September 2007
Problem of the Month
Leftovers Again?
by Dr. Michael Ecker
in GAMES magazine
What is the smallest whole number
that, when divided by 2, leaves a
remainder of 1; when divided by 3,
leaves a remainder of 2; and so
on, up to leaving a remainder of 9
when divided by 10?
Solution to the Problem:
The answer is 2519.
Since the remainder is always one less than
the divisor, the answer must be one less than a
multiple of all the divisors 2 through 10. The
smallest such number must be the least common
multiple of 2, 3, ... 9, 10, minus one. The least
common multiple is 2520, so the answer is 2519.
Correctly solved by:
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1. K. Sengupta
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Calcutta, India
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| 2. John Funk
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Ventura, California
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| 3. James Alarie
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Flint, Michigan
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| 4. David & Judy Dixon
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Bennettsville, South Carolina
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| 5. Richard K. Johnson
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La Jolla, California
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