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805+ (Hard)|   Exponents|   Remainders|      
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Chethan92
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What is the remainder when 5^91 is divided by 91? 06 Aug 2018, 09:53
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

18% (02:21) correct 82% (02:14) wrong based on 381 sessions

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What is the remainder when \(5^{91}\) is divided by 91?

A. 5
B. 8
C. 13
D. 34
E. 47
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E
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What is the remainder when 5^91 is divided by 91? 06 Aug 2018, 19:32
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Afc0892
What is the remainder when \(5^{91}\) is divided by 91?

1) 5
2) 8
3) 13
4) 34
5) 47
Show more


I doubt such a question will come in actuals..

Otherwise..
1) as per remainder theorem
\(91=7*13\) so LCM of (7-1,13-1) or (6,12) is 12
So \(5^{91}\) will give the same remainder as \(5^7\), since 91/12 gives 7 as remainder.
\(5^7=(5^3)^2*5=(125)^2*5\)
125 will leave 125-91=34 as remainder
So \(34*34*5=34*170\)
170 will leave a remainder of -12
\(34*(-12)= 68*(-6)=-23*-6=138\)
So remainder=138-91=47

2) if you do not know the theorem you will have to get the terms closer to multiple of 91
So \(5^{91}=(5^4)^{22}*5^3=625^{22}*5^3\)
91*7=637, so remainder is \(625-637=(-12)\)
So new term is \((-12)^{22}*5^3\)
As even power, it will be same as \(12^{22}*5^3=4^{22}*3^{22}*5^3\)

Also \(4^3=64\) and remainder is 64-91=-27
So \((4^3)^7*4*3^{22}*5^3=(-27)^7*3^{22}*5^3=(-3^3)^7*3^{22}*5^3=-(3^{43}*5^3*4)\)
Now \(3^6=729\) and 91*8=728 so remainder is 1
\(-(3^6)^7*3*5^3*4=-(1^7*3*5^3*4)=-1*3*5*100=-1*3*5*9=-135\)
So POSITIVE remainder will be \(91*2-135=182-135=47\)

E

you could have used this too to simplify
Now \(2^7*5=128*5=640\) remainder will be 640-637=3
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What is the remainder when 5^91 is divided by 91? 06 Aug 2018, 22:03
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91 = 7*13

Let us find out Rem[5^91/7] and Rem[5^91/13] - using binomial expansion
We will combine them later.

Rem [5^91/7]

= 5(5)^90 / 7
= 5( 125) ^ 30 / 7
=5 (126-1)^30 / 7

remainder = 5. (-1)^30 = 5 / 7
rem = 5



Rem [5^91/13]
= Rem 5. (5)^90 / 13
= 5. (25)^45 / 13
= 5. (26-1)^45)/ 13
remainder = 5. (-1)^45 / 13
= -5/13 = r= 8



So, our answer is a number which leaves a remainder of 5 when divided by 7 and it should leave a remainder of 8 when divided by 13.

simplifying this, we get N= 91K + 47
Hence 47 is the reaminder
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What is the remainder when 5^91 is divided by 91? 06 Aug 2018, 17:37
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This one is really tricky! I don't have an answer yet, but I'm wondering if the fact that \(5^7\) does have a remainder of 47, and that \(5^{91} = (5^7)^{13}\), is significant.

I'm also wondering if 91's proximity to 100 could be useful. If you were to divide any power of 5 (over \(5^1\)) by 100, you would always get a remainder of 25. So perhaps the answer has something to do with using 100 as an approximation and then adding to the remainder.

Very interested to hear others' thoughts!

Edit: formatting
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What is the remainder when 5^91 is divided by 91? 21 Feb 2020, 21:38
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chetan2u
Afc0892
What is the remainder when \(5^{91}\) is divided by 91?

1) 5
2) 8
3) 13
4) 34
5) 47


I doubt such a question will come in actuals..

Otherwise..
1) as per remainder theorem
\(91=7*13\) so LCM of (7-1,13-1) or (6,12) is 12
So \(5^{91}\) will give the same remainder as \(5^7\), since 91/12 gives 7 as remainder.
\(5^7=(5^3)^2*5=(125)^2*5\)
125 will leave 125-91=34 as remainder
So \(34*34*5=34*170\)
170 will leave a remainder of -12
\(34*(-12)= 68*(-6)=-23*-6=138\)
So remainder=138-91=47

2) if you do not know the theorem you will have to get the terms closer to multiple of 91
So \(5^{91}=(5^4)^{22}*5^3=625^{22}*5^3\)
91*7=637, so remainder is \(625-637=(-12)\)
So new term is \((-12)^{22}*5^3\)
As even power, it will be same as \(12^{22}*5^3=4^{22}*3^{22}*5^3\)

Also \(4^3=64\) and remainder is 64-91=-27
So \((4^3)^7*4*3^{22}*5^3=(-27)^7*3^{22}*5^3=(-3^3)^7*3^{22}*5^3=-(3^{43}*5^3*4)\)
Now \(3^6=729\) and 91*8=728 so remainder is 1
\(-(3^6)^7*3*5^3*4=-(1^7*3*5^3*4)=-1*3*5*100=-1*3*5*9=-135\)
So POSITIVE remainder will be \(91*2-135=182-135=47\)

E

you could have used this too to simplify
Now \(2^7*5=128*5=640\) remainder will be 640-637=3
Show more
chetan2u can you share some link for remainder theorem you used in 1)
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What is the remainder when 5^91 is divided by 91? 22 Jul 2020, 04:23
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5^91/91...Remainder?

