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 $11$ ?

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

Solutions

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.

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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 $q+r \equiv 0\pmod{11}$ . Another way to think about it is the number of possible values of q when r, the remainder, is $0$ . In this case, q itself has to be a multiple of $11$ . $\frac{999}{11} = 90$ . Then we'll need to subtract $9$ from $90$ since we only want multiples of $11$ greater than $100$ $(90-9=81)$

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.

Note

The part labeled "divisor trick" actually follows from the same observation we made in the previous step: $10\equiv (-1)\pmod{11}$ , therefore $10^{2k}\equiv 1$ and $10^{2k+1}\equiv (-1)$ for all $k$ . For a $5-$ digit number $\overline{abcde}$ we get $\overline{abcde}\equiv a\cdot 1 + b\cdot(-1) + c\cdot 1 + d\cdot(-1) + e\cdot 1 = a-b+c-d+e$ , as claimed.

Also note that in the "divisor trick" we actually want to assign the signs backwards - if we make sure that the last sign is a $+$ , the result will have the same remainder modulo $11$ as the original number.

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.)

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Question
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All AMC 10 Problems and Solutions

2003 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AMC 12 Problems and Solutions

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