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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.

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.

What is the remainder when \(32^{32^{32}}\) is divided by 7?

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

Many of you got the correct answer but wrong explanation and it is the biggest losing point of the learning as the same question is never going to come.

trueblue nailed it with correct explanation.

Here is my explanation and I would request the Quant guru of Gmat Club - Bunnel to correct and improve my explanation for the solutions.

\(32^{32^{32}}\) looks quite daunting, but remember one thing on exams like GMAT you will always get tricky and convoluted questions. But they can solved by simple Quant basics.

Always reduce your question.... Rof means remainder of

Rof \(32^{32^{32}}\) when divided by 7 = Rof \(4^{32^{32}}\) by 7 as 28+4 = 32

To calculate Rof \(4^{32^{32}}\) , we need to understand how to reduce it further.

Whenever we are solving remainder questions we always reduce it to the minimum value. To reduce the powers of 4 we need to find \(4^x\) such that when \(4^x\) is divided by 7 the remainder is either 1 or -1

If remainder of Rof \(4^{x}\) =1 , then we can reduce the Rof \(4^{y}\) multiple times until y>x.

Since Rof \(4^{3}\) when divided by 7 is 1, if we can reduce \(4^{32^{32}}\) to the form of \(4^{3k+r}\) we can easily eliminate redundant powers of 4.

To represent \(4^{32^{32}}\) as \(4^{3k+r}\) we need to represent \(32^{32}\) in the form of \(3k+r\).

Now to represent \(32^32\) in the form of \(3k+r\), we have to find the remainder when \(32^{32}\) is divided by 3. That will give the value of r.

Rof \(32^{32}\) when divided by 3 = Rof \(2^{32}\) = Rof \(2^{4*{8}}\)

Since Rof \(2^4\)when divided by 3 = 1, => Rof \(2^{4*{8}}\) = 1

Hence r = 1 => \(32^{32}\) = 3k+1 Now coming back to the main question.

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.

Is this even a GMAT Question...?????? because if it is I need to start worrying about Quants as well....

If this question is asked in the exam, it will be at 750+ level. So do not worry even if you are not able to solve this. Sometimes we learn concepts by understanding the explanation of difficult problem and use them while solving similar questions.

Give it a try and do not lose your confidence. When I will post the explanation do read and understand it carefully. _________________

I am using Euler's method, search on wikipedia if you need proof, else try to follow the steps :

HCF(32,7) = 1 "phi" 7 = 6 (it is the number of positive integers less than 7 and prime to 7.. In fact for any prime number "n", it will be "n-1").

=> 32^6 mod 7 = 1 (mod is same as "%")

(To make sure you understand it, please try for any number n!=7, n^6 mod 7 = 1)

So, we now need to express, 32^32 = 6x+k

i.e. 32^32 % 6 = ?

To make it easier, lets try to find out 16^32 % 3 and multiply the remainder by 4 (since 32 and 6 has a common factor 2, and also it is easier/helpful to get a remainder divided by a prime number)

Apply the same approach as shown above : HCF(16,3) = 1 "phi" 3 = 2

=> 16^2 mod 3 = 1 => 16^32 mod 3 = 1 => 32^32 mod 6 = 4*1 = 4

So, 32^32 mod 6 = 6y+4

Therefore; 32^32^32 mod 7 = 32^(6y+4) mod 7 = 32^4 mod 7 = (28+4)^4 mod 7 = 4^4 mod 7 = 4

(Hopefully, I didn't make any typo.... Let me know if there is any problem with understanding this)

Thanks

Last edited by anshumishra on 03 Sep 2010, 10:18, edited 1 time in total.

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!

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. _________________

Just quote my question and see how I have used the mathematical operation for the powers. This way you can represent them better as your solution is quite difficult to understand. _________________

Just quote my question and see how I have used the mathematical operation for the powers. This way you can represent them better as your solution is quite difficult to understand.

Got it. Here is the updated post.

\(32^{{32}^{32}\) -- \({{2^5}^{{2^5}^{2^5}}}\)

\({{2^5}^{2^5}}\) is \({2^{160}}\).

