perplexus.info :: Numbers : A permutation puzzle 2

A permutation puzzle 2 (Posted on 2022-05-21)

Each of M, N and P is a 4 digit non-leading zero positive integer such that:

M+N=P
The digits of each of M,N and P are distinct.
Each of M, N and P are permutations of one another.

Determine all possible triplets (M,N,P).

Submitted by K Sengupta

Rating: 5.0000 (1 votes)

Solution: (Hide)

With the further restriction M <=N, there are 21 valid cases.They are:
1503 + 3510 = 5013 1530 + 3501 = 5031 1089 + 8019 = 9108 1089 + 8091 = 9180 4095 + 4950 = 9045 4095 + 5409 = 9504 4590 + 4950 = 9540 1269 + 1692 = 2961 2691 + 6921 = 9612 1467 + 6147 = 7614 1467 + 6174 = 7641 1476 + 4671 = 6147 1746 + 4671 = 6417 2439 + 2493 = 4932 4392 + 4932 = 9324 2385 + 2853 = 5238 2538 + 3285 = 5823 2853 + 5382 = 8235 3285 + 5238 = 8523 4698 + 4986 = 9684 4896 + 4968 = 9864
If instead we look for 5 digit numbers, the number of solutions, again just looking at M < N, there are 522 solution, none with M = N. So 1044 allowing for those.
For an explanation, refer to the solution submitted by Larry in this location.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
No Subject	K Sengupta	2022-08-02 10:01:51
Analytical Note	Steve Herman	2022-05-22 07:53:01
Computer solution	Larry	2022-05-21 09:11:45

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