RMO 1995 Questions, Solutions, Discussions
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(RMO 1995 P3, AoPS, cf. Tournament of Towns) Prove that among any \(18\) consecutive three digit numbers there is at least one number which is divisible by the sum of its digits.
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- Show that one among any such consecutive integers is divisible by \(18\).
- Prove that its sum of digits, is a multiple of \(9\), and conclude that it is equal to one of \(9, 18, 27\).
- Show that the sum of its digits is not \(27\).
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Note that among \(18\) consecutive three digit numbers, there is an integer divisible by \(18\). Denote it by \(n = 100a + 10b + c\) with \(a, b, c\) denoting integers lying between \(0\) and \(9\). It follows that \(9\) divides \(n\), and hence \(9\) divides \(a + b + c\). This shows that \(a+b+c\) is equal to one of \(9, 18, 27\). Note that \(a + b + c = 27\) holds only if \(n = 999\). Since \(18\) divides \(n\), it follows that \(a + b + c \neq 27\), and hence, \(a + b + c\) is equal to one of \(9, 18\). This proves that \(a + b + c\) divides \(n\).
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