Call Me – Solution
Here is the logic behind the answer:
(copied verbatim from the puzzle magazine)
Each digit from 0 through 9 occurs once, so the sum of the 10 digits is 45. The smallest possible sum of five different digits is 10. The only two sums into which 45 can be split so that each sum is at least 10 and the larger sum is evenly divisible by the smaller are 30 and 15.
The sum of the first five digits must be either 15 or 30. The sum of 4, 2, and 7 is 13. If the sum of the first five digits is 15, their last two digits must be 0 and 2. But this is impossible because 427 already contains a 2.
So the sum of the first five digits is 30. The sum of their last two digits must be 17, so they must be 89 or 98. If the first five digits are 42789, the three digits in odd-numbered positions are 4 + 7 + 9 = 20, which means that the other two digits in odd-numbered positions must total 10. But the last five digits are 0, 1, 3, 5, and 6. It is impossible to total 10 with two of these.
So the first five digits must be 42798. The three digits in odd-numbered positions total 19, which is greater than 15, so the five digits in odd numbered positions must total 30. The other two digits in odd-numbered positions must total 11. The digits in the 7th and 9th positions must be either 5 … 6 or 6 … 5. The other three digits in the last half of the phone number are 0, 1, and 3, in some combination.
There are 12 possibilities for the last half of the phone number: 05163, 05361, 15063, 15360, 35061, 35160, 06153, 06351, 16053, 16350, 36051, 36150. The only one that is evenly divisible by 327 is 16350.
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