https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Bobjoebilly&feedformat=atom AoPS Wiki - User contributions [en] 2020-11-30T18:07:55Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_1&diff=137899 2020 AMC 8 Problems/Problem 1 2020-11-20T00:54:37Z <p>Bobjoebilly: /* Solution 5 */</p> <hr /> <div>Luka is making lemonade to sell at a school fundraiser. His recipe requires &lt;math&gt;4&lt;/math&gt; times as much water as sugar and twice as much sugar as lemon juice. He uses &lt;math&gt;3&lt;/math&gt; cups of lemon juice. How many cups of water does he need?<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 /> ==Solution 1==<br /> Luka will need &lt;math&gt;3\cdot 2=6&lt;/math&gt; cups of sugar and &lt;math&gt;6\cdot 4=24&lt;/math&gt; cups of water. The answer is &lt;math&gt;\boxed{\textbf{(E) } 24}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;W, S,&lt;/math&gt; and &lt;math&gt;L&lt;/math&gt; represent the number of cups of water, sugar, and lemon juice that Luka needs for his recipe, respectively. We are given that &lt;math&gt;W=4S&lt;/math&gt; and &lt;math&gt;S=2L&lt;/math&gt;. Since &lt;math&gt;L=3&lt;/math&gt;, it follows that &lt;math&gt;S=6&lt;/math&gt;, which in turn implies that &lt;math&gt;W=24 \implies\boxed{\textbf{(E) }24}&lt;/math&gt;.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ==Solution 3==<br /> We have that &lt;math&gt;\textsf{lemonade} : \textsf{water} : \textsf{lemon juice} = 8 : 2 : 1,&lt;/math&gt; so we have &lt;math&gt;3 \cdot 8 = \boxed{\textbf{(E) }24}.&lt;/math&gt;<br /> <br /> [pog]<br /> <br /> ==Solution 4==<br /> <br /> We are given that &lt;math&gt;4w:s&lt;/math&gt; and &lt;math&gt;2s=l&lt;/math&gt; which we combine to get &lt;math&gt;8w:2s:l&lt;/math&gt;. Letting all the variables equal &lt;math&gt;3&lt;/math&gt;, we find that the answer is &lt;math&gt;3\cdot 8=\boxed{\textbf{(E) }24} &lt;/math&gt;.<br /> <br /> -franzliszt<br /> <br /> ==Solution 5==<br /> <br /> Put the numbers in ratios &lt;math&gt;4w:s&lt;/math&gt; and &lt;math&gt;2s:lj&lt;/math&gt; when w = water, s = sugar, and lj = lemon juice. then since we know there is &lt;math&gt;3&lt;/math&gt; cups of lemon juice, do the math. &lt;math&gt;3\cdot2\cdot4=6\cdot4=\boxed{\textbf{(E) }24}.&lt;/math&gt;<br /> <br /> ~ bsu1<br /> <br /> ==Video Solution==<br /> https://youtu.be/FPC792h-mGE<br /> <br /> ~savannahsolver<br /> <br /> ==See also==<br /> {{AMC8 box|year=2020|before=First problem|num-a=2}}<br /> {{MAA Notice}}<br /> we just get 4*3*2=24$which is E<br /> -oceanxia</div> Bobjoebilly https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_1&diff=137898 2020 AMC 8 Problems/Problem 1 2020-11-20T00:54:18Z <p>Bobjoebilly: /* Solution 4 */</p> <hr /> <div>Luka is making lemonade to sell at a school fundraiser. His recipe requires &lt;math&gt;4&lt;/math&gt; times as much water as sugar and twice as much sugar as lemon juice. He uses &lt;math&gt;3&lt;/math&gt; cups of lemon juice. How many cups of water does he need?<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 /> ==Solution 1==<br /> Luka will need &lt;math&gt;3\cdot 2=6&lt;/math&gt; cups of sugar and &lt;math&gt;6\cdot 4=24&lt;/math&gt; cups of water. The answer is &lt;math&gt;\boxed{\textbf{(E) } 24}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;W, S,&lt;/math&gt; and &lt;math&gt;L&lt;/math&gt; represent the number of cups of water, sugar, and lemon juice that Luka needs for his recipe, respectively. We are given that &lt;math&gt;W=4S&lt;/math&gt; and &lt;math&gt;S=2L&lt;/math&gt;. Since &lt;math&gt;L=3&lt;/math&gt;, it follows that &lt;math&gt;S=6&lt;/math&gt;, which in turn implies that &lt;math&gt;W=24 \implies\boxed{\textbf{(E) }24}&lt;/math&gt;.