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:
| 1. K. Sengupta | Calcutta, India |
| 2. John Funk | Ventura, California |
| 3. James Alarie | Flint, Michigan |
| 4. David & Judy Dixon | Bennettsville, South Carolina |
| 5. Richard K. Johnson | La Jolla, California |