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Difference between revisions of "2003 AMC 12B Problems/Problem 18"
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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>.
 
<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>.
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Revision as of 10:26, 4 July 2013

Problem

Let $n$ be a 5-digit number, and let $q$ and $r$ be the quotient and remainder, respectively, when $n$ is divided by $100$. For how many values of $n$ is $q+r$ divisible by $11$?

$\mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090$

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 $11 \Rightarrow {B}$. These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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