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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 12 Jun 2014, 22:28
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]

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New post 14 Nov 2014, 06:26
gurpreetsingh wrote:
What is the remainder when \(32^{32^{32}}\) is divided by 7?

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

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

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]

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New post 19 Jul 2015, 12:15
trueblue 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.


Is the answer B?

Intuitively, i did it like this.

32^32^32 = (28+4)^32^32
As 28 is divisible by 7, we dont need to worry about that part. Hence for the purpose of remainder,
our equation boils down to 4^32^32

The cyclicity of 4 is 3 when divided by 7, hence we need to think about the value of 32^32 and what remainder it leaves when divided by 3.

Considering 32^32, it can be broken into (30+2)^32. Again 30^32 is divisible by 3. Hence we need to focus on 2^32.
2^32 can be written as (2*2)^31 = (3+1)^31. As 3^31 is also divisible by 3, we will be left with 1^31.
Thus 1 would be the remainder when 32^32 is divided by 3.

This implies that 4 will be the remainder when divided by 7.
Hence Answer is B.

Do let me know if i am wrong in my thinking.

Thanks.


Hello
I think 2^32 = (2^2)^16 =(4)^16 =(3+1)^16

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

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New post 13 Dec 2015, 07:36
32 can be written as 4k+4
k being multiple of 7
32^4k+4^4k+4
=32^4^4
=32^64
=32^9k+1
=32
32 div by 7 gives rem 4
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Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]

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New post 28 Jan 2016, 15:30
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VeritasPrepKarishma

Thank you, Karishma. Your explanations always amplify my understanding.
Quick question for you - in the last part, i.e.

4*(63+1)^x
Gives R4 when divided by 7.

Can you help me understand the binomial part -
Binomial dictates that every term except for the last will be divisible by 7.
However, can you help me understand why we are not multiplying every term we get with 4 and then trying to find the remainder?

i.e. my working after binomial is: 4*1 / 7 = R4

Why should it not be 4*(term 1/7)*(term 2/7)...*1 / 7 = Remainder

Would really appreciate if you could clarify this portion of my understanding
Thank you!!

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

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VeritasPrepKarishma wrote:

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.


Hi Karishma,

Is it always better to avoid subtraction when using binomial theorem? Here is my approach, and I found that if there is a negative sign in the equation, things can get a little bit complicated.

\(32^{32^{32}}\) = \((35-3)^{32^{32}}\)

Now we only care about \((-3)^{32^{32}} = 3^{32^{32}}\)

We know that 27 is 1 less than a multiple of 1 and that \(3^{3} = 27\)

\(32^{32} = (33-1)^{32}\)
Hence \(3^{32^{32}} = 3^{3x+1} = 3*3^{3x} = 3*27^{x} = 3*(28-1)^x\)

Here we have to check whether x is even or not in order to decide the value of the remainder (whether it is 3 or 4).
One additional step: Since 32 is even, \(32^{32}\) is even. \(32^{32} = 3x+1\) therefore \(x\) is odd.

We now can conclude that r = -3. Therefore r = 4.

My question is: 'Is it always safer to avoid subtraction when it comes to binomial theorem?'
Thank you Karishma!

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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 Feb 2016, 13:43
32^32^32
I took first 32^32 . 32/4 cyclicity i took . Then i took 32^32 again . Similarly i got the final figure as 32/7 . The remainder is 4.

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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 Feb 2016, 22:31
truongynhi wrote:
VeritasPrepKarishma wrote:

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.


Hi Karishma,

Is it always better to avoid subtraction when using binomial theorem? Here is my approach, and I found that if there is a negative sign in the equation, things can get a little bit complicated.

\(32^{32^{32}}\) = \((35-3)^{32^{32}}\)

Now we only care about \((-3)^{32^{32}} = 3^{32^{32}}\)

We know that 27 is 1 less than a multiple of 1 and that \(3^{3} = 27\)

\(32^{32} = (33-1)^{32}\)
Hence \(3^{32^{32}} = 3^{3x+1} = 3*3^{3x} = 3*27^{x} = 3*(28-1)^x\)

Here we have to check whether x is even or not in order to decide the value of the remainder (whether it is 3 or 4).
One additional step: Since 32 is even, \(32^{32}\) is even. \(32^{32} = 3x+1\) therefore \(x\) is odd.

We now can conclude that r = -3. Therefore r = 4.

My question is: 'Is it always safer to avoid subtraction when it comes to binomial theorem?'
Thank you Karishma!


Yes, if given a choice, one should prefer addition - not only is it easier, it's less prone to error. Here, (28+4) and (35 - 3) were equivalent so I used (28+4). Obviously, in a case such as (35 - 1), one would use subtraction.
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Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]

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New post 16 Feb 2016, 07:25
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.

Answer: B.

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

Hope it's clear.


4^{{32}^{32}}=4^{{(2^5)}^{32}}=4^{2^{160}}=2^{2^{161}}

how does it change to 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 16 Feb 2016, 07:57
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nishantdoshi 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.

