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 \(11\)?

- \(8180\)
- \(8181\)
- \(8182\)
- \(9190\)
- \(9000\)

**Key Concepts**

Number system

Probability

divisibility

## Check the Answer

But try the problem first…

Answer: \(8181\)

AMC-10A (2003) Problem 25

Pre College Mathematics

## Try with Hints

First hint

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……..

Second Hint

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……..

Final Step

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

## Other useful links

- https://www.cheenta.com/problem-based-on-triangle-prmo-2012-problem-7/
- https://www.youtube.com/watch?v=axSw4_SuKE8

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