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What is the remainder when 32^32^32 is divided by 7? 20 Oct 2016, 23:28
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ac556
Bunuel
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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 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.

Hope it's clear.

Hello Bunuel, I am having trouble understanding this bit - \(2^{161}\) is 2 in odd power, 2 in odd power gives remainder of 2 when divided by cyclicity number 3, so it's the second number in pattern.

Will it be possible for you to explain this in a bit more detail? Many thanks.
Show more

2^1 = 2 divided by 3 --> remainder 2;
2^3 = 8 divided by 3 --> remainder 2;
2^5 = 32 divided by 3 --> remainder 2;
...
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What is the remainder when 32^32^32 is divided by 7? 21 Mar 2020, 22:08
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VeritasKarishma



Definitely a good effort but how did you get this?

This is how I would do it using binomial theorem:

\(32^{32^{32}}\) divided by 7

\((28 + 4)^{32^{32}}\) divided by 7

We need to figure out \(4^{32^{32}}\) divided by 7
We know that 64 is 1 more than a multiple of 7 and that \(4^3 = 64\)

But how many 3s do we have in \(32^{32}\)?
\(32^{32} = (33 - 1)^{32}\) so when we divide \(32^{32}\) by 3, we get a remainder of 1.

\(4^{32^{32}} = 4^{3x+1} = 4*4^{3x} = 4*64^x = 4*(63+1)^x\)

When this is divided by 7, remainder is 4.
Show more

VeritasKarishma

Many Thanks for your explanation. Can you please help me with the question posted by Gurpreet, as one more example will help me get hold of the logic.

Remainder when 32^32^32 is divided by 9.

My solution,

We need to only think about 2^32^32

And, we know that 2^3 = 8 which is 9-1. So again we want to know can we write 32^32 as 3k+1.

=> (33-1)^32, now as the exponent is even we will have remainder as 1.
=> 2^32^32 = 2^(3k+1)
=> 2 x 2^3k = 2 x 8^k
=> 2 x (9-1)^k.

This is where I get stuck because I don't know whether k is even or odd.

If k is even, then we have 2 divided by 9
remainder = 2

and, if k is odd,

we have -1 x 2 = -2

and remainder = 9-2 = 7.

Can you please help me in this.

I tried doing this by (27+5)^32^32

and there too I get (126-1)^k and it again depends on k being even or odd.

Edit:

I thought about it and did the question as below, please let me know if the methodology is correct.

Remainder when 32^32^32 is divided by 9?

32^32^32 = (36-4)^32^32

Now, 36^x will always be divisible by 9 so we need to look at only 4^32^32

We also know that 4^3 = 64, which is 63+1. So we need to find whether 3 will leave any remainder in 32^32.

(33-1)^32/3 leaves a remainder of 1.

Therefore, we can write 4^32^32 as 4^(3k +1)

= 4 x 4^3k = 4 x 64^k
= 4 x (63+1)^k
= 4

Thus 32^32^32 leaves a remainder of 4 when divided by 9.

Is my solution correct?


Thanks
Saurabh
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What is the remainder when 32^32^32 is divided by 7? 22 Mar 2020, 01:51
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Sarjaria84
VeritasKarishma



Definitely a good effort but how did you get this?

This is how I would do it using binomial theorem:

\(32^{32^{32}}\) divided by 7

\((28 + 4)^{32^{32}}\) divided by 7

We need to figure out \(4^{32^{32}}\) divided by 7
We know that 64 is 1 more than a multiple of 7 and that \(4^3 = 64\)

But how many 3s do we have in \(32^{32}\)?
\(32^{32} = (33 - 1)^{32}\) so when we divide \(32^{32}\) by 3, we get a remainder of 1.

\(4^{32^{32}} = 4^{3x+1} = 4*4^{3x} = 4*64^x = 4*(63+1)^x\)

When this is divided by 7, remainder is 4.

VeritasKarishma

Many Thanks for your explanation. Can you please help me with the question posted by Gurpreet, as one more example will help me get hold of the logic.

Remainder when 32^32^32 is divided by 9.

My solution,

We need to only think about 2^32^32

And, we know that 2^3 = 8 which is 9-1. So again we want to know can we write 32^32 as 3k+1.

=> (33-1)^32, now as the exponent is even we will have remainder as 1.
=> 2^32^32 = 2^(3k+1)
=> 2 x 2^3k = 2 x 8^k
=> 2 x (9-1)^k.

This is where I get stuck because I don't know whether k is even or odd.

If k is even, then we have 2 divided by 9
remainder = 2

and, if k is odd,

we have -1 x 2 = -2

and remainder = 9-2 = 7.

