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Bunuel
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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
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Celestial


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

(A) 1
(B) 3
(C) 5
(D) 7
(E) 9


Kudos for a correct solution.
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Celestial09
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
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
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(137^13)^47
we can rewrite this as 137^(13*47) = 13^611
now let's look for the repetitive pattern
7^1 = units digit 7
7^2 = units digit 9
7^3 = units digit 3
7^4 = units digit 1
etc.

we can see that when raising 7 to a power, which is a multiple of 4, the units digit of the number is 1
we have number 612, which is a multiple of 4, and when 7 raised to 612 power, the units digit is 1. thus we can conclude that for 7 raised to the 611 power, the units digit is 3.
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Bunuel
The units digit of (137^13)^47 is:

(A) 1
(B) 3
(C) 5
(D) 7
(E) 9


Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

First of all, all we need is the last digit of the base, not 137, but just 7. Here’s the power sequence of the units of 7

7^1 has a units digit of 7
7^2 has a units digit of 9 (e.g. 7*7 = 49)
7^3 has a units digit of 3 (e.g. 7*9 = 63)
7^4 has a units digit of 1 (e.g. 7*3 = 21)
7^5 has a units digit of 7
7^6 has a units digit of 9
7^7 has a units digit of 3
7^8 has a units digit of 1
etc.

The period is 4, so 7 to the power of any multiple of 4 has a units digit of 1
7^12 has a units digit of 1
7^13 has a units digit of 7

So the inner parenthesis is a number with a units digit of 7.

Now, for the outer exponent, we are following the same pattern — starting with a units digit of 7. The period is still 4.
7^44 has a units digit of 1
7^45 has a units digit of 7
7^46 has a units digit of 9
7^47 has a units digit of 3

So the unit digit of the final output is 3.

Answer = B
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Bunuel
The units digit of (137^13)^47 is:

(A) 1
(B) 3
(C) 5
(D) 7
(E) 9


Simplifying, we have (137^13)^47 = 137^611. Now recall that the units digit of a number raised to a power is the same as the units digit of that number’s units digit raised to the same power. Therefore, we can just determine the units digit of 7^611.

Recall that the units digit pattern of 7 raised to a power is 7-9-3-1. We see that 7 raised to a power that is a multiple of 4 will result in a units digit of 1. Since 611/4 = 152 R 3, we see that the units digit of 7^611 (and hence 137^611) is 3 (i.e., the 3rd number in the units digit pattern).

Answer: B

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

(A) 1
(B) 3
(C) 5
(D) 7
(E) 9


Kudos for a correct solution.

\(7^4\) Will have units digit \(1\)

So, \(7^{13} = 7^{4*3 + 1}\) , Thus units digit will be 7

Again \(7^{47} = 7^{4*11 + 3} \)

\(7^3\) Will have units digit \(3\)

Hence, units digit of (137^13)^47 is (B) 3
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