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ==Solution 3==<br /> We have that &lt;math&gt;\textsf{lemonade} : \textsf{water} : \textsf{lemon juice} = 8 : 2 : 1,&lt;/math&gt; so we have &lt;math&gt;3 \cdot 8 = \boxed{\textbf{(E) }24}.&lt;/math&gt;<br /> <br /> [pog]<br /> <br /> ==Solution 4==<br /> <br /> We are given that &lt;math&gt;4w:s&lt;/math&gt; and &lt;math&gt;2s=l&lt;/math&gt; which we combine to get &lt;math&gt;8w:2s:l&lt;/math&gt;. Letting all the variables equal &lt;math&gt;3&lt;/math&gt;, we find that the answer is &lt;math&gt;3\cdot 8=\boxed{\textbf{(E) }24} &lt;/math&gt;.<br /> <br /> -franzliszt<br /> <br /> ==Solution 5==<br /> <br /> Put the numbers in ratios &lt;math&gt;4w:s&lt;/math&gt; and &lt;math&gt;2s:lj&lt;/math&gt; when w = water, s = sugar, and lj = lemon juice. then since we know there is &lt;math&gt;3&lt;/math&gt; cups of lemon juice, do the math. &lt;math&gt;3\cdot2\cdot4=6\cdot4=24&lt;/math&gt;<br /> <br /> ~ bsu1<br /> <br /> ==Video Solution==<br /> https://youtu.be/FPC792h-mGE<br /> <br /> ~savannahsolver<br /> <br /> ==See also==<br /> {{AMC8 box|year=2020|before=First problem|num-a=2}}<br /> {{MAA Notice}}<br /> we just get 4*3*2=24$ which is E<br /> -oceanxia</div> Bobjoebilly https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_1&diff=137897 2020 AMC 8 Problems/Problem 1 2020-11-20T00:54:05Z <p>Bobjoebilly: /* Solution 4 */</p> <hr /> <div>Luka is making lemonade to sell at a school fundraiser. His recipe requires &lt;math&gt;4&lt;/math&gt; times as much water as sugar and twice as much sugar as lemon juice. He uses &lt;math&gt;3&lt;/math&gt; cups of lemon juice. How many cups of water does he need?<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 /> ==Solution 1==<br /> Luka will need &lt;math&gt;3\cdot 2=6&lt;/math&gt; cups of sugar and &lt;math&gt;6\cdot 4=24&lt;/math&gt; cups of water. The answer is &lt;math&gt;\boxed{\textbf{(E) } 24}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;W, S,&lt;/math&gt; and &lt;math&gt;L&lt;/math&gt; represent the number of cups of water, sugar, and lemon juice that Luka needs for his recipe, respectively. We are given that &lt;math&gt;W=4S&lt;/math&gt; and &lt;math&gt;S=2L&lt;/math&gt;. Since &lt;math&gt;L=3&lt;/math&gt;, it follows that &lt;math&gt;S=6&lt;/math&gt;, which in turn implies that &lt;math&gt;W=24 \implies\boxed{\textbf{(E) }24}&lt;/math&gt;.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ==Solution 3==<br /> We have that &lt;math&gt;\textsf{lemonade} : \textsf{water} : \textsf{lemon juice} = 8 : 2 : 1,&lt;/math&gt; so we have &lt;math&gt;3 \cdot 8 = \boxed{\textbf{(E) }24}.&lt;/math&gt;<br /> <br /> [pog]<br /> <br /> ==Solution 4==<br /> <br /> We are given that &lt;math&gt;4w:s&lt;/math&gt; and &lt;math&gt;2s=l&lt;/math&gt; which we combine to get &lt;math&gt;8w:2s:l&lt;/math&gt;. Letting all the variables equal &lt;math&gt;3&lt;/math&gt;, we find that the answer is &lt;math&gt;3\cdot 8=\boxed{\textbf{(E) }24} 24&lt;/math&gt;.