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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

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.

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 .

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. _________________

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

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.

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.

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.

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

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.

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