Let S be a set of all six-digit integers.

Let S1 be a subset of S, including all members of S such that each consists

of distinct digits.

Let S2 be a subset of S1, including all members of S1 each with 5 being the difference between its largest digit and its lowest one.

Let S3 be a subset of S2, comprising all elements of S2 divisible by 143.

What is the cardinality of S3 ?

Explain your way of reasoning.

(In reply to

non-computer solution by xdog)

Here is my understanding of xdog's solution. Call it: xdog for dummies :-)

How do you use pairwise differences to find if a number is divisible by 11? I didn't know until i watched this very nice video that also explains why it works:

Now what to make of this line: "S3 has digits d to abs(d-5)" inclusive?

Ans: since there are 6 distinct digits with max-min=5, these can only be, in some order:

012345, 123456, 234567, 345678, or 456789. Fair enough.

Now the list any S3 number makes (if not 012345) will have the same pairwise differences (summed) if the digits were dropped down to the list 012345. This can be done for our examples above by subtracting 111111, or 222222 ... or 444444 from the number, because by doing so, we are effectively adding -1, -1, -1 and 1, 1, 1, to the differences, or 0. Likewise 2,2,2,-2,-2,-2 = 0 etc.

So finally, do any of the S3 permutations of 012345 have a sum of differences that make -11, 0 or 11? The largest sum of differences is 5+4+3 - 0-2-1 = 9 and the smallest -9.

So we are only left to ask about the sum of differences adding to 0. If we take pairwise differences of 3 even and 3 odd numbers, in any pairs, we will aways get an odd number, not 0. So none of S3's members are divisible by 11.

So maybe not within KISS, but I get it.

*Edited on ***July 7, 2020, 6:15 pm**