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Four generations (Posted on 2020-07-06) Difficulty: 3 of 5
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.)
Solution Better late than never. Analytic solution | Comment 8 of 9 |
Analytic solution, taking Ady's hint and running with it.
Every 6 digit number meeting the first 2 criteria could be scaled by adding or subtracting the same amount from each digit to make the smallest digit equal to 1.  This isn't necessary but helps me to visualize it.  All these numbers will be some permutation of 123456.

To be divisible by 11, the even digits minus the odd digits must be 11,0, or -11.
To prove that the cardinality of S3 is zero, all that is necessary is to show this can never be true for any permutation of 123456.

A permutation that maximizes that function is 142536.  (4+5+6) - (1+2+3) = 9, so there is no way to get to 11 or -11.
The sum of digits of 123456 is 21, so there is no way the function can be zero since there is no 3 digit subset of the 6 digits such that the sum of its digits would be 10.5.

  Posted by Larry on 2020-07-07 08:35:51
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