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A number when divided by 5 leaves a remainder 3

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by [email protected] » Thu Nov 17, 2011 12:07 am
IMO Answer is [spoiler](C)[/spoiler]

Explanation :

8 to the power 2 - remainder 4
8 to the power 3 - remainder 2
8 to the power 4 - remainder 1
8 to the power 5 - remainder 3
8 to the power 6 - remainder 4

so after every 4 power the cycle again start
8 to the power 100 meand 25 full cycle of 4 and hence the remainder will be 3
Ashish

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by shankar.ashwin » Thu Nov 17, 2011 12:14 am
3^100/5 - Remainder -?

3*1= 3 (remainder - 3)
3*3 = 9 (remainder - 4)
3*3*3 = 27 (remainder - 2)
27*3 = 81 (remainder - 1)

The pattern repeated every 4 cycles. So 3^100 would end in '1'. Remainder would be '1'. E IMO
Marysogh wrote:A number when divided by 5 leaves a remainder 3 , what would be the remainder when the number to the power 100 is divided by 5 ?
(a) 5
(b)4
(c)3
(d)2
(e)1
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by Anurag@Gurome » Thu Nov 17, 2011 12:49 am
Marysogh wrote:A number when divided by 5 leaves a remainder 3 , what would be the remainder when the number to the power 100 is divided by 5 ?
When 3 is divided by 5 the remainder is 3.
Hence, we need to find out the remainder when 3^100 is divided by 5.

3^100 = (3^4)^25 = 81^25 = Some large number with units digit 1

Hence, when 81^4 will be divided by 5, the remainder will be either 1 or 6.

The correct answer is E.
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