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SVP
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What is the remainder of 2 ^51 divided by 7 ? What is the [#permalink]
 07 Sep 2006, 23:20
What is the remainder of 2 ^51 divided by 7 ?
What is the remainder of 3^19 when divided by 10
a) 0
b) 1
c) 5
d) 7
e) 9
I have a question on those two exponents , i will post it after you solve it ?
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Senior Manager
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remainder of 2^0 / 7 = 1
2^1 / 7 = 2
2^2/ 7 = 4
and then the pattern repeats.
by this logic, remainder of 2^51/7 = 1
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Senior Manager
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for remainder of 3^19 / 10
unit digit of 3^0 = 1
3^1 = 3
3^2 = 9
3^3 = 7
then the pattern repeats.
so unit digit of 3^19 = 7
hence remainder of 3^19 / 10 = 7
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Manager
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2 ^ 51 = 8 ^ 17 = (7 + 1) ^ 17
We needn't expand this because the 18th ( the last term ) in this expansion will be 1 and the rest of the terms will have 7 as one of the factors. Hence the remainder would be 1.
The units digit in the expansion
3 ^ 1 = 3
3 ^ 2 = 9
3 ^ 3 = 27 ( units digit 7 )
3 ^ 4 = 81 ( units digit 1 )
3 ^ 5 = 153 ( units digit 3 )
........
There is a trend in the units digits and when a number is divided by 10 then the units digit would be the remainder.
Hope this helps.
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SVP
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jainvik7
Very comprehensive approach... very
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Manager
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Thanks,
iced_tea and yezz 
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Director
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IMO second Q can be re-phrased as; What is the last digit of 3^19 ? It is 7 .
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Current Student
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BG wrote: IMO second Q can be re-phrased as; What is the last digit of 3^19 ? It is 7 .
Same question here. I think it is 1.
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Intern
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It looks like we can use this approach
2 ^ 51 = 8 ^ 17 = (7 + 1) ^ 17
We needn't expand this because the 18th ( the last term ) in this expansion will be 1 and the rest of the terms will have 7 as one of the factors. Hence the remainder would be 1.
to solve for the 3^19 question as well
3^19 = 3^3*3^16 = 3^3*9^8 = 3^3*81^4 = 3^3*(80+1)^4
like before the (80+1)^4 only has one term that isnt divisible by 10, namely 1.
so we get the remainder as 1*3^3 = 27, but 27 is greater than 10 so the remainder is 27/10 which is 7.
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close
