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question on remainder..

Expert replies
by silvia928 » Mon May 26, 2008 10:58 pm
If n is a positive integer, what is the remainder when (7^[4n+3])(6^n) is divided by 10?

Ans. 8
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Source: — Problem Solving |

by ksh » Mon May 26, 2008 11:08 pm
We can go ahead with considering n to be 1(+ve int)

then the stimulus= 7^7*3*2/10
Now units digit of 7^7=3
and 3*3*2=18/10 will leave us remainder as 8

thats the ans !
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by netigen » Tue May 27, 2008 12:21 am
Silvia,

the way to solve these type of Qs is to realize that the power of 7 will have repeating unit digits for e.g.

unit digit of 7^1 = 7
unit digit of 7^2 = 9
unit digit of 7^3 = 3
unit digit of 7^4 = 1
unit digit of 7^5 = 7
unit digit of 7^6 = 9
unit digit of 7^7 = 3
unit digit of 7^8 = 1

We find that the unit digit repeats in cycles of 4

power of 7 is of the form 4n+3 which means that the unit digit will be 3

6^n will have unit digit = 6

so we get 6 x 3 = 8 in the units digit which will leave a remainder of 8
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by wawatan » Tue May 27, 2008 1:16 am
great explanation, netigen!
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