# 2003 AMC 10A Problems/Problem 25

## 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

### Simple Solution $11|q+r$ implies that $11|100q+r$, so $11|n$. Then, $n$ can range from $10010$ to $99990$ for a total of $\boxed{8181\Rightarrow \mathrm{(B)}}$ numbers.

### Solution 1

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 $\left\lfloor \frac{100}{11} \right\rfloor = 9$ 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 $\left\lfloor \frac{900}{11} \right\rfloor = 81$ 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 $900\cdot9+81\cdot1=8181 \Rightarrow B$.

### Solution 2

Let $n$ equal $\overline{abcde}$, where $a$ through $e$ are digits. Therefore, $q=\overline{abc}=100a+10b+c$ $r=\overline{de}=10d+e$

We now take $q+r\bmod{11}$: $q+r=100a+10b+c+10d+e\equiv a-b+c-d+e\equiv 0\bmod{11}$

The divisor trick for 11 is as follows:

"Let $n=\overline{a_1a_2a_3\cdots a_x}$ be an $x$ digit integer. If $a_1-a_2+a_3-\cdots +(-1)^{x-1} a_x$ is divisible by $11$, then $n$ is also divisible by $11$."

Therefore, the five digit number $n$ is divisible by 11. The 5-digit multiples of 11 range from $910\cdot 11$ to $9090\cdot 11$. There are $8181\Rightarrow \mathrm{(B)}$ divisors of 11 between those inclusive.

### Solution 3

Since $q$ is a quotient and $r$ is a remainder when $n$ is divided by $100$. So we have $n=100q+r$. Since we are counting choices where $q+r$ is divisible by $11$, we have $n=99q+q+r=99q+11k$ for some $k$. This means that $n$ is the sum of two multiples of $11$ and would thus itself be a multiple of $11$. Then we can count all the five digit multiples of $11$ as in Solution 2. (This solution is essentially the same as Solution 2, but it does not necessarily involve mods and so could potentially be faster.)