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2003 AMC 12A Problems/Problem 18

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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) \ } 9...

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 B.

See Also

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