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Hi Greg. The 5 digit number is divisible by 5, and only a number ending in zero or five is divisible by 5, so the last digit is a 5. We can now think of a four digit number with 1, 2, 3, 4 each used once. The last digit is a 4, so it must be an even number. The only choices are 2 and 4. As well, the first two digits end in an even number, because that's divisible by 2. So the 2nd and 4th digits are 2 and 4 That leaves 1 and 3 or 3 and 1 for the first and third digits. Can you write out the four possibilities and test each one? Then you can prove your hypothesis. Cheers,
Neat! I can prove that there are only two sixdigit number which uses each of the a) by the first digit (from the left) is divisible by 1, a) is no constraint at all. The two possible numbers are 123654 and 321654. Claude  


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