lt;/math&gt;.<br /> So the total is &lt;math&gt;24\cdot(6\cdot3+6\cdot8+3\cdot9)=2520&lt;/math&gt;<br /> And the remainder is &lt;math&gt;\boxed{520}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Mutinykids https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12A_Problems/Problem_15&diff=101713 2019 AMC 12A Problems/Problem 15 2019-02-10T15:00:35Z <p>Mutinykids: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Positive real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; have the property that<br /> &lt;cmath&gt;\sqrt{\log{a}} + \sqrt{\log{b}} + \log \sqrt{a} + \log \sqrt{b} = 100&lt;/cmath&gt;<br /> and all four terms on the left are positive integers, where log denotes the base 10 logarithm. What is &lt;math&gt;ab&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 10^{52} \qquad \textbf{(B) } 10^{100} \qquad \textbf{(C) } 10^{144} \qquad \textbf{(D) } 10^{164} \qquad \textbf{(E) } 10^{200} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since all four terms on the left are positive integers, from &lt;math&gt;\sqrt{\log{a}}&lt;/math&gt;, we know that both &lt;math&gt;\log{a}&lt;/math&gt; has to be perfect square and &lt;math&gt;a&lt;/math&gt; has to be a power of ten. The same applies to &lt;math&gt;b&lt;/math&gt; for the same reason. Setting &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; to &lt;math&gt;10^x&lt;/math&gt; and &lt;math&gt;10^y&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are the perfect squares, &lt;math&gt;ab = 10^{x+y}&lt;/math&gt;. By listing all the [https://artofproblemsolving.com/wiki/index.php/Perfect_square perfect squares] up to &lt;math&gt;14^2&lt;/math&gt; (as &lt;math&gt;15^2&lt;/math&gt; is larger than the largest possible sum of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; of &lt;math&gt;200&lt;/math&gt; from answer choice &lt;math&gt;E&lt;/math&gt;), two of those perfect squares must add up to one of the possible sums of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; given from the answer choices (&lt;math&gt;52&lt;/math&gt;, &lt;math&gt;100&lt;/math&gt;, &lt;math&gt;144&lt;/math&gt;, &lt;math&gt;164&lt;/math&gt;, or &lt;math&gt;200&lt;/math&gt;). <br /> <br /> Only a couple possible sums are seen: &lt;math&gt;16+36=52&lt;/math&gt;, &lt;math&gt;36+64=100&lt;/math&gt;, &lt;math&gt;64+100=164&lt;/math&gt;, &lt;math&gt;100+100=200&lt;/math&gt;, and &lt;math&gt;4+196=200&lt;/math&gt;. By testing each of these (by seeing whether &lt;math&gt;\sqrt{x} + \sqrt{b} + \frac{x}{2} + \frac{y}{2} = 100&lt;/math&gt;), only the pair &lt;math&gt;x = 64&lt;/math&gt; and &lt;math&gt;y=100&lt;/math&gt; work. Therefore, &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are &lt;math&gt;10^{64}&lt;/math&gt; and &lt;math&gt;10^{100}&lt;/math&gt;, and our answer is &lt;math&gt;\boxed{\textbf{(D) } 10^{164}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Given that &lt;math&gt;\sqrt{\log{a}}&lt;/math&gt; and &lt;math&gt;\sqrt{\log{b}}&lt;/math&gt; are both integers, &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; must be in the form &lt;math&gt;10^{m^2}&lt;/math&gt; and &lt;math&gt;10^{n^2}&lt;/math&gt;, respectively for some positive integers &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;\log \sqrt{a} = \frac{m^2}{2}&lt;/math&gt;. By substituting for a and b, the equation becomes &lt;math&gt;m + n + \frac{m^2}{2} + \frac{n^2}{2} = 100&lt;/math&gt;. After multiplying the equation by 2 and completing the square with respect to &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, the equation becomes &lt;math&gt;(m + 1)^2 + (n + 1)^2 = 202&lt;/math&gt;. Testing squares of positive integers that add to &lt;math&gt;202&lt;/math&gt;, &lt;math&gt;11^2 + 9^2&lt;/math&gt; is the only option. WLOG, let &lt;math&gt;m = 10&lt;/math&gt; and &lt;math&gt;n = 8&lt;/math&gt;. Plugging &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; to solve for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; gives