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# What is the remainder when [m]5^{91}[/m] is divided by 91?

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What is the remainder when [m]5^{91}[/m] is divided by 91?  [#permalink]

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06 Aug 2018, 08:53
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00:00

Difficulty:

95% (hard)

Question Stats:

13% (01:44) correct 87% (02:06) wrong based on 38 sessions

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What is the remainder when $$5^{91}$$ is divided by 91?

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

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Re: What is the remainder when [m]5^{91}[/m] is divided by 91?  [#permalink]

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06 Aug 2018, 09:11
Can someone help with this please
Bunuel chetan2u KarishmaB

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What is the remainder when [m]5^{91}[/m] is divided by 91?  [#permalink]

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06 Aug 2018, 16:37
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 [m]5^{91}[/m] is divided by 91?  [#permalink]

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06 Aug 2018, 18:32
1
Afc0892 wrote:
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
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Re: What is the remainder when [m]5^{91}[/m] is divided by 91?  [#permalink]

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06 Aug 2018, 21:03
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
Re: What is the remainder when [m]5^{91}[/m] is divided by 91? &nbs [#permalink] 06 Aug 2018, 21:03
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