2003 AMC 12B Problems/Problem 18: Difference between revisions

Revision as of 03:52, 9 June 2014

Problem

Let $x$ and $y$ be positive integers such that $7x^5 = 11y^{13}.$ The minimum possible value of $x$ has a prime factorization $a^cb^d.$ What is $a + b + c + d?$

$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$

Solution

Suppose $n = 100\cdot q + r = 99\cdot q + (q+r)$

Since $11|(q+r)$ and $11|99q$ , $11|n$

$10000 \leq n \leq 99999$ , so there are $\left\lfloor\frac{99999}{11}\right\rfloor-\left\lceil\frac{10000}{11}\right\rceil+1 = \boxed{8181}$ values of $q+r$ that are divisible by $11 \Rightarrow {B}$ .

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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 == Problem ==
-Let <math>n</math> be a 5-digit number, and let <math>q</math> and <math>r</math> be the quotient and remainder, respectively, when <math>n</math> is divided by <math>100</math>. For how many values of <math>n</math> is <math>q+r</math> divisible by <math>11</math>?
+Let <math>x</math> and <math>y</math> be positive integers such that <math>7x^5 = 11y^{13}.</math> The minimum possible value of <math>x</math> has a prime factorization <math>a^cb^d.</math> What is <math>a + b + c + d?</math>
-<math> \mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090 </math>
+<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34</math>
 == Solution ==

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2003 AMC 12B Problems/Problem 18: Difference between revisions

Revision as of 03:52, 9 June 2014

Problem

Solution

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