Question 1101535: Using only the number 87 how many times will you use it as a factor to get a product whose ones digit is 7?
Found 3 solutions by richwmiller, Alan3354, greenestamps:
Answer by richwmiller(17219)   (Show Source):
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! Using only the number 87 how many times will you use it as a factor to get a product whose ones digit is 7?
-----------
87^(4n+1), integer n >= 0
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
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.
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.
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, ...
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....
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.
I hope this explanation helps....
|
|
|
