Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]
30 Aug 2013, 23:13
Rem(32^32^32)=Rem(4^32)^32 now 4^3 = 64 = 63+1. hence when 64 to the power anything is divided by 7, the remainder will always be 1 so, Rem(32^32)/3 = Rem(33-1)^32/3 = 1. Hence Rem(4*64^k)/7 = 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.
Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]
04 Feb 2014, 21:40
Is there a post or resource about remainder cyclicities in general ? I know already about the units cyclicity but i am hoping for one about remainder cyclicities .
When we divide the power 5(32)(32) by three we get: + 2 remainder hence the remainder will be the second one in the series which is 4. _________________
Re: Tough remainder question [#permalink]
11 Jun 2014, 21:54
WoundedTiger 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.
Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]
12 Jun 2014, 21:28
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]
14 Nov 2014, 22:24
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.
Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]
09 Jul 2015, 11:39
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.
I am lost here 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.
I cannot figure out why the remainder of \(2^{161}\) equals that of \(2^{2^{161}}\) when the two values are divided by 7.
Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]
19 Jul 2015, 11:15
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!!
Re: What is the remainder when 32^32^32 is divided by 7? [#permalink]
11 Feb 2016, 06:56
Bunuel wrote:
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.
Hi Bunuel,
I fail to understand how \(4^{2^{160}}\) is equal to \(2^{2^{161}}\) Could you please shed some light on it?
Thank you!
gmatclubot
Re: What is the remainder when 32^32^32 is divided by 7?
[#permalink]
11 Feb 2016, 06:56
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...