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Difference between revisions of "2003 AMC 12B Problems/Problem 18"

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== Solution ==
 
== Solution ==
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Suppose <math>n = 100\cdot q + r = 99\cdot q + (q+r)</math>
 
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Since <math>11|(q+r)</math> and <math>11|99q</math>, <math>11|n</math>
 
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<math>10000 \leq n \leq 99999</math>, so there are <math>\left\lfloor\frac{99999}{11}\right\rfloor-\left\lceil\frac{10000}{11}\right\rceil+1 = \boxed{8181}</math> values of <math>q+r</math> that are divisible by <math>11 \Rightarrow {B}</math>.
 
  
 
== See Also==
 
== See Also==
 
{{AMC12 box|year=2003|ab=B|num-b=17|num-a=19}}
 
{{AMC12 box|year=2003|ab=B|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 04:52, 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

See Also

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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