us &lt;math&gt;a = 10^{100}&lt;/math&gt; and &lt;math&gt;b = 10^{64}&lt;/math&gt;. Therefore, &lt;math&gt;ab = \boxed{\textbf{(D) } 10^{164}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2019|ab=A|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Mutinykids https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12A_Problems/Problem_15&diff=101712 2019 AMC 12A Problems/Problem 15 2019-02-10T15:00:15Z <p>Mutinykids: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> Positive real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; have the property that<br /> &lt;cmath&gt;\sqrt{\log{a}} + \sqrt{\log{b}} + \log \sqrt{a} + \log \sqrt{b} = 100&lt;/cmath&gt;<br /> and all four terms on the left are positive integers, where log denotes the base 10 logarithm. What is &lt;math&gt;ab&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 10^{52} \qquad \textbf{(B) } 10^{100} \qquad \textbf{(C) } 10^{144} \qquad \textbf{(D) } 10^{164} \qquad \textbf{(E) } 10^{200} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since all four terms on the left are positive integers, from &lt;math&gt;\sqrt{\log{a}}&lt;/math&gt;, we know that both &lt;math&gt;\log{a}&lt;/math&gt; has to be perfect square and &lt;math&gt;a&lt;/math&gt; has to be a power of ten. The same applies to &lt;math&gt;b&lt;/math&gt; for the same reason. Setting &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; to &lt;math&gt;10^x&lt;/math&gt; and &lt;math&gt;10^y&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are the perfect squares, &lt;math&gt;ab = 10^{x+y}&lt;/math&gt;. By listing all the [https://artofproblemsolving.com/wiki/index.php/Perfect_square perfect squares] up to &lt;math&gt;14^2&lt;/math&gt; (as &lt;math&gt;15^2&lt;/math&gt; is larger than the largest possible sum of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; of &lt;math&gt;200&lt;/math&gt; from answer choice &lt;math&gt;E&lt;/math&gt;), two of those perfect squares must add up to one of the possible sums of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; given from the answer choices (&lt;math&gt;52&lt;/math&gt;, &lt;math&gt;100&lt;/math&gt;, &lt;math&gt;144&lt;/math&gt;, &lt;math&gt;164&lt;/math&gt;, or &lt;math&gt;200&lt;/math&gt;). <br /> <br /> Only a couple possible sums are seen: &lt;math&gt;16+36=52&lt;/math&gt;, &lt;math&gt;36+64=100&lt;/math&gt;, &lt;math&gt;64+100=164&lt;/math&gt;, &lt;math&gt;100+100=200&lt;/math&gt;, and &lt;math&gt;4+196=200&lt;/math&gt;. By testing each of these (by seeing whether &lt;math&gt;\sqrt{x} + \sqrt{b} + \frac{x}{2} + \frac{y}{2} = 100&lt;/math&gt;), only the pair &lt;math&gt;x = 64&lt;/math&gt; and &lt;math&gt;y=100&lt;/math&gt; work. Therefore, &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are &lt;math&gt;10^{64}&lt;/math&gt; and &lt;math&gt;10^{100}&lt;/math&gt;, and our answer is &lt;math&gt;\boxed{\textbf{(D) } 10^{164}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Given that &lt;math&gt;\sqrt{\log{a}}&lt;/math&gt; and &lt;math&gt;\sqrt{\log{b}}&lt;/math&gt; are both integers, &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; must be in the form &lt;math&gt;10^{m^2}&lt;/math&gt; and &lt;math&gt;10^{n^2}&lt;/math&gt;, respectively for some positive integers &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;\log \sqrt{a} = \frac{m^2}{2}&lt;/math&gt;. By substituting for a and b, the equation becomes &lt;math&gt;m + n + \frac{m^2}{2} + \frac{n^2}{2} = 100&lt;/math&gt;. After multiplying the equation by 2 and completing the square with respect to &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, the equation becomes &lt;math&gt;(m + 1)^2 + (n + 1)^2 = 202&lt;/math&gt;. Testing squares of positive integers that add to &lt;math&gt;202&lt;/math&gt;, &lt;math&gt;11^2 + 9^2&lt;/math&gt; is the only option. WLOG, let &lt;math&gt;m = 10&lt;/math&gt; and &lt;math&gt;n = 8&lt;/math&gt;. Plugging &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; to solve