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: PS-What is the remainder [#permalink]
24 Jan 2012, 11:31
2
This post received KUDOS
omerrauf wrote:
The answer should be C i.e. 2.
Here is how.
The expression is : \(3^{7^{11}}\)
Lets first resolve \(7^{11}\)
So \(7^2=49\)
So \(7^4=.....1\) where 1 is the last digit of the number
So \(7^8=.....1\) where 1 is the last digit of the number
So \(7^3=.....3\) where 3 is the last digit of the number
So \(7^{11}=7^3*7^8=........1*3=.............3\) where 3 is the last digit of the number
Now on to \(3^{7^{11}}=3^{......3}\) where 3 is the last digit of the exponent
We know the power is odd and its 3. Lets check out the last digit of some of the odd exponents for \(3\)
So \(3^1=3\) So \(3^3=27\) & \(3^5=....3\) & \(3^7=.....7\) & \(3^{11}=.........7\)
Recognize the patterns. It could be 3 or 7. Also, it is is 3 only in the case when the exponent is \(1\) or a multiple of \(5\)
We know the power is greater than \(1\) and not a multiple of \(5\) so the only possibility for the last digit is \(7\)
Now we know that \(\frac{7}{5}\) remainder is \(2\) Hence the answer must be C.
Well the answer posted above is correct, but the reasoning provided above is a little bit flawed. If you look at \(3^{13}\) it has '3' as the units digit in exponent and if you try to solve it the answer will have 3 as the units digit, which when divided by 5 will give 3 as the remainder.
Here's another approach which I believe you can use Understand that in this question all you need to find out is the units digit of the expression \(3^{7^{11}}\). In order to do so, you must reduce this term in the form of \(3^x\). \(3^x\) has a cyclicity of 4, i.e. the units digit of \(3^x\) repeats itself after four terms \(3^1=3\) --- so --- \(3^{4n+1}=...3\)------------------equation (1) \(3^2=9\) --- so --- \(3^{4n+2}=...9\)------------------equation (2) \(3^3=27\) --- so --- \(3^{4n+3}=...7\)------------------equation (3) \(3^4=81\) --- so --- \(3^{4n+4}=...1\)------------------equation (4) \(3^5=243\) --- so --- \(3^5=3^{4+1}=3^{4n+1}\) So we can write any power of 3 in the form of \(3^{4n+k}\). This way calculating the value of 'k', we can easily find the units digit. Here also we just need to write the power of 3 in 4n+k form. Lets concentrate on \(7^{11}\) If we divide this by 4, whatever we get will be the value of 'k' and our problem would be solved. Rewriting it as \((8-1)^{11}\) Divide \((8-1)^{11}\) by 4 to get the value of k. Here 8 will give the remainder 0 when divided by 4 and the only remainder we will get is from -1. Using the concept of negative remainders(which I'm assuming you know, incase you don't, feel free to ping me and I'll be happy to tell you) we'll get 3 as our final remainder Hence \(7^{11}\) can be written as 4n+3. So we can write our given expression \(3^{7^{11}}\) as \(3^{4n+3}\). Using equation (3) above, we can easily make out that our units digit will be 7. Dividing this by 5 will give 2 as the remainder.
PS: Forgive me for my poor formatting. I'm still learning
Re: PS-What is the remainder [#permalink]
24 Jan 2012, 12:15
1
This post received KUDOS
Expert's post
4
This post was BOOKMARKED
LM wrote:
What is the remainder of (3^7^11)/5
A. 0 B. 1 C. 2 D. 3 E. 4
Please explain what should be the approach in such kind of questions?
Also, someone please tell how I could use the formatting to display 3 raised to power 7 raised to power 11, divided by 5?
First of all I think that this question is a little bit out of the scope of the GMAT. But anyway:
The last digit of 3 in positive integer power repeats in pattern of 4: {3, 9, 7, 1}. So, basically we should find the remainder upon division 7^(11) by cyclicity of 4 (to see on which number in this pattern \(7^{11}\) falls on). \(7^{11}=(4+3)^{11}\), now if we expand this expression all terms but the last one will have 4 in them, thus will leave no remainder upon division by 4, the last term will be \(3^{11}\). Thus the question becomes: what is the remainder upon division \(3^{11}\) by 4: 3 divided by 4 yields remainder of 3; 3^2=9 divided by 4 yields remainder of 1; 3^3=27 divided by 4 yields remainder of 3; 3^4=81 divided by 4 yields remainder of 1.
