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perfect cube, not divisible by 10, erasing 3 last digit remains a perfect cube

Source: Greece JBMO TST 2001 p4

June 17, 2019
perfect cubenumber theory

Problem Statement

a) If positive integer NN is a perfect cube and is not divisible by 1010, then N=(10m+n)2N=(10m+n)^2 where m,n∈Nm,n \in N with 1≤n≤91\le n\le 9 b) Find all the positive integers NN which are perfect cubes, are not divisible by 1010, such that the number obtained by erasing the last three digits to be also also a perfect cube.
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