for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; gives us &lt;math&gt;a = 10^{100}&lt;/math&gt; and &lt;math&gt;b = 10^{64}&lt;/math&gt;. Therefore, &lt;math&gt;ab = \boxed{\textbf{(D) } 100^{164}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2019|ab=A|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Mutinykids https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_23&diff=100117 2008 AMC 12B Problems/Problem 23 2019-01-05T19:38:20Z <p>Mutinykids: /* Solution */</p> <hr /> <div>==Problem 23==<br /> The sum of the base-&lt;math&gt;10&lt;/math&gt; logarithms of the divisors of &lt;math&gt;10^n&lt;/math&gt; is &lt;math&gt;792&lt;/math&gt;. What is &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A)}\ 11\qquad \text{(B)}\ 12\qquad \text{(C)}\ 13\qquad \text{(D)}\ 14\qquad \text{(E)}\ 15&lt;/math&gt;<br /> <br /> __TOC__<br /> ==Solutions==<br /> === Solution 1 ===<br /> Every factor of &lt;math&gt;10^n&lt;/math&gt; will be of the form &lt;math&gt;2^a \times 5^b , a\leq n , b\leq n&lt;/math&gt;. Using the logarithmic property &lt;math&gt;\log(a \times b) = \log(a)+\log(b)&lt;/math&gt;, it suffices to count the total number of 2's and 5's running through all possible &lt;math&gt;(a,b)&lt;/math&gt;. For every factor &lt;math&gt;2^a \times 5^b&lt;/math&gt;, there will be another &lt;math&gt;2^b \times 5^a&lt;/math&gt;, so it suffices to count the total number of 2's occurring in all factors (because of this symmetry, the number of 5's will be equal). And since &lt;math&gt;\log(2)+\log(5) = \log(10) = 1&lt;/math&gt;, the final sum will be the total number of 2's occurring in all factors of &lt;math&gt;10^n&lt;/math&gt;.<br /> <br /> There are &lt;math&gt;n+1&lt;/math&gt; choices for the exponent of 5 in each factor, and for each of those choices, there are &lt;math&gt;n+1&lt;/math&gt; factors (each corresponding to a different exponent of 2), yielding &lt;math&gt;0+1+2+3...+n = \frac{n(n+1)}{2}&lt;/math&gt; total 2's. The total number of 2's is therefore &lt;math&gt;\frac{n \cdot(n+1)^2}{2} = \frac{n^3+2n^2+n}{2}&lt;/math&gt;. Plugging in our answer choices into this formula yields 11 (answer choice &lt;math&gt;\mathrm{(A)}&lt;/math&gt;) as the correct answer.<br /> <br /> <br /> ===Solution 2===<br /> <br /> We are given &lt;cmath&gt; \log_{10}d_1 + \log_{10}d_2 + \ldots + \log_{10}d_k = 792 &lt;/cmath&gt; The property &lt;math&gt;\log(ab) = \log(a)+\log(b)&lt;/math&gt; now gives &lt;cmath&gt; \log_{10}(d_1 d_2\cdot\ldots d_k) = 792 &lt;/cmath&gt; The product of the divisors is (from elementary number theory) &lt;math&gt;a^{d(n)/2}&lt;/math&gt; where &lt;math&gt;d(n)&lt;/math&gt; is the number of divisors. Note that &lt;math&gt;10^n = 2^n\cdot 5^n&lt;/math&gt;, so &lt;math&gt;d(n) = (n + 1)^2&lt;/math&gt;. Substituting these values with &lt;math&gt;a = 10^n&lt;/math&gt; in our equation above, we get &lt;math&gt;n(n + 1)^2 = 1584&lt;/math&gt;, from whence we immediately obtain &lt;math&gt;\framebox[1.2\width]{(A)}&lt;/math&gt; as the correct answer.<br /> <br /> === Solution 3 ===<br /> For every divisor &lt;math&gt;d&lt;/math&gt; of &lt;math&gt;10^n&lt;/math&gt;, &lt;math&gt;d \le \sqrt{10^n}&lt;/math&gt;, we have &lt;math&gt;\log d + \log \frac{10^n}{d} = \log 10^n = n&lt;/math&gt;. There are &lt;math&gt;\left \lfloor \frac{(n+1)^2}{2} \right \rfloor&lt;/math&gt; divisors of &lt;math&gt;10^n = 2^n \times 5^n&lt;/math&gt; that are &lt;math&gt;\le \sqrt{10^n}&lt;/math&gt;. After casework on the parity of &lt;math&gt;n&lt;/math&gt;, we find that the answer is given by &lt;math&gt;n \times \frac{(n+1)^2}{2} = 792 \Longrightarrow n = 11\ \mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2008|ab=B|num-b=22|num-a=24}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Mutinykids https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_24&diff=98927 2018 AMC 8 Problems/Problem 24 2018-11-22T15:35:11Z <p>Mutinykids: /* Solution */</p> <hr /> <div>==Problem 24==<br /> In the cube &lt;math&gt;ABCDEFGH&lt;/math&gt; with opposite vertices &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;E,&lt;/math&gt; &lt;math&gt;J&lt;/math&gt; and &lt;math&gt;I&lt;/math&gt; are the midpoints of edges &lt;math&gt;\overline{FB}&lt;/math&gt; and &lt;math&gt;\overline{HD},&lt;/math&gt; respectively. Let &lt;math&gt;R&lt;/math&gt; be the ratio of the area of the cross-section &lt;math&gt;EJCI&lt;/math&gt; to the area of one of the faces of the cube. What is &lt;math&gt;R^2?