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Four generations (Posted on 2020-07-06)

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.

No Solution Yet Submitted by Ady TZIDON

Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)

non-computer solution

| Comment 5 of 9 |

Integers of S3 have digits d through (d-5) inclusive.

If a member is divisible by 143=11*13 it's divisible by 11.

That means the sum of the differences of pairs of digits will equal a multiple of 11.

The set of digits can be replaced with (0,1,2,3,4,5) without changing divisibility status, and it takes only a couple of minutes handwork to verify the impossibility of divisibility, so S3 contains no members.

Edited on July 6, 2020, 4:01 pm

Posted by xdog on 2020-07-06 15:53:37

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