So, 3 in odd power yields remainder of 1 upon division by 4 --> \(3^{11}\) yields remainder of 3 --> finally, we have that \(3^{7^{11}}\) will have the same last digit as \(3^3\), which is 7. Thus as \(3^{7^{11}}\) has the last digit of 7 then divided by 5 it will yield remainder of 2.
Answer: C.
LM wrote:
Also, someone please tell how I could use the formatting to display 3 raised to power 7 raised to power 11, divided by 5?
Mark \frac{3^{7^{11}}}{5} and press (m) button: \(\frac{3^{7^{11}}}{5}\) _________________
Re: PS-What is the remainder [#permalink]
26 Jan 2012, 00:41
1
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
LM wrote:
Helps but can you give some link or details about concept of negative remainders.
Here are some links that discuss divisibility and remainders. The third link discusses negative remainders but I think it would make more sense if you first go through the first two links.
Re: PS-What is the remainder [#permalink]
26 Jan 2012, 03:52
Expert's post
LM wrote:
Helps but can you give some link or details about concept of negative remainders.
If you are preparing for the GMAT you probably shouldn't waste you valuable time on the out of the scope questions like this or on the concepts that aren't tested. _________________
Re: What is the remainder of (3^7^11)/5 [#permalink]
26 Jan 2012, 21:11
Expert's post
Such questions are definitely not GMAT's style but the concept of negative remainders is interesting and useful in certain situations. Besides its good to understand it as a part of the theory of divisibility and remainders. It is the complementary concept of positive remainders. So go ahead and check out the posts. They are all GMAT relevant. _________________
Re: What is the remainder of (3^7^11)/5 [#permalink]
26 Jan 2012, 23:46
Expert's post
VeritasPrepKarishma wrote:
Such questions are definitely not GMAT's style but the concept of negative remainders is interesting and useful in certain situations. Besides its good to understand it as a part of the theory of divisibility and remainders. It is the complementary concept of positive remainders. So go ahead and check out the posts. They are all GMAT relevant.
I agree with Karishma that the concept of negative remainders is interesting and useful in certain situations. Having said that I'd like to point out two issues:
1. Every divisibility/remainder question on the GMAT can be (easily) solved without this concept;
2. General/common definition of a remainder is that it's more than or equal to zero and less than divisor. Reffer to OG12:
If x and y are positive integers, there exist unique integers q and r, called the quotient and remainder, respectively, such that y = xq + r and 0 <= r < x.
Therefore, if time is an issue in your preparation, you should probably skip this concept (even though it's not hard at all) and concentrate more on orthodox approaches. _________________
Re: What is the remainder of (3^7^11)/5 [#permalink]
25 Mar 2012, 20:22
Expert's post
1
This post was BOOKMARKED
Cmplkj123 wrote:
I solved this 1 as follows:
3^7^11/5=???
7^11= 7^3 * 7^3* 7^3*7^2= 343*343*343*49
Last digit of above will be 3*3*3*9 = 3
eqn reduces to 3^3/5= 27/5=2
Guys , havse solved the questn with unit digit method.
Please correct me if i m wrong.
If i m rite, i deserve the kudo !!!
Actually, the logic is not entirely correct. \(3^{ab...3}\) i.e. 3 to a power that ends in 3 will not necessarily give you a remainder of 2 when divided by 5. e.g. \(3^{13}\) when divided by 5 gives 3 as the remainder. \(3^{33}\) when divided by 5 gives 3 as the remainder. etc
You need to find the unit's digit of \(3^x\) where \(x = 7^{11}\). Since 3 has a cyclicity of 4, you need to figure out the remainder when x is divided by 4. \(x = 7^{11} = (8-1)^{11}\) so remainder will be -1 i.e. 3 (for explanation of this, check out: http://www.veritasprep.com/blog/2011/05 ... ek-in-you/
So basically we have \(3^{4m + 3}\). Since 3 has a cyclicity of 4 {3, 9, 7, 1} , the unit's digit here will be 7. When you divide this by 5, the remainder will be 2. _________________
Re: PS-What is the remainder [#permalink]
23 Sep 2013, 11:11
VeritasPrepKarishma wrote:
LM wrote:
Helps but can you give some link or details about concept of negative remainders.