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(6cm);<br /> pair A,B,C,D,EE,F,G,H,I,J;<br /> C = (0,0);<br /> B = (-1,1);<br /> D = (2,0.5);<br /> A = B+D;<br /> G = (0,2);<br /> F = B+G;<br /> H = G+D;<br /> EE = G+B+D;<br /> I = (D+H)/2; J = (B+F)/2;<br /> filldraw(C--I--EE--J--cycle,lightgray,black);<br /> draw(C--D--H--EE--F--B--cycle); <br /> draw(G--F--G--C--G--H);<br /> draw(A--B,dashed); draw(A--EE,dashed); draw(A--D,dashed);<br /> dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); dot(G); dot(H); dot(I); dot(J);<br /> label(&quot;$A$&quot;,A,E);<br /> label(&quot;$B$&quot;,B,W);<br /> label(&quot;$C$&quot;,C,S);<br /> label(&quot;$D$&quot;,D,E);<br /> label(&quot;$E$&quot;,EE,N);<br /> label(&quot;$F$&quot;,F,W);<br /> label(&quot;$G$&quot;,G,N);<br /> label(&quot;$H$&quot;,H,E);<br /> label(&quot;$I$&quot;,I,E);<br /> label(&quot;$J$&quot;,J,W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{5}{4} \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{25}{16} \qquad \textbf{(E) } \frac{9}{4}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Note that &lt;math&gt;EJCI&lt;/math&gt; is a rhombus.<br /> Let the side length of the cube be &lt;math&gt;s&lt;/math&gt;. By the Pythagorean theorem, &lt;math&gt;EC= \sqrt 3s&lt;/math&gt; and &lt;math&gt;JI=\sqrt 2s&lt;/math&gt;. Since the area of a rhombus is half the product of it's diagonals, so the area of the cross section is &lt;math&gt;\frac{\sqrt 6s^2}{2}&lt;/math&gt;. &lt;math&gt;R = \frac{\sqrt 6}2&lt;/math&gt;. Thus &lt;math&gt;R^2 = \boxed{\textbf{(C) } \frac{3}{2}}&lt;/math&gt;<br /> <br /> ==Note==<br /> In the 2008 AMC 10A, Question 21 was nearly identical to this question.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=23|num-a=25}}<br /> <br /> {{MAA Notice}}</div> Mutinykids https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_8&diff=90832 2018 AMC 10A Problems/Problem 8 2018-02-09T22:42:06Z <p>Mutinykids: /* Solution 3 */</p> <hr /> <div>Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent coins, and 25-cent coins. He has 3 more 10-cent coins than 5-cent coins, and the total value of his collection is 320 cents. How many more 25-cent coins does Joe have than 5-cent coins?<br /> <br /> &lt;math&gt;\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;x&lt;/math&gt; be the number of 5-cent coins that Joe has. Therefore, he must have &lt;math&gt;(x+3)&lt;/math&gt; 10-cent coins and &lt;math&gt;(23-(x+3)-x)&lt;/math&gt; 25-cent coins. Since the total value of his collection is 320 cents, we can write<br /> &lt;cmath&gt;5x+10(x+3)+25(23-(x+3)-x)= 320 \Rightarrow 5x+10x+30+500-50x= 320 \Rightarrow 35x= 210 \Rightarrow x= 6&lt;/cmath&gt; <br /> Joe has 6 5-cent coins, 9 10-cent coins, and 8 25-cent coins. Thus, our answer is <br /> &lt;math&gt;8-6=\boxed{2}&lt;/math&gt;<br /> <br /> ~Nivek<br /> <br /> ==Solution 2==<br /> Let n be the number of 5 cent coins Joe has, d be the number of 10 cent coins, and q the number of 25 cent coins. We are solving for q - n.<br /> <br /> We know that the value of the coins add up to 320 cents.<br /> Thus, we have 5n + 10d + 25q = 320. Let this be (1).<br /> <br /> We know that there are 23 coins.<br /> Thus, we have n + d + q = 23. Let this be (2).<br /> <br /> We know that there are 3 more dimes than nickels, which also means that there are 3 less nickels than dimes.<br /> Thus, we have d - 3 = n.