Friday puzzler -- three numbers

Re: Friday puzzler -- three numbers

Larry Barrett

The unused digit is 4.

There are 'rules' to help you decide if a given number is divisible by a smaller number. We all know that if a given number is even, it is divisible by 2. And if a given number ends in 0 or 5 it is divisible by 5. Since all three of Alex's numbers are divisible by 11, a good starting point is apply the 'rule of 11'. I did not know this rule, but Google is your friend.

Re: Friday puzzler -- three numbers

Alex Y

Sounds like you have it (the four is not used in the correct answer).

I'll be interested to hear how you used the rule for divisibility by 11 to come up with your solution. I was able to get it with some targeted trial and error.

Re: Friday puzzler -- three numbers

Larry Barrett

The stated problem required all three 3 digit numbers to be divisible by 11; and then one or more to be divisible by 2, 3, 4, 5, 6, 7, 8, and 9.

The rule of eleven, combined with the problem statement that the three numbers are comprised of 9 digits (which means no repeating of digits) is a starting point. It does get you to a relatively small set of 3 digit numbers which must include the 3 digit numbers that satisfy the problem statement. For example, 902, 913, 924, 935, 946, 957, 968 are in the set. Once you see how the rule of eleven works, (9 -0 +2= 11, 9 -1 +3 =11, etc) you can repeat this starting with 8xy, 7xy, etc. to find all 3 digit numbers with non-repeating digits that are divisible by 11. There are 7 3 digit numbers in each set except for the set starting with 5; there are 8 in this set.

It was mostly trial and error from this point. I looked for the numbers within these sets that were divisible by 7 (924, 847, 693, 539, 462, 308, 385, 231, 154) and the numbers divisible by 5 (935, 825, 715, 605, 495, 385, 275, 165 - interesting pattern here), figuring that they were likely to not have many other divisors and thus might be part of the solution.

Finding a number within the set divisible by as many of the remaining divisors (2, 3, 4, 6, 8, 9) as possible was really trial and error. There were a few candidates - 924, 792, 528, 396, 132 all looked promising. I looked at each one, trying to find a 'divisible by 7' and 'divisible by 5' number that would work. The 3 digit numbers I found are 792, 308, 165.

Re: Friday puzzler -- three numbers

Alex Y

Nicely done. Here is a slightly different approach, but still some trial and error and lucky guesses in the process

If at least one of the numbers is divisible by 8 or 9, it is also divisible by 2,3, and 4, so we only have to test divisibility by 5,6,7,8,and 9

With three numbers, one or some are multiples of more than one of these 5 numbers.

The multiple of 5 is odd (otherwise it would be a multiple of 2,5,and 11, and any three-digit multiple of 110 has the same digit in the first two places. So any number with a 5 in any other spot is out.

Any three-digit multiple of 99 has 9 as the middle digit and first and last digits that add to 9. So eliminate any number with a nine in any other location.

Interesting observations, and narrowing the field, but not getting to a solution. Focusing on the overlap issue, if 8 and 9 were divisors of the same number, that would take care of 6 as well

9*8*11=792

Looking for multiples of 7 and 5 that don't overlap digits, start with the 5 multiples, since we can eliminate the even multiples

55*3 = 165

55*5 = 275

55*7 = 385

55*9 = 495

55*11 = 605

55*13 = 715

55*15 = 825

55*17= 935

Looking at the multiples of 77, we find most share a digit with 792 or violate the rule about 5 being the last digit. The only ones that don't are 308 and 385

385 is a multiple of both 5 and 7, but there is no way to form a multiple of eleven out of the remaining digits of 1,4,6,and 0.

So if we use 308 as the multiple of 77, that leaves us with 165 as the multiple of 5.

But that is not fully satisfying, since it requires luck to get at the 792 starting point. A reasoned starting guess, but still just a guess.