Here are some links that discuss divisibility and remainders. The third link discusses negative remainders but I think it would make more sense if you first go through the first two links.
Re: PS-What is the remainder [#permalink]
23 Sep 2013, 20:09
Expert's post
email2vm wrote:
VeritasPrepKarishma wrote:
LM wrote:
Helps but can you give some link or details about concept of negative remainders.
Here are some links that discuss divisibility and remainders. The third link discusses negative remainders but I think it would make more sense if you first go through the first two links.
Re: PS-What is the remainder [#permalink]
24 Sep 2013, 06:30
Expert's post
email2vm wrote:
VeritasPrepKarishma wrote:
LM wrote:
Helps but can you give some link or details about concept of negative remainders.
Here are some links that discuss divisibility and remainders. The third link discusses negative remainders but I think it would make more sense if you first go through the first two links.
Re: PS-What is the remainder [#permalink]
24 Sep 2013, 13:10
1
This post received KUDOS
Expert's post
email2vm wrote:
Hi Karishma,
Here is what I did...but couldn't conclude to the correct answer. Please check the below procedure
a^b^c= a^bc so 3^7^11=>3^77
(5-2)^77/5
all I have to worry is about (-2)^77
now for every 2^4 i have remainder 1 so finally i have (-2)/5 (since 77/4 remainder is 1)
....how to solve further. Please assist.
Let me try to explain......................
What is the remainder of \(\frac{3^{7^{11}}}{5}\)
Remainder of \(\frac{3^1}{5}\) is 3
Remainder of \(\frac{3^2}{5}\) is 4
Remainder of \(\frac{3^3}{5}\) is 2
Remainder of \(\frac{3^4}{5}\) is 1
Remainder of \(\frac{3^5}{5}\) is 3
Here we should recognize the cyclic pattern of remainders. as the power increases remainder continues to move on in a pattern 3421 3421 3421.... so on.
Now if.......
\(7^{11}\) is completely divisible by 4, then the pattern will stop on 1
if \(7^{11}\) is divisible by 4 with remainder 1, then the pattern will stop on 3
if \(7^{11}\) is divisible by 4 with remainder 2, then the pattern will stop on 4
And if \(7^{11}\) is divisible by 4 with remainder 3, then the pattern will stop on 2
So we basically have to find the remainder when \(7^{11}\) divided by 4
Rule :- The expression \(\frac{A * B * C}{M}\) will give the same remainder as \(\frac{Ar * Br * Cr}{M}\) where Ar, Br, Cr are the remainders of A, B, C when divided by 'M' individually.
\(7^{11}\) can be simplified as 49*49*49*49*49*7
Remainder of \(\frac{49*49*49*49*49*7}{4}\) will be the same as that of \(\frac{1*1*1*1*1*7}{4}\) or that of \(\frac{7}{4}\)
Remainder of \(\frac{7}{4}\) is 3
Since \(7^{11}\) divisible by 4 with remainder 3, the pattern will stop on 2 and thus the remainder of\(\frac{3^{7^{11}}}{5}\) will be 2
Re: What is the remainder of (3^7^11)/5 [#permalink]
09 Mar 2014, 19:24
hi guys ,
I used wilson reminder therom and found the answer ,
so 3^4/5=1 now we have to make 7^11 in the format of 4x+something so we divide 7^11 by 4 or 3^11 by 4 which will give me reminder of 3 hence 7^11 can be written as 4x+3 ie (3^4x+3)/5 so 3^4x/5 will be 1 3^3 /5 will give reminder 2 ie our answer
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Perhaps known best for its men’s basketball team – winners of five national championships, including last year’s – Duke University is also home to an elite full-time MBA...
Hilary Term has only started and we can feel the heat already. The two weeks have been packed with activities and submissions, giving a peek into what will follow...