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You have 5 numbers

1, 9, 3, 5, and 7

you need to arrange these number like

Rules
create 2 pairs from numbers above keep one of the number in the middle then apply this law
1st pair * last pair - middle number. after arranging and applying the above rules the result should be a number where whose all digits are same.

example if I arrange it like below

79(1st-pair) 5(middle) 31(2nd-pair)
79 * 31 - 5 = 2444 (1st pair * last pair - middle number)

it should be 4444 instead of 2444 ( it can be a right solution if instead of 2444 I get 4444)

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  • $\begingroup$ Do we have to list all solutions, or at least one? (I don't know if there are multiple, but I found one) $\endgroup$ – the default. Oct 18 '19 at 14:44
  • $\begingroup$ @someone I think there is just 1 solution the other is mirror :) let's see if you come up with more $\endgroup$ – Sayed Mohd Ali Oct 18 '19 at 14:45
  • $\begingroup$ is it pair * pair - digit or pair * digit - pair? the rules says the first, but the example demonstrates the second $\endgroup$ – Omega Krypton Oct 18 '19 at 14:47
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The only solutions are

39, 1, 57 ($39\cdot 57 - 1 = 2222$) and 57, 1, 39 ($57\cdot 39 - 1 = 2222$ too, because it's pretty much the same)

After finding these, I have brute forced this with the following ugly Mathematica script to prove no other solutions exist:

Print@Select[Permutations[{1,3,5,7,9}],Equal@@IntegerDigits[(#[[1]]*10+#[[2]])*(#[[4]]*10+#[[5]])-#[[3]]]&]

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I have found one answer

$(39 \times 57) - 1 = 2222$

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