<br /> <br /> Plugging d-3 into the other two equations for n, (1) becomes 2d + q - 3 = 23 and (2) becomes 15d + 25q - 15 = 320. (1) then becomes 2d + q = 26, and (2) then becomes 15d + 25q = 335.<br /> <br /> Multiplying (1) by 25, we have 50d + 25q = 650 (or 25^2 + 25). Subtracting (2) from (1) gives us 35d = 315, which means d = 9.<br /> <br /> Plugging d into d - 3 = n, n = 6.<br /> <br /> Plugging d and q into the (2) we had at the beginning of this problem, q = 8.<br /> <br /> Thus, the answer is 8 - 6 = 2.<br /> ==Solution 3==<br /> So you set the number of 5-cent coins as x, the number of 10-cent coins as x+3, and the number of quarters y.<br /> <br /> You make the two equations:<br /> &lt;cmath&gt;5x+10(x+3)+25y=320 \Rightarrow 15x+25y+30=320 \Rightarrow 15x+25y=290&lt;/cmath&gt;<br /> &lt;cmath&gt;x+x+3+y=23 \Rightarrow 2x+3+y=23 \Rightarrow 2x+y=20&lt;/cmath&gt;<br /> <br /> From there, you multiply the second equation by 25 to get<br /> &lt;cmath&gt;50x+25y=500&lt;/cmath&gt;<br /> <br /> You subtract the first equation from the multiplied second equation to get<br /> &lt;cmath&gt;35x=210 \Rightarrow x=6&lt;/cmath&gt;<br /> You can plug that value into one of the equations to get &lt;cmath&gt;y=8&lt;/cmath&gt;<br /> So, the answer is &lt;math&gt;8-6=\boxed{2}&lt;/math&gt;, which is (C)<br /> <br /> - mutinykids<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Mutinykids https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_8&diff=90831 2018 AMC 10A Problems/Problem 8 2018-02-09T22:41:48Z <p>Mutinykids: /* Solution 3 */</p> <hr /> <div>Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent coins, and 25-cent coins. He has 3 more 10-cent coins than 5-cent coins, and the total value of his collection is 320 cents. How many more 25-cent coins does Joe have than 5-cent coins?<br /> <br /> &lt;math&gt;\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;x&lt;/math&gt; be the number of 5-cent coins that Joe has. Therefore, he must have &lt;math&gt;(x+3)&lt;/math&gt; 10-cent coins and &lt;math&gt;(23-(x+3)-x)&lt;/math&gt; 25-cent coins. Since the total value of his collection is 320 cents, we can write<br /> &lt;cmath&gt;5x+10(x+3)+25(23-(x+3)-x)= 320 \Rightarrow 5x+10x+30+500-50x= 320 \Rightarrow 35x= 210 \Rightarrow x= 6&lt;/cmath&gt; <br /> Joe has 6 5-cent coins, 9 10-cent coins, and 8 25-cent coins. Thus, our answer is <br /> &lt;math&gt;8-6=\boxed{2}&lt;/math&gt;<br /> <br /> ~Nivek<br /> <br /> ==Solution 2==<br /> Let n be the number of 5 cent coins Joe has, d be the number of 10 cent coins, and q the number of 25 cent coins. We are solving for q - n.<br /> <br /> We know that the value of the coins add up to 320 cents.<br /> Thus, we have 5n + 10d + 25q = 320. Let this be (1).<br /> <br /> We know that there are 23 coins.<br /> Thus, we have n + d + q = 23. Let this be (2).<br /> <br /> We know that there are 3 more dimes than nickels, which also means that there are 3 less nickels than dimes.<br /> Thus, we have d - 3 = n.<br /> <br /> Plugging d-3 into the other two equations for n, (1) becomes 2d + q - 3 = 23 and (2) becomes 15d + 25q - 15 = 320. (1) then becomes 2d + q = 26, and (2) then becomes 15d + 25q = 335.<br /> <br /> Multiplying (1) by 25, we have 50d + 25q = 650 (or 25^2 + 25). Subtracting (2) from (1) gives us 35d = 315, which means d = 9.<br /> <br /> Plugging d into d - 3 = n, n = 6.<br /> <br /> Plugging d and q into the (2) we had at the beginning of this problem, q = 8.<br /> <br /> Thus, the answer is 8 - 6 = 2.<br /> ==Solution 3==<br /> So you set the number of 5-cent coins as x, the number of 10-cent coins as x+3, and the number of quarters y.