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

ANSWER is B NOT A,


Here is why the answer is B....


Number 7 has 4 cycle in its repeating power :

7^1 =7

7^2= ..9

7^3 = ...3

7^4= ....1

from this point the unit digit of 7^5 is 7 so it is repeating so keep in mind that we have 4 cycles that are repeating ...

so here lets apply this to the problem, first look at the initial term without considering the power 47 : ( 137^13 ) , here ONLY unit digit matters so we have : ( 7^13 ) by applying above cycle

pattern , we see that 13 is 3*4 +1 , so we should cover 4 FULL cycle and the 13th term is our answer . by viewing to the cycle we see that after 4 full cycle the 13th number is conform to the first number in 7 power so 13th's unit digit is 7...


Now lets apply the power 47 ... we have a number some thing like this : (.....7 ) ^47 , once again we must apply our pattern : 47 = 11*4 +3 , so the third term is equal to 47th term

and we see here the third term is 3 .... so our answer is 3 , so we can say that the unit digit of ( 137^13) ^47 is 3 .... answer B...

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

(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.

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

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

\(7^4\) 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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