Again \({{2^{160}}^{32}}\) is \({2^{5120}}\)

(2^x)/7 has a cyclicity or repeatability of 3. That is \({2^1}/7 - 2\), \({2^2}/7- 4\), \({2^3}/7 - 1\) .....

Hence \({2^{5120}}/7\) boils down to the same as \({2^{2}}/7\) which is 4.

Answer is 4 (B). _________________

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Just quote my question and see how I have used the mathematical operation for the powers. This way you can represent them better as your solution is quite difficult to understand.

Got it. Here is the updated post.

\(32^{{32}^{32}\) -- \({{2^5}^{{2^5}^{2^5}}}\)

\({{2^5}^{2^5}}\) is \({2^{160}}\).

Again \({{2^{160}}^{32}}\) is \({2^{5120}}\)

(2^x)/7 has a cyclicity or repeatability of 3. That is \({2^1}/7 - 2\), \({2^2}/7- 4\), \({2^3}/7 - 1\) .....

Hence \({2^{5120}}/7\) boils down to the same as \({2^{2}}/7\) which is 4.

Answer is 4 (B).

Take my advice when ever you solve your question always check the scope and domain.

do you think \(32^{32^{32}}\) can be equal to \({2^{5120}}\) ?

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

as per your calculation

\(2^{5*{32^{32}}}\) should be equal to \({2^{5120}}\) => \(5*{32^{32}}\) should be equal to 5120. ? This is wrong.

Just cross check always whether you are making right moves or not. I hope you know your mistake now? _________________

Just quote my question and see how I have used the mathematical operation for the powers. This way you can represent them better as your solution is quite difficult to understand.

Got it. Here is the updated post.

\(32^{{32}^{32}\) -- \({{2^5}^{{2^5}^{2^5}}}\)

\({{2^5}^{2^5}}\) is \({2^{160}}\).

Again \({{2^{160}}^{32}}\) is \({2^{5120}}\)

(2^x)/7 has a cyclicity or repeatability of 3. That is \({2^1}/7 - 2\), \({2^2}/7- 4\), \({2^3}/7 - 1\) .....

Hence \({2^{5120}}/7\) boils down to the same as \({2^{2}}/7\) which is 4.

Answer is 4 (B).

Take my advice when ever you solve your question always check the scope and domain.

do you think \(32^{32^{32}}\) can be equal to \({2^{5120}}\) ?

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

as per your calculation

\(2^{5*{32^{32}}}\) should be equal to \({2^{5120}}\) => \(5*{32^{32}}\) should be equal to 5120. ? This is wrong.

Just cross check always whether you are making right moves or not. I hope you know your mistake now?

Not sure if I got what you mentioned, but upon analyzing my approach I found the error.

\({{2^5}^{2^5}} is {2^{160}}\). -- This is correct.

[highlight]Again \({{2^{160}}^{32}} is {2^{5120}}\)[/highlight] -- This is the incorrect part.

The above line should have been \({32^{2^{160}}}\).

Now if I understand correctly, this was the only error in my approach. Further proceeding to solve the question....

{32^1}/7 - 4, {32^2}/7 - 2, {32^3}/7 - 1, {32^4}/7 - 4 .... hence the repeatability is 3.

Now \(32^{2^{160}}\) should be the same as the power of 32 which is \({2^{160}}\) divided by 3.

\({2^{160}}\) divided by 3 is the same as \({2^{1}}\) divided by 3 -- which is 2.

Now combining them \(32^{2^{160}}/7\) is the same as \({32^{2^1}}/7\)

which is 2. Final answer should be 2 (C).

Hope I have answered correctly this time. I will await the final solution and the explanation. _________________

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anshumishra : Its good if you know Euler theorem. The theorem is Fermat's little theorem that uses Euler theorem.

Guys you do not have to learn all these theorems for Gmat, if you know then its good if not then also you can solve this question using basics of remainders.Do not panic.

When 32^32 is divided by 6 the remainder is 4 not 2. Please check your solution.

32^32 when divided by 6 gives remainder same as when 2^32 is divided by 6

=> 2^32 mod 6 = \((2^{5*{6}} )* (2^2)\) mod 6 = 32^6 * 4 mod 6

= 2^6 *4 mod 6 = 2^5 * 2^3 mod 6 = 2*2 = 4 _________________

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