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
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 \(900 \times 9 +81 \times 1=8181\)
Other useful links
- https://www.cheenta.com/problem-based-on-triangle-prmo-2012-problem-7/
- https://www.youtube.com/watch?v=axSw4_SuKE8