Units Digit Problem

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by HSPA » Tue Mar 15, 2011 11:54 pm
powers of 6: 6,36,216... ending 6 always
powers of 7: 7,49,343,XXX1.... ending in 7,9,3,1,7... repeat
powers of 9: 9,81,729... ending in 9,1,9,1..repeat

Now using all unit digit endings (9^3+7^4) = 9+1 = 10
Now 6-0 = 6

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by ankur.agrawal » Wed Mar 16, 2011 12:41 am
6983manish wrote:What is the units digit of 6^15 - 7^4 - 9^3?
A) 8
B) 7
C) 6
D) 5
E) 4
I have seen multiples between the two numbers .For the first time I see a subtraction.

ok 6^15 ---> unit digit is 6

7^4 --> unit digit is 1

9^3 ---> unit digit is 9 . Hmm ?

( 6^15 - 7^4) - 9^3 = Using Unit digit endings

( 6-1) - 9 = 5-9 -----> take a carry so 15-9=6 ?

Is this the correct way to do it . Pls comment.