Remainder

[ani781]

When 51^25 is divided by 13, the remainder obtained is:
A. 12
B. 10
C. 2
D. 1
E. 0

OA after some discussion,but how to approach this kind of a problem ?

[sanjoy18]

Answer A..51 and 13 are co-prime number hence by using Euler theorem

reminder is 12

[vinay1983]

I think we can see that the units digit is 1, so 1 raised to any number of times will always be 1, so when 1 is divided by 13, we get 12, so has to be A

[ganeshrkamath]

When 51^25 is divided by 13, the remainder obtained is:
A. 12
B. 10
C. 2
D. 1
E. 0

OA after some discussion,but how to approach this kind of a problem ?

13*4 = 52
So 51 mod 13 = -1
51^25 mod 13 = (-1)^25 mod 13 = (-1) mod 13 = 12
Choose A

Cheers

[Brent@GMATPrepNow]

I think we can see that the units digit is 1, so 1 raised to any number of times will always be 1, so when 1 is divided by 13, we get 12, so has to be A

Be careful. The answer is, indeed, A but for a different reason. If we applied the above logic, the remainder when 51^2 is divided by 13 will be 12, but the remainder is actually 1.

Cheers,
Brent

[vipulgoyal]

When 51^25 is divided by 13, the remainder obtained is:
A. 12
B. 10
C. 2
D. 1
E. 0

OA after some discussion,but how to approach this kind of a problem ?13*4 = 52
So 51 mod 13 = -1
51^25 mod 13 = (-1)^25 mod 13 = (-1) mod 13 = 12
Choose A

----

I think GMAT doesnt suppose students to know mod, ne ways here we go
51^25/13
51/13 = {13(3) +12 }/13 = 12

[theCodeToGMAT]

Another approach:

Divide 25 by 4.. the remainder is "1"

So, rephrase the question: (51)^1/13 = 51/13 = 12

Answer [spoiler]{A}[/spoiler]