What is the remainder of (3^7^11)/5

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.

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 3 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.



Mark \frac{3^{7^{11}}}{5} and press (m) button: \(\frac{3^{7^{11}}}{5}\)
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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

Hope this helps.

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.

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/04 ... unraveled/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/04 ... y-applied/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/05 ... emainders/
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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.
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Links given by Karishma should help you.

Are you preparing for CAT? Because I don't think GMAT tests you on such difficult questions.

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.
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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.
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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: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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.
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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.

The first step is incorrect: a^b^c is not equal to a^bc

\(3^{7^{11}}\) is not \(3^{77}\)

Just like 2^4 is not 2*4, \(7^{11}\) is not 77. \(7^{11}\) is much much greater than 77
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\((a^m)^n=a^{mn}\)

\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\) (if exponentiation is indicated by stacked symbols, the rule is to work from the top down).


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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

Hope that helps!

I used the following way

3^7 = 2048. Now (2048)^11. Here last digit is 8 and cycle for this is 8,4,2,6. Now here 11th digit is 2. So when we divide we will get 2 as remainder.

First of all 3^7=2,187 not 2048.

Next 3^7^11 doe NOT equal to 2,187^11: \(a^m^n=a^{(m^n)}\) and not \((a^m)^n\) (if exponentiation is indicated by stacked symbols, the rule is to work from the top down).

Hope it helps.
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