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by gmatmillenium » Sun Jun 20, 2010 2:36 am
Princess23 wrote:What is the remainder of 7^365 divided by 5 and how did you go about it?
Hi,

As you know, to find the remainder of any division by 5, we primarily need to know the units digit of the numerator....

lets find the units digit in the following powers

7 to the power 1...7
7 to the power 2....9
7 to the power 3....3
7 to the power 4.....1
7 to the power 5....7

the units digit starts repeating the pattern after every 4th power....so the 364th power of 7 will have the units digit as 1 and 365th will have 7.....7 divided by 5 yields a remainder of 2.

is this the right answer?

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by jube » Sun Jun 20, 2010 3:11 am
When 7^1 is divided by 5, you get remainder 2
When 7^2 is divided by 5, you get remainder 4
When 7^3 is dvided by 5, you get remainder 3
When 7^4 is divided by 5, you get remainder 1
When 7^5 is divided by 5, you get remainder 2 again & the cycle repeats itself

Notice that you only need to know the unit digits for the above computation.

So whenever we divide powers of 7 by 5 the remainder will always have the pattern 2, 4, 3, 1 and then cycle back on itself

Now 365 will be 4(91) + 1 i.e. when we divide 7^365 with 5 we have to look at the remainder for 7^1 - so the answer is 2

Is that the OA?
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by Princess23 » Sun Jun 20, 2010 9:37 am
Thanks for the solutions. Much appreciated. My thankful wishes to all of you!
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