<br /> <br /> -franzliszt<br /> <br /> ==Solution 5==<br /> <br /> Put the numbers in ratios &lt;math&gt;4w:s&lt;/math&gt; and &lt;math&gt;2s:lj&lt;/math&gt; when w = water, s = sugar, and lj = lemon juice. then since we know there is &lt;math&gt;3&lt;/math&gt; cups of lemon juice, do the math. &lt;math&gt;3\cdot2\cdot4=6\cdot4=24&lt;/math&gt;<br /> <br /> ~ bsu1<br /> <br /> ==Video Solution==<br /> https://youtu.be/FPC792h-mGE<br /> <br /> ~savannahsolver<br /> <br /> ==See also==<br /> {{AMC8 box|year=2020|before=First problem|num-a=2}}<br /> {{MAA Notice}}<br /> we just get 4*3*2=24\$ which is E<br /> -oceanxia</div> Bobjoebilly https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_10&diff=128842 2015 AIME I Problems/Problem 10 2020-07-22T01:00:34Z <p>Bobjoebilly: /* Solution 8(godspeed) */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;f(x)&lt;/math&gt; be a third-degree polynomial with real coefficients satisfying<br /> &lt;cmath&gt;|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.&lt;/cmath&gt; Find &lt;math&gt;|f(0)|&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let &lt;math&gt;f(x)&lt;/math&gt; = &lt;math&gt;ax^3+bx^2+cx+d&lt;/math&gt;.<br /> Since &lt;math&gt;f(x)&lt;/math&gt; is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up.<br /> By drawing a coordinate axis, and two lines representing &lt;math&gt;12&lt;/math&gt; and &lt;math&gt;-12&lt;/math&gt;, it is easy to see that &lt;math&gt;f(1)=f(5)=f(6)&lt;/math&gt;, and &lt;math&gt;f(2)=f(3)=f(7)&lt;/math&gt;; otherwise more bends would be required in the graph. Since only the absolute value of f(0) is required, there is no loss of generalization by stating that &lt;math&gt;f(1)=12&lt;/math&gt;, and &lt;math&gt;f(2)=-12&lt;/math&gt;. This provides the following system of equations.<br /> &lt;cmath&gt; a + b + c + d = 12&lt;/cmath&gt;<br /> &lt;cmath&gt;8a + 4b + 2c + d = -12&lt;/cmath&gt;<br /> &lt;cmath&gt; 27a + 9b + 3c + d = -12&lt;/cmath&gt;<br /> &lt;cmath&gt;125a + 25b + 5c + d = 12&lt;/cmath&gt;<br /> &lt;cmath&gt;216a + 36b + 6c + d = 12&lt;/cmath&gt;<br /> &lt;cmath&gt;343a + 49b + 7c + d = -12&lt;/cmath&gt;<br /> Using any four of these functions as a system of equations yields &lt;math&gt;|f(0)| = \boxed{072}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Express &lt;math&gt;f(x)&lt;/math&gt; in terms of powers of &lt;math&gt;(x-4)&lt;/math&gt;:<br /> &lt;cmath&gt;f(x) = a(x-4)^3 + b(x-4)^2 + c(x-4) + d&lt;/cmath&gt;<br /> By the same argument as in the first Solution, we see that &lt;math&gt;f(x)&lt;/math&gt; is an odd function about the line &lt;math&gt;x=4&lt;/math&gt;, so its coefficients &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are 0. From there it is relatively simple to solve &lt;math&gt;f(2)=f(3)=-12&lt;/math&gt; (as in the above solution, but with a smaller system of equations):<br /> &lt;cmath&gt;a(1)^3 + c(1) = -12&lt;/cmath&gt;<br /> &lt;cmath&gt;a(2)^3 + c(2) = -12&lt;/cmath&gt;<br /> &lt;math&gt;a=2&lt;/math&gt; and &lt;math&gt;c=-14&lt;/math&gt;<br /> &lt;cmath&gt;|f(0)| = |2(-4)^3 - 14(-4)| = \boxed{072}&lt;/cmath&gt;<br /> <br /> ==Solution 3==<br /> Without loss of generality, let &lt;math&gt;f(1) = 12&lt;/math&gt;. (If &lt;math&gt;f(1) = -12&lt;/math&gt;, then take &lt;math&gt;-f(x)&lt;/math&gt; as the polynomial, which leaves &lt;math&gt;|f(0)|&lt;/math&gt; unchanged.) Because &lt;math&gt;f&lt;/math&gt; is third-degree, write<br /> &lt;cmath&gt;f(x) - 12 = a(x - 1)(x - b)(x - c)&lt;/cmath&gt;<br /> &lt;cmath&gt;f(x) + 12 = a(x - d)(x - e)(x - f)&lt;/cmath&gt;<br /> where &lt;math&gt;\{b, c, d, e, f \}&lt;/math&gt; clearly must be a permutation of &lt;math&gt;\{2, 3, 5, 6, 7\}&lt;/math&gt; from the given condition. Thus &lt;math&gt;b + c + d + e + f = 2 + 3 + 5 + 6 + 7 = 23.