What is the remainder when 54^124 is divided by 17?

Question

What is the remainder when  54^{124}  is divided by 17?

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

KarishmaB · Accepted Answer

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)

XSatishX · Answer

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.

Kinshook · Answer

Asked: What is the remainder when  54^{124}  is divided by 17?

Remainder when 54^{124} is divided by 17
= remainder when 3^{124} is divided by 17
= remainder when (-4)^{31} is divided by 17
= remainder when (-1)^{15}×(-4) is divided by 17
= 4

IMO A

Posted from my mobile device

akadiyan · Answer

What is the remainder when  54^{124}  is divided by 17?

54^{124}  =  3^{124}  mod 17

54^{124}  =  9^{62}  mod 17

54^{124}  =  81^{31}  mod 17

54^{124}  =  -4^{31}  mod 17

54^{124}  = ( -4^{1} ) * ( -4^{30} ) mod 17

54^{124}  = ( -4^{1} ) * ( 16^{15} ) mod 17

54^{124}  = (-4) * ( -1^{15} ) mod 17

54^{124}  = (-4*-1) mod 17

54^{124}  = 4 mod 17 =  4

Ans =  A

TarunKumar1234 · Answer

Given, (54^124)/17 or, /17 So, it will have same remainder as /17 or, /17 or, /17 or, /17 or, /17 (Trying to reach to -1 or +1) It has same remainder as /17 = (-4)*(-1) = 4. So, I think A.

wishmasterdj · Answer

Very efficient method, thanks!

chetan2u  what are the odds of such a question coming up, considering we have to calculate the cyclicity upto 16 levels, as per the first method?

sthahvi · Answer

I did not understand the method how are we going to -4 and its degrees, can someone explain the question and how it will be solved.  VeritasKarishma

rdrdrd1201 · Answer

hi anyone got similar questions like this?

Bunuel · Answer

6. Remainders 
     Theory   
 REMAINDERS 
 Remainders: Tips and hints 
 Divisibility and Remainders on the GMAT 
 Remainders on the GMAT 
 A Remainders Shortcut for the GMAT 
 All About Negative Remainders on the GMAT 
 A Tricky Question on Negative Remainders 
 Compilation of tips and tricks to deal with remainders 
 Cyclicity and the remainders on the GMAT 
 GMAT Question Patterns: Remainders 
     Questions   
 Units digits, exponents, remainders problems  
 DS Questions 
 PS Questions

8. Exponents and Roots of Numbers 
     Theory   
 Exponents and Roots on the GMAT: Tips and hints 
 Cyclicity and the remainders on the GMAT 
     Questions   
 Units digits, exponents, remainders problems 
 Tough and tricky DS exponents and roots questions with detailed solutions 
 Tough and tricky PS exponents and roots questions with detailed solutions 
 Exponents DS Questions 
 Exponents PS Questions 
 Roots DS Questions 
 Roots PS Questions

For other subjects:
 ALL YOU NEED FOR QUANT ! ! ! 
 Ultimate GMAT Quantitative Megathread

Hope it helps.

BrushMyQuant · Answer

We need to find what is the remainder when  54^{124}  is divided by 17

54^{124}  =  (51 + 3)^{124}  =  (17*3 + 3}^{124}

using binomial theorem if we open this then all terms except the last term will be a multiple of 17 => Remainder of all the terms except the last term with 17 will be zero

=> Remainder will be same as the remainder of the last term =  3^{124}  =  3^{4*31}  =  (3^4)^{31}  =  81^{31}  =  (85 -4)^{31}  =  (17*5 -4)^{31}

using binomial theorem if we open this then all terms except the last term will be a multiple of 17 => Remainder of all the terms except the last term with 17 will be zero

=> Remainder will be same as the remainder of the last term =  (-4)^{31}  = -4 *  4^{30}  = -4 *  4^{2*15}  = -4 *  16^{15}  = -4 *  (17 - 1)^{15}

using binomial theorem if we open  (17 - 1)^{15}  then all terms except the last term will be a multiple of 17 => Remainder of all the terms except the last term with 17 will be zero

=> Remainder will be same as the remainder of the last term * -4 =  (-1)^{15}  = -1 * -4 = 4

So,  Answer will be A 
Hope it helps!

Watch the following video to learn the Basics of Remainders

https://www.youtube.com/watch?v=P0S6j2uTaj4

wadhwakaran · Answer

54 when divided by 17 gives a remainder of 3
54^4 when divided by 17 gives a remainder of 3^4 i.e. 81/17 i.e. -4
54^8 when divided by 17 gives a remainder of -4*-4 = 16/17 i.e. -1
54^16 when divided by 17 gives a remainder of 1
124/16 give a remainder of 12
54^124 = 54^12 = 54^4*54^8= -4*-1=4
 option A

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What is the remainder when 54^124 is divided by 17?

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