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# Divisibility Problem from AMC 10A, 2003 | Problem 25

Try this beautiful problem from Number theory based on divisibility from AMC-10A, 2003. You may use sequential hints to solve the problem.

Try this beautiful problem from Number theory based on divisibility from AMC 10A, 2003.

## Number theory in Divisibility – AMC-10A, 2003- Problem 25

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 $11$?

• $8180$
• $8181$
• $8182$
• $9190$
• $9000$

Number system

Probability

divisibility

## Check the Answer

Answer: $8181$

AMC-10A (2003) Problem 25

Pre College Mathematics

## Try with Hints

Since $11$ divides $q+r$ so may say that $11$ divides $100 q+r$. Since $n$ is a $5$ digit number …soTherefore, $q$ can be any integer from $100$ to $999$ inclusive, and $r$ can be any integer from $0$ to $99$ inclusive.

can you finish the problem……..

Since $n$ is a five digit number then and $11 | 100q+r$ then $n$ must start from $10010$ and count up to $99990$

can you finish the problem……..

Therefore, the number of possible values of $n$ such that $900 \times 9 +81 \times 1=8181$