&lt;/math&gt; However, subtracting the two equations gives &lt;math&gt;-24 = a[(x - 1)(x - b)(x - c) - (x - d)(x - e)(x - f)]&lt;/math&gt;, so comparing &lt;math&gt;x^2&lt;/math&gt; coefficients gives &lt;math&gt;1 + b + c = d + e + f&lt;/math&gt; and thus both values equal to &lt;math&gt;\dfrac{24}{2} = 12&lt;/math&gt;. As a result, &lt;math&gt;\{b, c \} = \{5, 6 \}&lt;/math&gt;. As a result, &lt;math&gt;-24 = a (12)&lt;/math&gt; and so &lt;math&gt;a = -2&lt;/math&gt;. Now, we easily deduce that &lt;math&gt;f(0) = (-2) \cdot (-1) \cdot (-5) \cdot (-6) + 12 = 72,&lt;/math&gt; and so removing the without loss of generality gives &lt;math&gt;|f(0)| = \boxed{072}&lt;/math&gt;, which is our answer.<br /> <br /> <br /> ==Solution 4==<br /> The following solution is similar to solution 3, but assumes nothing. Let &lt;math&gt;g(x)=(f(x))^2-144&lt;/math&gt;. Since &lt;math&gt;f&lt;/math&gt; has degree 3, &lt;math&gt;g&lt;/math&gt; has degree 6 and has roots 1,2,3,5,6, and 7. Therefore, &lt;math&gt;g(x)=k(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)&lt;/math&gt; for some &lt;math&gt;k&lt;/math&gt;. Hence &lt;math&gt;|f(0)|=\sqrt{g(0)+144}=\sqrt{1260k+144}&lt;/math&gt;. Note that &lt;math&gt;g(x)=(f(x)+12)(f(x)-12)&lt;/math&gt;. Since &lt;math&gt;f&lt;/math&gt; has degree 3, so do &lt;math&gt;f(x)+12&lt;/math&gt; and &lt;math&gt;f(x)-12&lt;/math&gt;; and both have the same leading coefficient. Hence &lt;math&gt;f(x)+12=a(x-q)(x-r)(x-s)&lt;/math&gt; and &lt;math&gt;f(x)-12=a(x-t)(x-u)(x-v)&lt;/math&gt; for some &lt;math&gt;a\neq 0&lt;/math&gt; (else &lt;math&gt;f&lt;/math&gt; is not cubic) where &lt;math&gt;\{q,r,s,t,u,v\}&lt;/math&gt; is the same as the set &lt;math&gt;\{1,2,3,5,6,7\}&lt;/math&gt;. Subtracting the second equation from the first, expanding, and collecting like terms, we have that <br /> &lt;cmath&gt;24=a((t+u+v-(q+r+s))x^2-a(tu+uv+tv-(qr+qs+rs))x+a(tuv-qrs)&lt;/cmath&gt;<br /> which must hold for all &lt;math&gt;x&lt;/math&gt;. Since &lt;math&gt;a\neq 0&lt;/math&gt; we have that (1) &lt;math&gt;t+u+v=q+r+s&lt;/math&gt;, (2) &lt;math&gt;tu+uv+tv=qr+qs+rs&lt;/math&gt; and (3) &lt;math&gt;a(tuv-qrs)=24&lt;/math&gt;. Since &lt;math&gt;q+r+s+t+u+v&lt;/math&gt; is the sum of 1,2,3,5,6, and 7, we have &lt;math&gt;q+r+s+t+u+v=24&lt;/math&gt; so that by (1) we have &lt;math&gt;q+r+s=12&lt;/math&gt; and &lt;math&gt;t+u+v=12&lt;/math&gt;. We must partition 1,2,3,5,6,7 into 2 sets each with a sum of 12. Consider the set that contains 7. It can't contain 6 or 5 because the sum of that set would already be &lt;math&gt;\geq 12&lt;/math&gt; with only 2 elements. If 1 is in that set, the other element must be 4 which is impossible. Hence the two sets must be &lt;math&gt;\{2,3,7\}&lt;/math&gt; and &lt;math&gt;\{1,5,6\}&lt;/math&gt;. Note that each of these sets happily satisfy (2). By (3), since the sets have products 42 and 30 we have that &lt;math&gt;|a|=\frac{24}{|tuv-qrs|}=\frac{24}{12}=2&lt;/math&gt;. Since &lt;math&gt;a&lt;/math&gt; is the leading coefficient of &lt;math&gt;f(x)&lt;/math&gt;, the leading coefficient of &lt;math&gt;(f(x))^2&lt;/math&gt; is &lt;math&gt;a^2=|a|^2=2^2=4&lt;/math&gt;. Thus the leading coefficient of &lt;math&gt;g(x)&lt;/math&gt; is 4, i.e. &lt;math&gt;k=4&lt;/math&gt;. Then from earlier, &lt;math&gt;|f(0)|=\sqrt{g(0)+144}=\sqrt{1260k+144}=\sqrt{1260\cdot4+144}=\sqrt{5184}=72&lt;/math&gt; so that the answer is &lt;math&gt;\boxed{072}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> By drawing the function, WLOG let &lt;math&gt;f(1)=f(5)=f(6)=12&lt;/math&gt;. Then, &lt;math&gt;f(2)=f(3)=f(7)&lt;/math&gt;. Realize that if we shift &lt;math&gt;f(x)&lt;/math&gt; down 12, then this function &lt;math&gt;f(x)-12&lt;/math&gt; has roots &lt;math&gt;1,5,6&lt;/math&gt; with leading coefficient &lt;math&gt;-2&lt;/math&gt; because &lt;math&gt;f(2)-12=-24=-2(1)(-3)(-4)&lt;/math&gt;. Therefore &lt;math&gt;f(x)=-2(x-1)(x-5)(x-6)+12&lt;/math&gt;, and then &lt;math&gt;|f(0)|=60+12=\boxed{072}&lt;/math&gt;.<br /> <br /> ==Solution 6 (Finite differences)==<br /> Because a cubic must come in a &quot;wave form&quot; with two &quot;humps&quot; (Called points of inflections) , we can see that &lt;math&gt;f(1)=f(5)=f(6)&lt;/math&gt;, and &lt;math&gt;f(2)=f(3)=f(7)&lt;/math&gt;. By symmetry, &lt;math&gt;f(4)=0&lt;/math&gt;. Now, WLOG let &lt;math&gt;f(1)=12&lt;/math&gt;, and &lt;math&gt;f(2)=f(3)=-12&lt;/math&gt;. Then, we can use finite differences to get that the third (constant) difference is &lt;math&gt;12&lt;/math&gt;, and therefore &lt;math&gt;f(0)=12+(24+(24+12))=\boxed{072}&lt;/math&gt;.<br /> <br /> ==Solution 7 (Like solution 1 without annoying systems)==<br /> We can rewrite our function as two different cubics, &lt;math&gt;f(x)=k(x-a)(x-b)(x-c)\pm12=k(x-d)(x-e)(x-f)\mp12&lt;/math&gt;. Note that &lt;math&gt;k&lt;/math&gt; is the same in both because they must be equal, so their leading terms must be. We must then, following Vieta's, choose our roots such that &lt;math&gt;a+b+c=d+e+f&lt;/math&gt; and verify that the other Vieta's formulas hold. Additionally, a cubic must only cross the x-axis thrice, restricting our choices for roots further. Choosing &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=5&lt;/math&gt;, &lt;math&gt;c=6&lt;/math&gt;, &lt;math&gt;d=2&lt;/math&gt;, &lt;math&gt;e=3&lt;/math&gt;, &lt;math&gt;f=7&lt;/math&gt; yields: &lt;cmath&gt;kx^3-12kx^2+41kx-30k\pm12=kx^3-12kx^2+41kx-42k\mp12&lt;/cmath&gt; For the constant terms to have a difference of 24 (&lt;math&gt;|\pm12-\mp12|&lt;/math&gt;), we must have &lt;math&gt;k=\pm2&lt;/math&gt;, so the constant term of our polynomial is &lt;math&gt;\pm72&lt;/math&gt;, the absolute value of which is &lt;math&gt;\boxed{072}&lt;/math&gt;. -- Solution by eiis1000<br /> <br /> ==Solution 8(godspeed)==<br /> Trivially, assume the function can be written as &lt;math&gt;f(x) = (x-4)(g(x))&lt;/math&gt;, where &lt;math&gt;g(x)&lt;/math&gt; is quadratic. Then we set up the following equations using the coefficients of &lt;math&gt;g(x)&lt;/math&gt;.<br /> <br /> &lt;math&gt;a + b + c = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;4a + 2b + c = -6&lt;/math&gt;<br /> <br /> &lt;math&gt;9a + 3b + c = -12&lt;/math&gt;<br /> <br /> Subtract the first equation from the second and the second from the third to get a system of two variables. Solve and plug it in to get &lt;math&gt;g(x) = 2x^2-16x+18&lt;/math&gt;.