Number Theory #8
Does there exist an integer such that each of the ten digits appears exactly once as a digit in exactly one of the numbers ?
(Adapted from a past year Singapore Mathematical Olympiad question)
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1 solution
The problem states to find if a number "A" exists, where each number from "0 to 9" (ten digits in all), are present in A, A^2 and A^3.
This means, first of all, that there are only ten digits between A, A^2 and A^3, which means that A must be a two-digit number, A^2 must be a three-digit number and A^3 must be a five-digit number.
(If A is a one-digit number, A^2 and A^3 will not have enough digits. If A is a three-digit number, A^2 and A^3 will have too many digits. Therefore , A must be a two-digit number. If A^2 is a four digit number, A^3 will have too many digits. Therefore, A^2 should have three digits which leaves A^3 with five digits. )
This means that numbers between 22 and 31, inclusive, are possible values for A. (21^3 only has four digits and 32^2 has four digits.)
From 22 to 31, numbers that end with 0,1,5 and 6 are eliminated, as they will end in the same digit when squared. This will repeat digits and each digit must only occur once. 22 repeats the same digits within itself, so that is eliminated. 24 ends with 4 when cubed and 29 ends with 9 when cubed, so those are eliminated as well. That leaves us with 23, 27, 28. If any of the remaining numbers are tried, they will also repeat some digits, meaning that a number like A does not exist.