Charmaine has written down three 3-digit positive integers which between them contains exactly 9 distinct digits.
Among these three positive integers:
- There is at least one that is divisible by 2.
- There is at least one that is divisible by 3.
- There is at least one that is divisible by 4.
- There is at least one that is divisible by 5.
- There is at least one that is divisible by 6.
- There is at least one that is divisible by 7.
- There is at least one that is divisible by 8.
- There is at least one that is divisible by 9.
In addition, it is known that all the three positive integers are divisible by 11.
What are the three 3-digit integers posited by Charmaine?
Note: Adapted from Enigma # 1776 which appeared in 'New Scientist' in 2013.
A three digit number that is divisible by 11 either has its middle digit equal to the sum of the other two, or the middle digit is zero and the other two digits sum to 11.
Assume that all three of Charmine's numbers are the first type, so no occurrence of zero. Then the sum of the nine digits is even; but also we would use each digit 1-9 exactly once and that sum is 45 which is odd. A contradiction. So then one of Charmine's numbers is the kind with a zero as the middle digit.
One of these numbers must be a multiple of 9. The number with the zero cannot be it, since those digits must add to 11. Then it is one of the other two numbers.
Then for the middle digit to equal the sum of the other two digits and the sum of all three be a multiple of 9, the middle digit is 9.
So far I have numbers A0B, C9D and EFG, with A+B=11 and C+D=9. C=7 and D=2 would be very convenient as 792 is a multiple of 72 and would cover all the divisibility clauses except divisibility by 5 or 7.
With 2 and 7 out of the pool, then 5+6=11 and 3+8=11 are options for A and B.
Case 605 is divisible by 5, leaving digits 1,3,4,8 for the third number. That leaves only 341 and 143 as possibilities for the third number but neither is divisible by 7
Case 308 is divisible by 7, leaving digits 1,4,5,6 for the third number. For divisibility by 5 we need the digit 5 at the end, which allows one possible third number as 165.
Then the three 3-digit integers posited by Charmaine can be 792, 308, and 165.
Answer
