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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]  30 Aug 2013, 23:13
Rem(32^32^32)=Rem(4^32)^32
now 4^3 = 64 = 63+1. hence when 64 to the power anything is divided by 7, the remainder will always be 1
so, Rem(32^32)/3 = Rem(33-1)^32/3 = 1.
Hence Rem(4*64^k)/7 = 4
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Re: Tough remainder question [#permalink]  02 Sep 2013, 00:29
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

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.

Similar problem: remainder-99724.html?hilit=expand%20this,%20all%20terms#p768816

Hope it's clear.

Hello Bunuel,

Could you please explain the portion highlighted in red.
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Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]  31 Jan 2014, 19:15
gurpreetsingh wrote:
What is the remainder when $$32^{32^{32}}$$ is divided by 7?

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

Check the solution here : tough-remainder-question-100316.html#p774893

Consider the following:

when (32)^1, unit digit =2.
when (32)^2, unit digit = 4.
when (32)^3, unit digit = 8.
when (32)^4, unit digit = 6.
when (32)^5, unit digit = 2.

Hence (32)^x, where x is an integer has a cyclicity of 4.

=> (32)^32 will have 2 as unit digit; this is because (32/4) = 8. Hence, original expression becomes:

{(32)^2}/7 = ?

From above, when (32)^2, unit digit = 4.

Expression becomes 4/7 which has a remainder of 4.

Took me 1:15 mins to solve.
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Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]  04 Feb 2014, 21:40
Is there a post or resource about remainder cyclicities in general ? I know already about the units cyclicity but i am hoping for one about remainder cyclicities .
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Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]  05 Feb 2014, 00:09
gurpreetsingh wrote:
What is the remainder when $$32^{32^{32}}$$ is divided by 7?

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

Check the solution here : tough-remainder-question-100316.html#p774893

2^5(32)(32)

2^1/7 = Remainder = 2
2^2 /7 = Remainder = 4
2 ^3 /7 = Remainder= 1
16/7 = Remainder = 2
32/7 = Remainder 4
the process continues

When we divide the power 5(32)(32) by three we get: + 2 remainder hence the remainder will be the second one in the series which is 4.
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Re: Tough remainder question [#permalink]  04 Apr 2014, 12:03
ramana 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

I got B

$$32^{32^{32}}$$ can be reduced as {32^32} ^ 32 = 1024 ^ 32

and 1024 = 2^10

= (2)^10*32 -> 2^320

R[ (2^x)/7 ] is cyclical,2^320/7 is same as 2^2/7 and the answer is 4!

correct me if am wrong

$$32^{32^{32}} = 32^{1.461501637331 *10^{48}}$$

You do top down when there is no parentheses and so you do 32^32 and than you take 32 to the power of what you got for 32^32
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Re: Tough remainder question [#permalink]  04 Apr 2014, 23:27
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

Is the remainder 5?

9 x 3 + 5 =32
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Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]  30 Apr 2014, 05:01
Found this in .30 sec, hope I'm right though.

I (In the last part - 32^32^32) find the cyclicity of 32 (or, since it ends with a 2, find the cyclicity of 2)
2...4...8...6 .

II We can see that 32 is evenly divisible by 4 so the units digit of 32^32^32 is something with a 2.

III Squaring 32 (32^2) ends with a units digit of 4

IV dividing 4 by 7 leaves a remainder of 4.

Hence, B.

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Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]  08 May 2014, 07:02
is this a GMAT type question?
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Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]  08 May 2014, 07:14
Expert's post
satsymbol wrote:
is this a GMAT type question?

Concepts tested are relevant for the GMAT, though question itself is harder than one can expect on the real test.
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Re: Tough remainder question [#permalink]  11 Jun 2014, 21:54
WoundedTiger 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

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.

Similar problem: remainder-99724.html?hilit=expand%20this,%20all%20terms#p768816

Hope it's clear.

Hello Bunuel,

Could you please explain the portion highlighted in red.

Hi Bunuel,

Indeed that part in red is a bit confusing...
Shouldn't we get at the end 2^2^2?

Can you please go over that part in a little more detail?
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Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]  12 Jun 2014, 14:12
B.
Boils down to 2^8/7 R=4
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Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]  12 Jun 2014, 21:28
Brunel,

Can we solve it by expanding 32 as (35-3) ?

Here is how I approached :
32^32^32= (35-3) ^32^32

All other terms are divisible by 7 , barring last term 3^32^32.
Here is the cyclicity of

3^1 when divided by 7 gives a remainder of 3
3^2 when divided by 7 gives a remainder of 2
3^3 when divided by 7 gives a remainder of 6
3^4 when divided by 7 gives a remainder of 4
3^5 when divided by 7 gives a remainder of 5
3^6 when divided by 7 gives a remainder of 1

3^7 when divided by 7 gives a remainder of 3
3^8 when divided by 7 gives a remainder of 2
3^9 when divided by 7 gives a remainder of 6
3^10 when divided by 7 gives a remainder of 4
3^11 when divided by 7 gives a remainder of 5
3^12 when divided by 7 gives a remainder of 1
..........

The power of 3 i.e 32^32 is equivalent to 2^160 which will be equivalent to remainder when 2^160 is divided 6

2^160 /6 = 2^159 /3 = (3-1)^159 / 3

Now, all terms in (3-1)^159 will be divisible by 3, except for the last term (-1)^159 => -1 => (-1) * 3 +2 . Therefore, the remainder is 2. This, needs to be multiplied by 2, since we had reduced 2^160/6 to 2^159/3. Therefore the final remainder is 4

Therefore, the final value comes out to be 3^4 divided 7, that gives 4 as a remainder.

Is this approach correct ?
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Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]  14 Nov 2014, 05:26
gurpreetsingh wrote:
What is the remainder when $$32^{32^{32}}$$ is divided by 7?

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

Check the solution here : tough-remainder-question-100316.html#p774893

I got it.
Start solving from the top-most power of 32.
Just imagine now 2^32 instead of 32^32.
So as per cycle, 2^4 = 6.
So here 2^32 = 6 as a unit digit.
Now,
32^6 means 2^6, which means 2^2 = 4.
So the remainder is 4.
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Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]  14 Nov 2014, 22:24
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

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.

Similar problem: remainder-99724.html?hilit=expand%20this,%20all%20terms#p768816

Hope it's clear.

Bunuel is seriously a stud.
Re: What is the remainder when 32^32^32 is divided by 7?   [#permalink] 14 Nov 2014, 22:24

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