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

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

But try the problem first...

Answer: \(8181\)

Source

Suggested Reading

AMC-10A (2003) Problem 25

Pre College Mathematics

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 \(900 \times 9 +81 \times 1=8181\)

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

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