For integer values of n how many triplets exist such that -77<n<77 and each member of the triplet (n,n+1,n+2) has the same last digit as it's cube.
Provide the reasoning of getting your answer.
Numbers with last digit 4,5,6,9,0,1 have the same last digit as their cube. Note these come in triples. So for negatives, n ends in 6 or 1. For positives, n ends in 4 or 9. In between there's the triple -1,0,1.
I count 31 total: 16 negative and 16 positive minus one so the in-between isn't counted twice.
Edit: I think I got the right answer with wrong reasoning.
There are 16 negative (eight ending in 6 and eight ending in 1) and 15 positive (eight ending in 4 and seven ending in 9.) Still 31, though.
Edited on August 22, 2023, 9:34 am
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Posted by Jer
on 2023-08-22 09:22:52 |
Solution