Answer: B.

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

Hope it's clear.


4^{{32}^{32}}=4^{{(2^5)}^{32}}=4^{2^{160}}=2^{2^{161}}

how does it change to 161?????


Hi,
I'll try to explain to you..
\(4^{2^{160}}= {2^{2*2^{160}}={2^{2^{160+1}}=2^{2^{161}}\)

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

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New post 16 Mar 2016, 07:38
MeghaP wrote:


4^32^32 can also be solved by binomial theorem which would result in the remainder one.

2^2^32^32 = 2^3^16^32 =
8^16^32 = (7+1)^16^32
and hence the remainder will be one.

I am not sure where I am going wrong (text in red above is not correct). Please suggest.!! :beat


You are not breaking down \(4^{{32}^{32}}\) correctly.

It should rather be,

\(4^{{32}^{32}} = (2^2)^{{32}^{32}} = (2^2)^{{(2^5)}^{32}} =(2^2)^{{2^{160}} = 2^{2^{161}}\)

Now, follow what Bunuel has done in his solution at what-is-the-remainder-when-32-32-32-is-divided-by-100316.html#p774902

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

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New post 17 Mar 2016, 00:11
Looked ...Re looked ...RE RE looked...
NO where to go
chetan2u you gotta help with this..
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Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]

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Chiragjordan wrote:
Looked ...Re looked ...RE RE looked...
NO where to go
chetan2u you gotta help with this..
Regards



Hi,

Before I explain you this Q, I will just explain the Binomial theorem--



\((a + b)^n = nC0 a^n + nC1 a^{n − 1}b + nC2 a^{n − 2}b^2 + nC3 a^{n − 3}b^3+.....nC(n-1)b^{n − 1}a + nC0 b^n\)
so if you look at this whatever be the power 'n' be, its all terms will be div by a, except one term nC0 b^n, same is the case with div by b..

EXAMPLE--
\((x+5)^4 = x^4 + 20x^3 + 150x^2 + 500x + 625\)
here all terms except 625 will be div by x and all terms except x^4 wil be div by 5..

lets use this to hep in finding remainders..



When \({{32}^{32}}^{32}\) is div by 7..
we know 8=7+1..
lets convert in this form..

\({{2^5}^{32}}^{32}\)=\({{(2^3*2^2)}^{32}}^{32}\)=\({{2^3}^{32}}^{32}*{{2^2}^{32}}^{32}\)..

now we have two terms
1)\({{2^3}^{32}}^{32}\)=\({{8}^{32}}^{32}={{7+1}^{32}}^{32}\)
so here all terms in the expansion will be div by 7 except the one with only 1s, that is \(nCn7^0*{1^{32}}^{32}\)
So remainder will be 1 for this..


2)\({{2^2}^{32}}^{32}\)..
similarly make the power inside as 2^3 instead of 2^2 and work ahead..
Or
4^1=4 it will leave the remainder of 4
4^2=16, this will leave 2
4^3=64, this will leave 1

so get above in this form
\({{2^2}^{32}}^{32}\).= \({{4}^{30}}^{32}\)*\({{4}^{2}}^{32}\).

again \({{4}^{30}}^{32}\) will leave remainder 1, and \({{4}^{2}}^{32}\) will leave 2^32 as remainder

\(2^{32}= 2^{30}*2^2\)..
\(2^{30} = 8^{10}\), so here too remainder is 1..
2^2 will leave 4 as remainder

ans 4

You may find this lengthy, since I have explained all terms for your understanding..
but If you get hang of it, it will be easier and faster..
Ofcourse there are always shortcuts as per each Q but we should know the standard method and WHY/ of each Q.

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

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ac556 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.

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.


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? [#permalink]

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New post 27 Nov 2016, 11:48
[quote="gurpreetsingh"]What is the remainder when \(32^{32^{32}}\) is divided by 7?

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

\(32 = 4 (mod 7)\)

(4^32^32)/7 (I wasn't able to make it proper looking composite power. Sorry.)

\(7\) is prime, \(GCF (4; 7) = 1\). So \(4^6 = 1 (mod 7)\)

\(\frac{32^{32}}{6} = \frac{(2^5)^{32}}{6} = \frac{2^{160}}{6} = \frac{2^{159}}{3}\)

\(2 = -1 (mod 3)\)

\(\frac{(-1)^{159}}{3} = \frac{-1}{3} = -1 (mod 3) = 2 (mod 3)\)

Multiplying by cancelled factor \(2\): = \(4 (mod 6)\)

\(\frac{1 * 4^4}{7} = \frac{2^8}{7}\)

\(2^3 = 1 (mod 7)\)

\(\frac{(2^3)^2*2^2}{7} = \frac{1*4}{7}\)

Our remainder is \(4\).

Answer B.

Kudos [?]: 388 [0], given: 40

What is the remainder when 32^32^32 is divided by 7?   [#permalink] 27 Nov 2016, 11:48

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

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