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What is the remainder when 32^32^32 is divided by 7?

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Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

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New post 29 Apr 2020, 07:40
gurpreetsingh wrote:
Similar question to test what you have learnt from the previous post.

What is the remainder when \(32^{32^{32}}\) is divided by 9?

A. 7
B. 4
C. 2
D. 0
E. 1



32^{32^{32}}=(27+5)^{32^{32}}
Solving this binomial power only the last term won't be divisible by 9, so let's consider only this term:
5^{32^{32}}=5^{2^{5*32}}=5^{2^{160}}



Let's consider ciclicity when a power of 5 is divided by 9 (NO WAY I COULD DO THIS ON TEST DAY!!!):
- 5^1 divided by 9 gives a reminder of 5
- 5^2 divided by 9 gives a reminder of 7
- 5^3 divided by 9 gives a reminder of 8
- 5^4 divided by 9 gives a reminder of 4
- 5^5 divided by 9 gives a reminder of 2
- 5^6 divided by 9 gives a reminder of 1

- 5^7 divided by 9 gives a reminder of 5
- 5^8 divided by 9 gives a reminder of 7
- 5^9 divided by 9 gives a reminder of 8
- 5^10 divided by 9 gives a reminder of 4
...

Therefore ciclicity is 6.
Let's divide our exponent 2^{160} by 6 and figure out the reminder.



Ciclicty for power of 2 divided by 6 is:
- 2^1 divided by 6 gives a reminder of 2
- 2^2 divided by 6 gives a reminder of 4

- 2^3 divided by 6 gives a reminder of 2
- 2^4 divided by 6 gives a reminder of 4
...

For even exponents the reminder is always 4 (use it to go back in the previous ciclicity and pick the right number).

THE REMINDER IS 4

CORRECT ANS IS B!!
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Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

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New post 11 May 2020, 07:54
Bunuel wrote:
gurpreetsingh wrote:
What is the remainder when \(32^{32^{32}}\) is divided by 7?

A. 5
B. 4
C. 2
D. 0
E. 1

If we use the above approach I'd work with prime as a base.

\(32^{{32}^{32}}=(28+4)^{{32}^{32}}\) now if we expand this, all terms but the last one will have 28 as a multiple and thus will be divisible by 7. The last term will be \(4^{{32}^{32}}=4^{{(2^5)}^{32}}=4^{2^{160}}=2^{2^{161}}\). So we should find the remainder when \(2^{2^{161}}\) is divided by 7.

Hope it's clear.


VeritasKarishma

Can you pls explain how we got the highlighted portion, 4 can be written as 2^2 and that would translate to 2^4^160

Pls explain
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Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

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New post 11 May 2020, 08:48
GDT wrote:
Bunuel wrote:
gurpreetsingh wrote:
What is the remainder when \(32^{32^{32}}\) is divided by 7?

A. 5
B. 4
C. 2
D. 0
E. 1

If we use the above approach I'd work with prime as a base.

\(32^{{32}^{32}}=(28+4)^{{32}^{32}}\) now if we expand this, all terms but the last one will have 28 as a multiple and thus will be divisible by 7. The last term will be \(4^{{32}^{32}}=4^{{(2^5)}^{32}}=4^{2^{160}}=2^{2^{161}}\). So we should find the remainder when \(2^{2^{161}}\) is divided by 7.

Hope it's clear.


VeritasKarishma

Can you pls explain how we got the highlighted portion, 4 can be written as 2^2 and that would translate to 2^4^160

Pls explain


\(4^{2^{160}}=(2^2)^{2^{160}}= 2^{2*2^{160}}=2^{2^{160+1}}=2^{2^{161}}\)
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Re: What is the remainder when 32^32^32 is divided by 7?  [#permalink]

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New post 12 May 2020, 00:47
Bunuel wrote:
GDT wrote:
Bunuel wrote:
gurpreetsingh wrote:
What is the remainder when \(32^{32^{32}}\) is divided by 7?

A. 5
B. 4
C. 2
D. 0
E. 1

If we use the above approach I'd work with prime as a base.

