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
Related Program
AMC – AIME Boot Camp… for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