&lt;math&gt;|f(0)| = |18*-4| = | -72| = \boxed{072}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AIME box|year=2015|n=I|num-b=9|num-a=11}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Bobjoebilly https://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=123467 User:Piphi 2020-06-02T17:24:50Z <p>Bobjoebilly: /* User Count */</p> <hr /> <div>{{User:Piphi/Template:Header}}<br /> &lt;br&gt;<br /> &lt;div style=&quot;border:2px solid black; -webkit-border-radius: 10px; background:#dddddd&quot;&gt;<br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;User Count&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;If this is your first time visiting this page, edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;<br /> &lt;center&gt;&lt;font size=&quot;100px&quot;&gt;106&lt;/font&gt;&lt;/center&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#cccccc;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;About Me&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;A legendary AoPSer.&lt;br&gt;<br /> <br /> My main project on the AoPS wiki is [[AoPS_Administrators#Current_Admins | a list of all the AoPS admins]], everyone is welcome to add more admins to the list by clicking [https://artofproblemsolving.com/wiki/index.php?title=AoPS_Administrators&amp;action=edit&amp;section=1 here]. I've also added a lot of the info in the [[Reaper Archives]].<br /> <br /> You can also check out Greed Control Game 19 statistics that I found, [https://artofproblemsolving.com/community/c19451h2126212 here]&lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#bbbbbb;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;[[User:Piphi/Asymptote|Asymptote]]&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;For a complete list of my Asymptote drawings, go [[User:Piphi/Asymptote|here]].&lt;/div&gt;<br /> &lt;/div&gt;</div> Bobjoebilly https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_6&diff=120992 2011 AMC 10B Problems/Problem 6 2020-04-14T23:10:10Z <p>Bobjoebilly: </p> <hr /> <div>== Problem==<br /> <br /> On Halloween Casper ate &lt;math&gt;\frac{1}{3}&lt;/math&gt; of his candies and then gave &lt;math&gt;2&lt;/math&gt; candies to his brother. The next day he ate &lt;math&gt;\frac{1}{3}&lt;/math&gt; of his remaining candies and then gave &lt;math&gt;4&lt;/math&gt; candies to his sister. On the third day he ate his final &lt;math&gt;8&lt;/math&gt; candies. How many candies did Casper have at the beginning?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 30 \qquad\textbf{(B)}\ 39 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 66 &lt;/math&gt;<br /> <br /> == Solution ==<br /> Let &lt;math&gt;x&lt;/math&gt; represent the amount of candies Casper had at the beginning.<br /> &lt;cmath&gt;\begin{align*}<br /> \frac{2}{3} \left(\frac{2}{3} x - 2\right) - 4 - 8 &amp;= 0\\<br /> \frac{2}{3} x - 2 &amp;= 18\\<br /> \frac{2}{3} x &amp;= 20\\<br /> x &amp;= \boxed{\textbf{(A)} 30}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> <br /> == Solution 2 ==<br /> We work backwards. If he had 8 candies at the end, then before he gave candies to his sister he had 12 candies. This means that at the end of Halloween he had 18 candies, so before he gave candies to his brother he had 20 candies. Therefore, at the start he had &lt;math&gt;\boxed{\textbf{(A)} 30}&lt;/math&gt;<br /> <br /> ~bobjoebilly<br /> <br /> == See Also==<br /> <br /> {{AMC10 box|year=2011|ab=B|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Bobjoebilly