The LCM of the divisors is 2^4 * 3^2 * 5 * 7 * 11 * 13 * 17
divisibility by 3^2 is automatic as the digit sum is 45
divisibility by 2 and 5 requires the last digit be 0
divisibility by 11 requires every other digit to be part of two sets of 5 digits with sums of 17 and 28
the partitions are:
01259 34678
01268 34579
01349 25678
01358 24679
01367 24589
01457 23689
02348 15679
02357 14689
02456 13789
05689 12347
04789 12356
(this narrows it down to 5!*4!=2880 possibilities)
Next, consider 16. 2 is taken care of with the zero at the end. The three digits before this must be divisible by 8. There are 125 possibilities which narrow down to 56 when you exclude 0's and repeated digits (I won't list them)
Here's how I might proceed: For each possible last four digits, cross-check to see what partitions are possible.
One of them is 1520. Only the next t the last partition has 05 and 12 separated.
_ _ _ _ _ _ 1 5 2 0
3,4,7 go in the first, third and fifth spaces
6,8,9 go in the second, fourth and sixth spaces
3!*3!=36 numbers need to be checked for divisibility by 7, 13, and 17.
OK, that's too much work...
Maybe this can be simplified by noting 7*11*13=1001 and finding a divisibility rule for 1001, but that still leaves 17.
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Posted by Jer
on 2017-12-29 12:04:42 |
Start of impractical analytical solution
