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| An Atypical Product Problem (Posted on 2007-01-03) |
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Determine all possible positive integers, two or more digits long, like ABC...XYZ, such that ABC...XY0Z is a multiple of it.
Note: digits may be repeated.
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Submitted by K Sengupta
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Rating: 4.0000 (1 votes)
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Solution:
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15, 18, 45 and 10*k ( where k is a positive integer)
EXPLANATION:
Let n denote an integer assuming the values in accordance with provisions of the problem.
Let m be the number obtained by appending the 0, n must also divide 10n and accordingly, n must divide 10n – m.
If the last digit of n is d, then 10n - m = 9d. So n must divide 9d. In particular, n must be a 2 digit number.
For example if d = 9, we need a two digit number ending in 9 that divides 81.
There are none. Similarly, we check d = 8 giving n = 18; d = 7 no solutions; d = 6, no solutions, d = 5 giving n = 15 or 45; d = 4 no solutions; d = 3 no solutions; d = 2 no solutions ; d = 1 no solutions.
Finally if d = 0, then any number will satisfy the conditions of the problem.
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SOLUTION SOURCE: Problem A1 of 17th Mexican MO, 2003
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