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Apr 2808:00 AM PDT
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11 Jan 2021, 08:03
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A
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Difficulty:
95%
(hard)
Question Stats:
37% (02:33) correct
63%
(02:30)
wrong
based on 259
sessions
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What is the remainder when \(54^{124}\) is divided by 17?
A. 4
B. 5
C. 7
D. 13
E. 15
A. 4
B. 5
C. 7
D. 13
E. 15
Find the remainder when 54^124 is divided by 17.
a)4
b)5
c)13
d)15
a)4
b)5
c)13
d)15
10 Mar 2021, 01:24
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Bunuel
First check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... ek-in-you/
\(54^{124} = (51 + 3)^{124}\)
Since 51 is divisible by 17, remainder when divided by 17 will be \(3^{124}\).
\(3^{124} = (3^4)^{31} = 81^{31} = (85 - 4)^{31}\)
Since 85 is divisible by 17, remainder will be \((-4)^{31}\).
\((-4)^{31} = -4*(4)^{30} = -4*(16)^{15} = -4*(17 - 1)^{15}\)
Remainder on division by 17 will be -4*-1 = 4
Answer (A)
General Discussion
XSatishX
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11 Jan 2021, 08:50
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To begin, find the remainder when 54 is divided by 17.
R(54/17) = 3. So our question can be reduced to what is the remainder when 3^124 is divided by 17.
Let us look for the cyclicity then.
R(3/17) = 3
R(9/17)=9
R(27/17)=10
R(81/17)=13
R(243/17) = 5
R(729/17)=15
R(2187/17)=11
R(6561/17)=16
R(19683/17)=14
R(59049/17)=8
R(177147/17)=7
R(531441/17)=4
R(1594323)=12
R(4782969/17)=2
R(14348907/17)=6
R(43046721/17)=1
R(387420489/17)=3
Finally, we have a cycle with cyclicity 16. So Remainder of 3^124 divided by 17 will be equal to 3^Remainder of 124/16 divided by 17.
R(124/16) = 12
So R(3^12)/17 =4
Hence Answer will be 4
Now a trick to solve this kind of question. The maximum cyclicity a unit digit has is 4 and the maximum cyclicity a remainder can have is 16. So instead of doing this calculation, we can safely use 16 as cyclicity and all the remainders with cyclicity of 4 and 8 will also have a cyclicity of 16.
R(54/17) = 3. So our question can be reduced to what is the remainder when 3^124 is divided by 17.
Let us look for the cyclicity then.
R(3/17) = 3
R(9/17)=9
R(27/17)=10
R(81/17)=13
R(243/17) = 5
R(729/17)=15
R(2187/17)=11
R(6561/17)=16
R(19683/17)=14
R(59049/17)=8
R(177147/17)=7
R(531441/17)=4
R(1594323)=12
R(4782969/17)=2
R(14348907/17)=6
R(43046721/17)=1
R(387420489/17)=3
Finally, we have a cycle with cyclicity 16. So Remainder of 3^124 divided by 17 will be equal to 3^Remainder of 124/16 divided by 17.
R(124/16) = 12
So R(3^12)/17 =4
Hence Answer will be 4
Now a trick to solve this kind of question. The maximum cyclicity a unit digit has is 4 and the maximum cyclicity a remainder can have is 16. So instead of doing this calculation, we can safely use 16 as cyclicity and all the remainders with cyclicity of 4 and 8 will also have a cyclicity of 16.