<br /> <br /> You make the two equations:<br /> &lt;cmath&gt;5x+10(x+3)+25y=320 \Rightarrow 15x+25y+30=320 \Rightarrow 15x+25y=290&lt;/cmath&gt;<br /> &lt;cmath&gt;x+x+3+y=23 \Rightarrow 2x+3+y=23 \Rightarrow 2x+y=20&lt;/cmath&gt;<br /> <br /> From there, you multiply the second equation by 25 to get<br /> &lt;cmath&gt;50x+25y=500&lt;/cmath&gt;<br /> <br /> You subtract the first equation from the multiplied second equation to get<br /> &lt;cmath&gt;35x=210 \Rightarrow x=6&lt;/cmath&gt;<br /> You can plug that value into one of the equations to get &lt;cmath&gt;y=8&lt;/cmath&gt;<br /> So, the answer is &lt;math&gt;8-6=\boxed{2}&lt;/math&gt;, which is (C)<br /> <br /> ~ - mutinykids ~<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Mutinykids https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_8&diff=90830 2018 AMC 10A Problems/Problem 8 2018-02-09T22:35:32Z <p>Mutinykids: /* Solution 3 */</p> <hr /> <div>Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent coins, and 25-cent coins. He has 3 more 10-cent coins than 5-cent coins, and the total value of his collection is 320 cents. How many more 25-cent coins does Joe have than 5-cent coins?<br /> <br /> &lt;math&gt;\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;x&lt;/math&gt; be the number of 5-cent coins that Joe has. Therefore, he must have &lt;math&gt;(x+3)&lt;/math&gt; 10-cent coins and &lt;math&gt;(23-(x+3)-x)&lt;/math&gt; 25-cent coins. Since the total value of his collection is 320 cents, we can write<br /> &lt;cmath&gt;5x+10(x+3)+25(23-(x+3)-x)= 320 \Rightarrow 5x+10x+30+500-50x= 320 \Rightarrow 35x= 210 \Rightarrow x= 6&lt;/cmath&gt; <br /> Joe has 6 5-cent coins, 9 10-cent coins, and 8 25-cent coins. Thus, our answer is <br /> &lt;math&gt;8-6=\boxed{2}&lt;/math&gt;<br /> <br /> ~Nivek<br /> <br /> ==Solution 2==<br /> Let n be the number of 5 cent coins Joe has, d be the number of 10 cent coins, and q the number of 25 cent coins. We are solving for q - n.<br /> <br /> We know that the value of the coins add up to 320 cents.<br /> Thus, we have 5n + 10d + 25q = 320. Let this be (1).<br /> <br /> We know that there are 23 coins.<br /> Thus, we have n + d + q = 23. Let this be (2).<br /> <br /> We know that there are 3 more dimes than nickels, which also means that there are 3 less nickels than dimes.<br /> Thus, we have d - 3 = n.<br /> <br /> Plugging d-3 into the other two equations for n, (1) becomes 2d + q - 3 = 23 and (2) becomes 15d + 25q - 15 = 320. (1) then becomes 2d + q = 26, and (2) then becomes 15d + 25q = 335.<br /> <br /> Multiplying (1) by 25, we have 50d + 25q = 650 (or 25^2 + 25). Subtracting (2) from (1) gives us 35d = 315, which means d = 9.<br /> <br /> Plugging d into d - 3 = n, n = 6.<br /> <br /> Plugging d and q into the (2) we had at the beginning of this problem, q = 8.<br /> <br /> Thus, the answer is 8 - 6 = 2.<br /> ==Solution 3==<br /> So you set the number of 5-cent coins as x, the number of 10-cent coins as x+3, and the number of quarters y.<br /> <br /> You make the two equations:<br /> &lt;cmath&gt;5x+10(x+3)+25y=320 \Rightarrow 15x+25y+30=320 \Rightarrow 15x+25y=290&lt;/cmath&gt;<br /> &lt;cmath&gt;x+x+3+y=23 \Rightarrow 2x+3+y=23 \Rightarrow 2x+y=20&lt;/cmath&gt;<br /> <br /> From there, you multiply the second equation by 25 to get<br /> &lt;cmath&gt;50x+25y=500&lt;/cmath&gt;<br /> <br /> You subtract the first equation from the multiplied second equation to get<br /> &lt;cmath&gt;35x=210 \Rightarrow x=6&lt;/cmath&gt;<br /> You can plug that value into one of the equations to get &lt;cmath&gt;y=8&lt;/cmath&gt;<br /> So, the answer is &lt;math&gt;8-6=\boxed{2}&lt;/math&gt;, which is (C)<br /> <br /> - mutinykids<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Mutinykids https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_8&diff=90829 2018 AMC 10A Problems/Problem 8 2018-02-09T22:35:07Z <p>Mutinykids: /* Solution 3 */</p> <hr /> <div>Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent coins, and 25-cent coins. He has 3 more 10-cent coins than 5-cent coins, and the total value of his collection is 320 cents. How many more 25-cent coins does Joe have than 5-cent coins?