Question 1101535
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If you only care about the ones digit, then you don't need to keep any of the other digits.  So, to begin with, you can simplify the problem to using 7 as a factor instead of 87.<br>
7 as a factor 2 times: 7*7=49; you only care about the ones digit, which is 9.
7 as a factor 3 times: start with the 9, which is the ones digit of 7^2, and multiply by 7; the result is 63.  You only care about the ones digit, which is 3.
7 as a factor 4 times: start with the 3 and multiply by 7 again; the result is 21; keep only the ones digit, which is 1.
7 as a factor 5 times: multiply the 1 times 7; you get a ones digit of 7 again.<br>
You should be able to see that, after that, the cycle will repeat.  The ones digit for successive factors of 7 (or 87, or 192308437) are
7, 9, 3, 1, 7, 9, 3, 1, 7, ...<br>
You can see that you get a ones digit of 7 with either 7 by itself, or with 5 factors of 7, or with 9 factors of 7, or....<br>
That's why one of the other tutors, without any explanation, said the number of factors could be any number of the form 4n+1, where n is any integer greater than or equal to 0.<br><br>
I hope this explanation helps....