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.
If the last digit is ... It's cube ends in ...
1 1
2 8
3 7
4 4
5 5
6 6
7 3
8 2
9 9
0 0
The sequences 9,0,1 and 4,5,6 work and these triplets start with 9 and 4 respectively. For consideration on the negative side, we're interested in 1,0,9 and 6,5,4 respectively beginning with 1 and 6. A special consideration is the transition from negative to positive with 1, 0, 1.
On the negative side, each of the 8 decades contains numbers ending 1 and in 6, thereby starting 16 triplets; in fact this includes the sign-transitional case.
On the positive side, numbers ending in 4 occur in all 8 decades, but those ending in 9 occur in only 7 of the decades, thus totaling 15 triplets.
The total is 16 + 15 = 31.
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Posted by Charlie
on 2023-08-22 09:35:24 |
solution
