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

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The units digit of (137^13)^47 is: [#permalink]

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New post 05 Mar 2015, 10:00
2
4
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

66% (00:56) correct 34% (00:53) wrong based on 266 sessions

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Re: The units digit of (137^13)^47 is: [#permalink]

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New post 05 Mar 2015, 13:13
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.
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Re: The units digit of (137^13)^47 is: [#permalink]

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New post 05 Mar 2015, 14:52
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1
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...
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The units digit of (137^13)^47 is: [#permalink]

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New post 05 Mar 2015, 15:48
1
2
7^1=7
7^2=9
7^3=3
7^4=1
7^5=7
that means the cycle repeats after 4

(137^13)^47 = 137^611. 611/4 =152 and has a reminder 3.

So in 137^611 the unit digit will be 3
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Re: The units digit of (137^13)^47 is: [#permalink]

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New post 06 Mar 2015, 09:59
1
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
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Re: The units digit of (137^13)^47 is: [#permalink]

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New post 06 Mar 2015, 10:10
1
(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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Re: The units digit of (137^13)^47 is: [#permalink]

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New post 08 Mar 2015, 14:57
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.


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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Re: The units digit of (137^13)^47 is: [#permalink]

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New post 12 May 2017, 04:48
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Re: The units digit of (137^13)^47 is:   [#permalink] 12 May 2017, 04:48
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