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

Revision as of 03:53, 24 May 2013

Problem

Let $n$ be a $5$ -digit number, and let $q$ and $r$ be the quotient and the 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

When a $5$ -digit number is divided by $100$ , the first $3$ digits become the quotient, $q$ , and the last $2$ digits become the remainder, $r$ .

Therefore, $q$ can be any integer from $100$ to $999$ inclusive, and $r$ can be any integer from $0$ to $99$ inclusive.

For each of the $9\cdot10\cdot10=900$ possible values of $q$ , there are at least $\lfloor \frac{100}{11} \rfloor = 9$ possible values of $r$ such that $q+r \equiv 0\pmod{11}$ .

Since there is $1$ "extra" possible value of $r$ that is congruent to $0\pmod{11}$ , each of the $\lfloor \frac{900}{11} \rfloor = 81$ values of $q$ that are congruent to $0\pmod{11}$ have $1$ more possible value of $r$ such that $q+r \equiv 0\pmod{11}$ .

Therefore, the number of possible values of $n$ such that $q+r \equiv 0\pmod{11}$ is $900\cdot9+81\cdot1=8181 \Rightarrow\boxed{(B)}$ .

@@ Line 17: / Line 17: @@
 == See Also ==
-*[[2003 AMC 12A Problems]]
-*[[2003 AMC 12A Problems/Problem 17|Previous Problem]]
-*[[2003 AMC 12A Problems/Problem 19|Next Problem]]
-[[Category:Introductory Number Theory Problems]]
+==See Also==
+{{AMC12 box|year=2003|ab=A|num-b=17|num-a=19}}

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

Revision as of 03:53, 24 May 2013

Contents

Problem

Solution

See Also

See Also