\(32^{{32}^{32}}=(28+4)^{{32}^{32}}\) now if we expand this, all terms but the last one will have 28 as a multiple and thus will be divisible by 7. The last term will be \(4^{{32}^{32}}=4^{{(2^5)}^{32}}=4^{2^{160}}=2^{2^{161}}\). So we should find the remainder when \(2^{2^{161}}\) is divided by 7.

Hope it's clear.


VeritasKarishma

Can you pls explain how we got the highlighted portion, 4 can be written as 2^2 and that would translate to 2^4^160

Pls explain


\(4^{2^{160}}=(2^2)^{2^{160}}= 2^{2*2^{160}}=2^{2^{160+1}}=2^{2^{161}}\)


Thank you Bunuel! It's clear now
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What is the remainder when 32^32^32 is divided by 7?  [#permalink]

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New post Updated on: 27 May 2020, 08:15
Bunuel wrote:
gurpreetsingh wrote:
What is the remainder when \(32^{32^{32}}\) is divided by 7?

A. 5
B. 4
C. 2
D. 0
E. 1

Please do not just post the answer, do explain as well.

I will post the Answer and the explanation after some replies.


If we use the above approach I'd work with prime as a base.

\(32^{{32}^{32}}=(28+4)^{{32}^{32}}\) now if we expand this, all terms but the last one will have 28 as a multiple and thus will be divisible by 7. The last term will be \(4^{{32}^{32}}=4^{{(2^5)}^{32}}=4^{2^{160}}=2^{2^{161}}\). So we should find the remainder when \(2^{2^{161}}\) is divided by 7.

2^1 divided by 7 yields remainder of 2;
2^2 divided by 7 yields remainder of 4;
2^3 divided by 7 yields remainder of 1;

2^4 divided by 7 yields remainder of 2;
2^5 divided by 7 yields remainder of 4;
2^6 divided by 7 yields remainder of 1;
...

The remainder repeats the pattern of 3: 2-4-1.

So we should find \(2^{161}\) (the power of 2) is 1st, 2nd or 3rd number in the above pattern of 3. \(2^{161}\) is 2 in odd power, 2 in odd power gives a remainder of 2 when divided by cyclicity number 3, so it's the second number in pattern. Which means that remainder of \(2^{2^{161}}\) divided by 7 would be the same as \(2^2\) divided by 7. \(2^2\) divided by 7 yields remainder of 4.

Answer: B.

Similar problem: http://gmatclub.com/forum/remainder-997 ... ms#p768816

Hope it's clear.


I am confused about the highlighted part.
Why didn't you consider the remainder of 1 as well??

I'm really having a hard trying to figure out how we were able to find \(2^{161}\) and used it to find our remainder.
Please help, thanks :)

Originally posted by D4kshGargas on 12 May 2020, 12:54.
Last edited by D4kshGargas on 27 May 2020, 08:15, edited 2 times in total.
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What is the remainder when 32^32^32 is divided by 7?  [#permalink]

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New post 27 May 2020, 07:49
Quote:
I am confused about the highlighted part.
Why didn't you consider the remainder of 1 as well??

I'm really having a hard to figure out how we were able to find \(2^{161}\) and use to find our remainder.
Please help, thanks :)
[/quote]

Hi D4kshGargas

I will try to clear your doubts.
a.(About highlighted part)-
Here if you see cycle of reminder as mentioned in earlier post are (2,4,1) which repeats again and again.
Thus
2^1 divided by 7 yields remainder of 2;
2^2 divided by 7 yields remainder of 4;
2^3 divided by 7 yields remainder of 1;
2^4 divided by 7 yields remainder of 2;
2^5 divided by 7 yields remainder of 4;
2^6 divided by 7 yields remainder of 1;
Now to find reminder of \(2^{161}\) when divided by 7, Divide 161 by 3 to get the reminder to know the value from above cycle.
(why 3 ? cycle repeats after 3 values)
Here reminder is 1 hence our final reminder will be 2.

Now 2^2 divided by 7 yields remainder of 4.

I hope it helps :)
Feel free to ask more doubts if any.
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What is the remainder when 32^32^32 is divided by 7?   [#permalink] 27 May 2020, 07:49

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