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AMC 8 Math Olympiad USA Math Olympiad

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 \(11\)?

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

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

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