Can you please help me in this.

I tried doing this by (27+5)^32^32

and there too I get (126-1)^k and it again depends on k being even or odd.

Edit:

I thought about it and did the question as below, please let me know if the methodology is correct.

Remainder when 32^32^32 is divided by 9?

32^32^32 = (36-4)^32^32

Now, 36^x will always be divisible by 9 so we need to look at only 4^32^32

We also know that 4^3 = 64, which is 63+1. So we need to find whether 3 will leave any remainder in 32^32.

(33-1)^32/3 leaves a remainder of 1.

Therefore, we can write 4^32^32 as 4^(3k +1)

= 4 x 4^3k = 4 x 64^k
= 4 x (63+1)^k
= 4

Thus 32^32^32 leaves a remainder of 4 when divided by 9.

Is my solution correct?


Thanks
Saurabh
Show more

I don't understand the first step of your solution (highlighted). From where do you get 2^32^32?

The solution doesn't change much (from the solution when the divisor was 7):

\(32^{32^{32}}\) divided by 9

\((36 - 4)^{32^{32}}\) divided by 9

Since the exponent is even, we need to find the remainder of \(4^{32^{32}}\)

We need to figure out \(4^{32^{32}}\) divided by 9
We know that 64 is 1 more than a multiple of 9 and that \(4^3 = 64\)

But how many 3s do we have in \(32^{32}\)?
\(32^{32} = (33 - 1)^{32}\) so when we divide \(32^{32}\) by 3, we get a remainder of 1.

\(4^{32^{32}} = 4^{3x+1} = 4*4^{3x} = 4*64^x = 4*(63+1)^x\)

When this is divided by 9, remainder is 4.
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What is the remainder when 32^32^32 is divided by 7? 29 Apr 2020, 08:40
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gurpreetsingh
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
Show more


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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What is the remainder when 32^32^32 is divided by 7? 11 May 2020, 08:54
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Bunuel
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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.
Show more

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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What is the remainder when 32^32^32 is divided by 7? 11 May 2020, 09:48
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Bunuel
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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
Show more

\(4^{2^{160}}=(2^2)^{2^{160}}= 2^{2*2^{160}}=2^{2^{160+1}}=2^{2^{161}}\)
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What is the remainder when 32^32^32 is divided by 7? 24 Jul 2020, 00:08
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I broke down the question to 2^30/7
Now if we directly use the cyclicity of 2, we get 4 as the remainder.
But if i break it in terms of 8 ie 2 ^3 like (2^3)^10/7, we get remainder as 1. How would we know which is right and which is wrong?
Bunuel chetan2u VeritasKarishma
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What is the remainder when 32^32^32 is divided by 7? 26 Jul 2020, 22:04
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TaizKaran
I broke down the question to 2^30/7
Now if we directly use the cyclicity of 2, we get 4 as the remainder.
But if i break it in terms of 8 ie 2 ^3 like (2^3)^10/7, we get remainder as 1. How would we know which is right and which is wrong?
Bunuel chetan2u VeritasKarishma
Show more

TaizKaran:

Units digit cyclicity DOES NOT help you with remainders except in cases when the divisor is 2/5/10.

Note that
26/7 Remainder = 5,
36/7 Remainder = 1,
46/7 Remainder = 4
So even though they all have same units digit of 6, the remainders are different in each case.

You need to use binomial as done by you here "(2^3)^10/7, we get remainder as 1"

See:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/1 ... questions/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/1 ... ns-part-2/
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What is the remainder when 32^32^32 is divided by 7? 23 Aug 2020, 21:27
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Hi Bunuel chetan2u,

I have a question with respect to the simplification of \((4)^{2^1^6^0^}\). I understand how you got \((2)^{2^1^6^1}\).
However, I don't get the same result when I simplify \((4)^{2^1^6^0^}\) as follows: \((2 * 2)^{2^1^6^0}\) => \((2)^{2^1^6^0}\) * \((2)^{2^1^6^0}\) => \((2)^{2^(^1^6^0+^1^6^0^)}\)

Please can you help me understand my mistake?

Thanks a lot
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What is the remainder when 32^32^32 is divided by 7? 23 Aug 2020, 23:30
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VIVA1060
Hi Bunuel chetan2u,

I have a question with respect to the simplification of \((4)^{2^1^6^0^}\). I understand how you got \((2)^{2^1^6^1}\).
However, I don't get the same result when I simplify \((4)^{2^1^6^0^}\) as follows: \((2 * 2)^{2^1^6^0}\) => \((2)^{2^1^6^0}\) * \((2)^{2^1^6^0}\) => \((2)^{2^(^1^6^0+^1^6^0^)}\)

Please can you help me understand my mistake?

