Idea:
Factor 111111:3×7×11×13×37
Factor 222222:3×7×11×13×37 * 2, etc
Consider:
The divisors of 111111 with 2 to 4 digits are: 11,13,21,33,37,39,77,91,111,143,231,259,273,407,429,481,777,1001,1221,1443, 2849,3003, 3367, 5291, 8547. 9 times a single digit divisor or 11 is less than 100, so those divisors can be ignored.Those with 5 digits can also be ignored because they are too big.
None of the digits K can feature in any of the divisors, so K=1 can be ruled out, as the smallest one-less number greater than 100, 259*429=111111, so that there are insufficient digits to form WHAT.
Say K=2. Possible divisors are: 77,91,111,143,231,259,273,407,429,481,777,1001,1221,1443, 2849,3003,3367, with one being multiplied by 2. We can rule out those that obviously replicate digits, either alone or multiplied by 2: 111,777,1001,1221,1443,3003,3367, leaving 91,143,231,259,273,407,429,481,2849. 222222/2849 = 78, and 481*2= 962, so 2 can also be eliminated.
At this point, it might be worth setting up a table to evaluate the outcomes more quickly:
The divisors are down the centre. The RHS (table A) shows the result when the divisors are applied to 111111,222222, etc.
The LHS (table B) shows the multiples 2x, 3x and 4x of the divisors which can be used to generate all other solutions (2*2=4, 2*3=6, etc.)
We can eliminate a lot of the table easily, e.g.
1. Candidate is too small or too large
2. Candidate contains a repeated digit, e.g. 858.
3. Candidate is in Table A and greater than 4 and contains a digit matching K
etc.
In the event it turns out that 2*77*4329=154*4329 satisfies the requirements of the puzzle, as indicated by Larry.
Edited on April 30, 2020, 9:27 am
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Posted by broll
on 2020-04-30 09:22:02 |