Solution:
=5^91/91
=5^91/7*13

=Points to observe
5 (num) and 91(deno) are co-primes
Need to factor the denominator,

=We need to develop 2 equations (as 2 factors are there in denom) based on divisibility of individual denominators..

Part:1
=5^91/7
=Based on Euler's theorem, Euler number for prime number at denom. will be 6. (7-1=6)
=dividing 91/6= 6K+1
=so 5^1/7= 5
=5 remainder.
=Equation1 N1= 7A+5

Part:2
=5^91/13
=Based on Euler's theorem, Euler number for prime number at denom. will be 12. (13-1=12)
=dividing 91/13= 12K+7
=so 5^7/13
=25*25*25*5/13
=8 Remainder
= Equation2 N2= 13B+8

Fill the values in Equation N1 (N1= 7A+5), Values are= 5,12,19,26,33,40,47
Fill the values in Equation N2 (N2= 13B+8), Values are= 8,21,34,47

First value satisfying both the equations will be remainder, i.e. 47.

Thanks!
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What is the remainder when 5^91 is divided by 91? 05 Jul 2022, 02:24
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Bunuel

can you help with this ? can such qs be asked in the gmat
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What is the remainder when 5^91 is divided by 91? 10 Jul 2022, 10:57
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chetan2u
Afc0892
What is the remainder when \(5^{91}\) is divided by 91?

1) 5
2) 8
3) 13
4) 34
5) 47


I doubt such a question will come in actuals..

Otherwise..
1) as per remainder theorem
\(91=7*13\) so LCM of (7-1,13-1) or (6,12) is 12
So \(5^{91}\) will give the same remainder as \(5^7\), since 91/12 gives 7 as remainder.
\(5^7=(5^3)^2*5=(125)^2*5\)
125 will leave 125-91=34 as remainder
So \(34*34*5=34*170\)
170 will leave a remainder of -12
\(34*(-12)= 68*(-6)=-23*-6=138\)
So remainder=138-91=47

2) if you do not know the theorem you will have to get the terms closer to multiple of 91
So \(5^{91}=(5^4)^{22}*5^3=625^{22}*5^3\)
91*7=637, so remainder is \(625-637=(-12)\)
So new term is \((-12)^{22}*5^3\)
As even power, it will be same as \(12^{22}*5^3=4^{22}*3^{22}*5^3\)

Also \(4^3=64\) and remainder is 64-91=-27
So \((4^3)^7*4*3^{22}*5^3=(-27)^7*3^{22}*5^3=(-3^3)^7*3^{22}*5^3=-(3^{43}*5^3*4)\)
Now \(3^6=729\) and 91*8=728 so remainder is 1
\(-(3^6)^7*3*5^3*4=-(1^7*3*5^3*4)=-1*3*5*100=-1*3*5*9=-135\)
So POSITIVE remainder will be \(91*2-135=182-135=47\)

E

you could have used this too to simplify
Now \(2^7*5=128*5=640\) remainder will be 640-637=3
Show more


Can you please share some link for examples on remainder theorem used in 1
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What is the remainder when 5^91 is divided by 91? 12 Jul 2022, 10:44
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rahulkashyap
91 = 7*13

Let us find out Rem[5^91/7] and Rem[5^91/13] - using binomial expansion
We will combine them later.

Rem [5^91/7]

= 5(5)^90 / 7
= 5( 125) ^ 30 / 7
=5 (126-1)^30 / 7

remainder = 5. (-1)^30 = 5 / 7
rem = 5



Rem [5^91/13]
= Rem 5. (5)^90 / 13
= 5. (25)^45 / 13
= 5. (26-1)^45)/ 13
remainder = 5. (-1)^45 / 13
= -5/13 = r= 8



So, our answer is a number which leaves a remainder of 5 when divided by 7 and it should leave a remainder of 8 when divided by 13.

simplifying this, we get N= 91K + 47
Hence 47 is the reaminder
Show more

How did you arrive at the equation N= 91K + 47 from this statement - is a number which leaves a remainder of 5 when divided by 7 and it should leave a remainder of 8 when divided by 13. Could you pls explain in detail?
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What is the remainder when 5^91 is divided by 91? 10 Sep 2022, 03:23
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Re(5^91/ 91)
Split 91, the divisor to 13 and 7

Separately find out out the remainders with 13 and 7.
With 7,
E(7)=6, therefore 5^91 actually reduces to Rem(5^1/ 7)----------Remainder is 5; the no is of the form 7a+5
Similarly with 13,
E913)=12, therefore 5^91 actually reduces to Rem(5^7/ 7)----------Remainder is 8; the no is of the form 13b+8
Congruence of these two number types gives 47 the answer. (7a+5=13b+8)
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What is the remainder when 5^91 is divided by 91? 11 Mar 2025, 08:20
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