<br /> <br /> &lt;math&gt;\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;x&lt;/math&gt; be the number of 5-cent coins that Joe has. Therefore, he must have &lt;math&gt;(x+3)&lt;/math&gt; 10-cent coins and &lt;math&gt;(23-(x+3)-x)&lt;/math&gt; 25-cent coins. Since the total value of his collection is 320 cents, we can write<br /> &lt;cmath&gt;5x+10(x+3)+25(23-(x+3)-x)= 320 \Rightarrow 5x+10x+30+500-50x= 320 \Rightarrow 35x= 210 \Rightarrow x= 6&lt;/cmath&gt; <br /> Joe has 6 5-cent coins, 9 10-cent coins, and 8 25-cent coins. Thus, our answer is <br /> &lt;math&gt;8-6=\boxed{2}&lt;/math&gt;<br /> <br /> ~Nivek<br /> <br /> ==Solution 2==<br /> Let n be the number of 5 cent coins Joe has, d be the number of 10 cent coins, and q the number of 25 cent coins. We are solving for q - n.<br /> <br /> We know that the value of the coins add up to 320 cents.<br /> Thus, we have 5n + 10d + 25q = 320. Let this be (1).<br /> <br /> We know that there are 23 coins.<br /> Thus, we have n + d + q = 23. Let this be (2).<br /> <br /> We know that there are 3 more dimes than nickels, which also means that there are 3 less nickels than dimes.<br /> Thus, we have d - 3 = n.<br /> <br /> Plugging d-3 into the other two equations for n, (1) becomes 2d + q - 3 = 23 and (2) becomes 15d + 25q - 15 = 320. (1) then becomes 2d + q = 26, and (2) then becomes 15d + 25q = 335.<br /> <br /> Multiplying (1) by 25, we have 50d + 25q = 650 (or 25^2 + 25). Subtracting (2) from (1) gives us 35d = 315, which means d = 9.<br /> <br /> Plugging d into d - 3 = n, n = 6.<br /> <br /> Plugging d and q into the (2) we had at the beginning of this problem, q = 8.<br /> <br /> Thus, the answer is 8 - 6 = 2.<br /> ==Solution 3==<br /> So you set the number of 5-cent coins as x, the number of 10-cent coins as x+3, and the number of quarters y.<br /> <br /> You make the two equations:<br /> &lt;cmath&gt;5x+10(x+3)+25y=320 \Rightarrow 15x+25y+30=320 \Rightarrow 15x+25y=290&lt;/cmath&gt;<br /> &lt;cmath&gt;x+x+3+y=23 \Rightarrow 2x+3+y=23 \Rightarrow 2x+y=20&lt;/cmath&gt;<br /> <br /> From there, you multiply the second equation by 25 to get<br /> &lt;cmath&gt;50x+25y=500&lt;/cmath&gt;<br /> <br /> You subtract the first equation from the multiplied second equation to get<br /> &lt;cmath&gt;35x=210 \Rightarrow x=6&lt;/cmath&gt;<br /> You can plug that value into one of the equations to get &lt;cmath&gt;y=8&lt;/cmath&gt;<br /> So, the answer is &lt;math&gt;8-6=\boxed{2}&lt;/math&gt;, which is (C)<br /> - mutinykids<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Mutinykids https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_8&diff=90828 2018 AMC 10A Problems/Problem 8 2018-02-09T22:34:24Z <p>Mutinykids: /* Solution 3 */</p> <hr /> <div>Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent coins, and 25-cent coins. He has 3 more 10-cent coins than 5-cent coins, and the total value of his collection is 320 cents. How many more 25-cent coins does Joe have than 5-cent coins?<br /> <br /> &lt;math&gt;\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;x&lt;/math&gt; be the number of 5-cent coins that Joe has. Therefore, he must have &lt;math&gt;(x+3)&lt;/math&gt; 10-cent coins and &lt;math&gt;(23-(x+3)-x)&lt;/math&gt; 25-cent coins. Since the total value of his collection is 320 cents, we can write<br /> &lt;cmath&gt;5x+10(x+3)+25(23-(x+3)-x)= 320 \Rightarrow 5x+10x+30+500-50x= 320 \Rightarrow 35x= 210 \Rightarrow x= 6&lt;/cmath&gt; <br /> Joe has 6 5-cent coins, 9 10-cent coins, and 8 25-cent coins. Thus, our answer is <br /> &lt;math&gt;8-6=\boxed{2}&lt;/math&gt;<br /> <br /> ~Nivek<br /> <br /> ==Solution 2==<br /> Let n be the number of 5 cent coins Joe has, d be the number of 10 cent coins, and q the number of 25 cent coins. We are solving for q - n.<br /> <br /> We know that the value of the coins add up to 320 cents.<br /> Thus, we have 5n + 10d + 25q = 320. Let this be (1).<br /> <br /> We know that there are 23 coins.<br /> Thus, we have n + d + q = 23. Let this be (2).<br /> <br /> We know that there are 3 more dimes than nickels, which also means that there are 3 less nickels than dimes.