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
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How you got 2^800? read the question carefully.

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.
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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?
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Last two digits of 32^32 is 76 not 4.

Also last digit of 32^4 is 6 and it is not same as 32^1
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I got B

32^{32^{32}} can be reduced as {32^32} ^ 32 = 1024 ^ 32

and 1024 = 2^10

= (2)^10*32 -> 2^320

R[ (2^x)/7 ] is cyclical,2^320/7 is same as 2^2/7 and the answer is 4!

correct me if am wrong

Check this : tough-remainder-question-100316.html#p773920

You are making the same mistake. Also 1024 - 32^2 not 32^32.
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32^32^32 % 7 = ?

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

Thanks Gurpreet, I have edited my post where i missed to multiply by 4 instead of 2.
I have made "4" as bold 4.

Thanks

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

Reason : Rof 4^{xy} = Rof 4^x * 4^y = Rof 4^{x} * Rof 4^{y}

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

Hence r = 1 => 32^{32} = 3k+1

Now coming back to the main question.

Rof 4^{32^{32}} = Rof 4^{3k+1} = Rof 4^{3k} * Rof 4^1 = 1*4 = 4

Hence B.

Whenever you see such question, always apply the above rule.
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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.
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