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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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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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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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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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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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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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11 Feb 2016, 09:00
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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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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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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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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.

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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16 Feb 2016, 07:57
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Expert's post
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.

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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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.!!

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

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17 Mar 2016, 02:07
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Expert's post
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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20 Oct 2016, 23:28
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Expert's post
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.

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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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 $$4$$.

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

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