<br /> Thus, we have d - 3 = n.<br /> <br /> Plugging d-3 into the other two equations for n, (1) becomes 2d + q - 3 = 23 and (2) becomes 15d + 25q - 15 = 320. (1) then becomes 2d + q = 26, and (2) then becomes 15d + 25q = 335.<br /> <br /> Multiplying (1) by 25, we have 50d + 25q = 650 (or 25^2 + 25). Subtracting (2) from (1) gives us 35d = 315, which means d = 9.<br /> <br /> Plugging d into d - 3 = n, n = 6.<br /> <br /> Plugging d and q into the (2) we had at the beginning of this problem, q = 8.<br /> <br /> Thus, the answer is 8 - 6 = 2.<br /> ==Solution 3==<br /> So you set the number of 5-cent coins as x, the number of 10-cent coins as x+3, and the number of quarters y.<br /> <br /> You make the two equations:<br /> &lt;cmath&gt;5x+10(x+3)+25y=320 \Rightarrow 15x+25y+30=320 \Rightarrow 15x+25y=290&lt;/cmath&gt;<br /> &lt;cmath&gt;x+x+3+y=23 \Rightarrow 2x+3+y=23 \Rightarrow 2x+y=20&lt;/cmath&gt;<br /> <br /> From there, you multiply the second equation by 25 to get<br /> &lt;cmath&gt;50x+25y=500&lt;/cmath&gt;<br /> <br /> You subtract the first equation from the multiplied second equation to get<br /> &lt;cmath&gt;35x=210 \Rightarrow x=6&lt;/cmath&gt;<br /> You can plug that value into one of the equations to get &lt;cmath&gt;y=6&lt;/cmath&gt;<br /> So, the answer is &lt;math&gt;8-6=\boxed{2}&lt;/math&gt;, which is &lt;cmath&gt;\Rightarrow (E)&lt;/cmath&gt;<br /> - mutinykids<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Mutinykids https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_8&diff=90826 2018 AMC 10A Problems/Problem 8 2018-02-09T22:25:47Z <p>Mutinykids: /* Solution 2 */</p> <hr /> <div>Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent coins, and 25-cent coins. He has 3 more 10-cent coins than 5-cent coins, and the total value of his collection is 320 cents. How many more 25-cent coins does Joe have than 5-cent coins?<br /> <br /> &lt;math&gt;\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;x&lt;/math&gt; be the number of 5-cent coins that Joe has. Therefore, he must have &lt;math&gt;(x+3)&lt;/math&gt; 10-cent coins and &lt;math&gt;(23-(x+3)-x)&lt;/math&gt; 25-cent coins. Since the total value of his collection is 320 cents, we can write<br /> &lt;cmath&gt;5x+10(x+3)+25(23-(x+3)-x)= 320 \Rightarrow 5x+10x+30+500-50x= 320 \Rightarrow 35x= 210 \Rightarrow x= 6&lt;/cmath&gt; <br /> Joe has 6 5-cent coins, 9 10-cent coins, and 8 25-cent coins. Thus, our answer is <br /> &lt;math&gt;8-6=\boxed{2}&lt;/math&gt;<br /> <br /> ~Nivek<br /> <br /> ==Solution 2==<br /> Let n be the number of 5 cent coins Joe has, d be the number of 10 cent coins, and q the number of 25 cent coins. We are solving for q - n.<br /> <br /> We know that the value of the coins add up to 320 cents.<br /> Thus, we have 5n + 10d + 25q = 320. Let this be (1).<br /> <br /> We know that there are 23 coins.<br /> Thus, we have n + d + q = 23. Let this be (2).<br /> <br /> We know that there are 3 more dimes than nickels, which also means that there are 3 less nickels than dimes.<br /> Thus, we have d - 3 = n.<br /> <br /> Plugging d-3 into the other two equations for n, (1) becomes 2d + q - 3 = 23 and (2) becomes 15d + 25q - 15 = 320. (1) then becomes 2d + q = 26, and (2) then becomes 15d + 25q = 335.<br /> <br /> Multiplying (1) by 25, we have 50d + 25q = 650 (or 25^2 + 25). Subtracting (2) from (1) gives us 35d = 315, which means d = 9.<br /> <br /> Plugging d into d - 3 = n, n = 6.<br /> <br /> Plugging d and q into the (2) we had at the beginning of this problem, q = 8.<br /> <br /> Thus, the answer is 8 - 6 = 2.<br /> ==Solution 3==<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Mutinykids https://artofproblemsolving.com/wiki/index.php?title=User:Mutinykids&diff=89229 User:Mutinykids 2017-12-28T16:47:35Z <p>Mutinykids: Created page with &quot;Hi my name is jeff. i play diep.io&quot;</p> <hr /> <div>Hi my name is jeff. i play diep.io</div> Mutinykids