2005 PMWC Problems/Problem I15
Problem
The sum of the two three-digit integers, and , is divisible by . What is the largest possible product of and ?
Solution
A number is divisible by 18 iff it is divisible by 2 and 9. Divisibility by 2 is already satisfied, so we need the number to be divisible by 9; the divisibility rule for 9 states that we only need the sum of the digits to be divisible by 9. The units digit is 6; so the sum, , of the digits of , satisfies . The only reasonable values for .
- . It quickly becomes apparent that , which gives us .
- . Suppose . Then gives either , and we carry over the one to . So the sum of the digits of must add up to , which quickly shows us that this isn't possible.
- Hence, . Greedy algorithm: if , then , so the sum of the digits of must be 12. So . The other possible pairs are .
Quickly taking the product of these, we find that gives us the largest product of .
See also
2005 PMWC (Problems) | ||
Preceded by Problem I14 |
Followed by Problem T1 | |
I: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 T: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 |