Celestial09 wrote:

IMO it should be 1 (A)

this is how I would have done in exam :

power repetitive of 7..

7^1 =7

7^2 = 9

7^3 = 3

7^4 = 1

7^5=7

7^6=9

7^7=3

this means remainders of power ending with 1 gives unit digit 7

remainders of power ending with 2 gives unit digit 9

13/7 gives remainder of 6... hence unit digit left is 9

9^1=9

9^2=1

9^3=9

this means remainders of power ending with 1 gives unit digit 9

remainders of power ending with 2 gives unit digit 1

repetitive nature is 2

47/9 gives remainder 2

hence unit digit is 1

unit digit is 1

Kudos please if my solution is right and having appropriate method

Thanks

Celestial

Bunuel wrote:

The units digit of (137^13)^47 is:

(A) 1

(B) 3

(C) 5

(D) 7

(E) 9

Kudos for a correct solution.

hi

Celestial09,

your approach is correct but you have gone wrong in your observation of repetitive nature of power of 7.. it is 7,9,3,1,7,9,3,1.. so every term after multiple of 4 starts a new repetition..

this is the standard repetitive nature of powers , they repeat after every 4th term(2,3,7,8).... with some after every 2nd term example 9,4 and few each term is same... eg 1,5,6,0...

so if 7 has 7,9,3,1...13 leaves a remainder of 1 when divided by 4.. so we have xy..7^47..

now 47 leaves a remainder of 3, so last digit is 3..ans B

_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372

2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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