Thanks a lot
Show more


\((2)^{2^{160}}*(2)^{2^{160}}=2^{(2^{160}+2^{160})}=(2)^{(2*2^{160})}=2^{2^{161}}\)
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What is the remainder when 32^32^32 is divided by 7? 06 Oct 2020, 02:24
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\(32 ^{32}\) is always even.

and 32 divided by 7 will give remainder as 4.

\(4^{even power}\) , remainder will be 4.

Answer B
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What is the remainder when 32^32^32 is divided by 7? 03 Sep 2024, 08:21
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Bunuel

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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 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: https://gmatclub.com/forum/remainder-99 ... ms#p768816

Hope it's clear.
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­Hello Bunuel, could you explain how \( 4^{2^{160}}=2^{2^{161}} \) this happens?
Many thanks!
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What is the remainder when 32^32^32 is divided by 7? 03 Sep 2024, 08:36
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lnyngayan


­Hello Bunuel, could you explain how \( 4^{2^{160}}=2^{2^{161}} \) this happens?
Many thanks!
Show more
­
\(4^{2^{160}}=(2^2)^{2^{160}}= 2^{2*2^{160}}=2^{2^{160+1}}=2^{2^{161}}\)­
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What is the remainder when 32^32^32 is divided by 7? 17 Sep 2024, 07:32
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We need to find the remainder when \(32^{32^{32}}\) is divided by 7?

\(32^{32^{32}}\)= \(32^{2^{5*32}}\) = \(32^{2^{160}}\)

We solve these problems by using Binomial Theorem, where we split the number into two parts, one part is a multiple of the divisor(7) and a big number, other part is a small number.

=> \(32^{2^{160}}\) = \((28+4)^{2^{160}}\)

Watch this video to MASTER BINOMIAL Theorem

Now, if we use Binomial Theorem to expand this then all the terms except the last term will be a multiple of 7
=> All terms except the last term will give remainder of 0 when divided by 7
=> Problem is reduced to what is the remainder when the last term (i.e. 2^160 C 2^160 * 28^0 * \(4^{2^{160}}\)) is divided by 7
=> Remainder of \(4^{2^{160}}\) is divided by 7
=> Remainder of \(2^{2^1 * 2^{160}}\) is divided by 7
=> Remainder of \(2^{2^{161}}\) is divided by 7

To solve this problem we need to find the cycle of remainder of power of 2 when divided by 7

Remainder of \(2^1\) (=2) by 7 = 2
Remainder of \(2^2\) (=4) by 7 = 4
Remainder of \(2^3\) (=8) by 7 = 1
Remainder of \(2^4\) (=16) by 7 = 2
Remainder of \(2^5\) (=32) by 7 = 4
Remainder of \(2^6\) (=64) by 7 = 1

=> Cycle is 3

We need to find Remainder of \(2^{161}\) by 3

Remainder of \(2^1\) (=2) by 3 = 2
Remainder of \(2^2\) (=4) by 3 = 1
Remainder of \(2^3\) (=8) by 3 = 2
Remainder of \(2^4\) (=16) by 3 = 1

=> Cycle is 2

=> Remainder of \(2^{161}\) by 3 = Remainder of \(2^{1}\) by 3 = 2
=> Remainder of \(2^{2^{161}}\) is divided by 7 = Remainder of 2^2 by 7 = 4

So, Answer will be B
Hope it Helps!

Watch following video to MASTER Remainders by 2, 3, 5, 9, 10 and Binomial Theorem

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What is the remainder when 32^32^32 is divided by 7? 12 Jul 2025, 05:43
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Honestly, This question wasn't difficult at all IMHO. It should not be in "Out of Scope- Too Hard" section of Gmatclub.
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What is the remainder when 32^32^32 is divided by 7? 12 Jul 2025, 11:11
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What is the remainder when \(32^{32^{32}}\) is divided by 7?

The remainder when \(32^{32^{32}}\) is divided by 7
= The remainder when \((-3)^{32^{32}}\) is divided by 7
= The remainder when \(3^{32^{32}}\) is divided by 7
= The remainder when \({3^{3*10+2}}^{32}\) is divided by 7
= The remainder when \(3^{2*32}\) is divided by 7
= The remainder when \(3^{3*21+1}\) is divided by 7
= The remainder when \((-1)^{21}*3\) is divided by 7
= The remainder when -3 is divided by 7
=7-3 = 4

IMO B
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