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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.

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]

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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.

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!!

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 \(3^{3} = 27\)

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!

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 \(3^{3} = 27\)

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.
_________________

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.

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.

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. _________________

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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