Not necessarily. You did get the correct answer but with a 'lucky' approach. 3^4 also has a unit digit 1 but remainder of 3^4(=81) when you divide it by 26 is 3 and not 1.

So for this and other similar questions try to find a number closest to either the divisor or a multiple of the divisor.

You can clearly see that 3^3=27=26+1. Use this to your advantage as follows:

\(3^{84} = (3^3)^{28} = (27)^{28} = (26+1)^{28}\)

Applying the binomial theorem to expand \((a+b)^n = a^n*b^0+a^{n-1}*b^1....+a^1*b^{n-1}+a^0*b^n\), you get

\((26+1)^{28} = 26^{28}+26^{27}*1^1...26^1*1^{27}+26^0*1^{28}\), so apart from the last term, all other terms have 26 and thus you will get a remainder of 0 with all but 1 terms in this expanded form.

Thus the remainder will be = remainder of \(\frac{26^0*1^{28}}{26}\) = 1

B is the correct answer.

Hope this helps.
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3^84/26 = 27^26/26

27 is divided by 26, the remainder is 1
=> 27^26 is divided by 26, the remainder is still 1

Ans B

Please correct if i am wrong

Thanks

Dear experts, do you think this is a valid method ?
3^84 has a unit digit 1, so the remainder is 1 Answer(B)

No, it is not a correct approach. You can use units digit to guide you in finding the remainder when dividing by 2, 5, or 10 for example. For example \(\frac{12}{5}\) and \(\frac{122222}{5}\) have the same remainder. The way I approach these types of remainder questions, is I first look at the divisor. If it is a small divisor that is neither 2,5, nor 10, I will choose between looking at the cyclicity of the base raised to a positive exponent divided by the divisor and binomial expansion.

For example consider finding the remainder for \(\frac{2^{86}}{7}\). Since the divisor is not 2,5, or 10, looking at the units digit wouldn't work. If I were to use the cyclicity method,
I would start by listing,
\(\frac{2^{1}}{7}\)yields remainder of 2

\(\frac{2^{2}}{7}\)yields remainder of 4

\(\frac{2^{3}}{7}\)yields remainder of 1

\(\frac{2^{4}}{7}\)yields remainder of 2

There is a cyclicity of 3, since 86/3 yields remainder of 2 then the remainder of \(\frac{2^{86}}{7}\), will be the same remainder of \(\frac{2^{2}}{7}\),which as remainder of 4

Alternately, using binomial expansion method:

\(\frac{2^{86}}{7}\) --> \(\frac{(2^3)^{28}*2^2}{7}\) --> \(\frac{(7+1)^{28}* 2^2}{7}\)

Then I will look at the individual remainders of each of the 2 terms when dividing by 7. For example let's say you wanted to find the remainder of \(\frac{(18*23)}{7}\). Mod(\(\frac{18}{7}\)) is 4. Mod(\(\frac{23}{7}\)) is 2. Combine 4 and 2 by multiplication because the operation in question is multiplication and then find the remainder. So the remainder of \(\frac{(4*2)}{7}\) is 1 and this is the answer.

This can be done with negative remainders too. For example let's say you wanted to find the remainder of \(\frac{(23-18)}{7}\). Combine 2 and 4 by subtraction and then find the remainder. So the remainder of \(\frac{(2-4)}{7}\) is 5. When you have a negative remainder, find the remainder first as if the numerator was positive and then subtract from the divisor.
If you had positive 2 on top, the remainder would have been 2, but you have negative 2, so subtract positive 2 from 7 and you get a remainder of 5.

Back to the question. The remainder of \(\frac{(7+1)^{28}}{7}\) is 1; every term will be divisible by 7 except the last term,which is \(1^{28}.\) See Engr2012's above explanation regarding binomial expansion. The remainder of \(\frac{2^2}{7}\) is 4. Combine 1 and 4 by multiplication and then find the remainder. So the remainder of \(\frac{(1*4)}{7}\) is 4.

Since the divisor in the original question is 26, the cyclicity would be too big. The cyclicity method wouldn't be a good idea. I would have done this question exactly as Engr2012 did.

Thanks altogether, I've just tried this approach in 2 or 3 questions and did get a right answer, what made me think that this one is the fastest approach... but I saw my mistake now.

\(\frac{2^{86}}{7}\) isn't it the same using cyclicity method...?
\(2^{86}/7\) has a unit's digit of 4 , so the remainder is 4

Engr2012, here we use cyclicity approach and it's valid. I just don't understand where is the limit where we can use or don't this approach ??
what-is-the-remainder-of-126493.html

First off, 1 method "working" for a particular question does not mean that it is the most efficient / repeatable method. \(2^{86}/7\) luckily gave you the correct answer but you want to develop and understand a method that can be applied to all such questions. Cyclicity is not such a method.

Cyclicity is to be used for finding out what will be the unit's digit but for questions similar to \(2^{86}/7\), I will use the binomial theorem route.

\(2^{86}/7\) = \(2^2*(2^6)^{14}\) =\(4*(64)^{14}\) = \(4*(63+1)^{14}\)

Again, all terms of (63+1)^14 will have 63 (a multiple of 7) except the last one (=1^14), giving you

\(4*1^{14}\) = 4 (as the last term).

Thus the remainder when you divide 4/7=4.

Idea behind 2^6 is to find the closest multiple of 7 (such that the number = multiple of 7 + 1), alternately you could have used 2^3 as well as it will give you 8 = 7-1. Concept remains the same.

Similarly, to generalize this concept, 2^n needs to be expressed in the form of (divisor+1)^n in order to use the binomial theorem.

Hope this helps.
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Hi ENGR2012, first of all , thanks for your time. After investigating about this topic, I must say that cyclicity is a valid approach (as bhaskar438 have already stated) only if we have 2,5 or 10 as a divisor. But I'll better use binomial theorem to be 100% sure.

GMAT is all about picking the best method in terms of execution and time you spend on a particular question. If you are comfortable to add the extra bit of learning that 2,5,10 maintain cyclicity then by all means do it but make sure to not experiment in the GMAT.

I never said that cyclicity is wrong. For me, binomial theorem application is more straightforward.

Pick your method and stick to it. Pattern recognition and application of a fixed method of attack are of paramount importance in GMAT.
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Hi Brainlab,

I was actually talking about 3 separate methods: units digit method, cycle method, and binomial expansion method. A specific units digit method for divisors 2,5,10. In the cyclicity method you are only looking at the cycle of the units digit to see when it stops repeating. You are not looking solely at the units digit to find the remainder as you would for divisors 2,5,10. In the previous question \(\frac{2^{86}}{7}\) was similar to finding the remainder for \(\frac{2^{2}}{7}\) because I found out the remainder should be the second item out of the three items in the cycle. Only use the cyclicity method when the divisor is small.

Sometimes the most efficient method is using binomial expansion first and then the cyclicity method as Bunuel as done in this question: what-is-the-remainder-when-18-22-10-is-divided-by-99724.html#p768816

I can't think of a better method to find the remainder for \(\frac{18^{220}}{7}\) than the method he did. Also the cyclicity method is different from units digit method (which is for divisors 2,5, or 10)

I dont think such a question will be asked in GMAT (not right now, even Bunuel mentions this in the first line of his solution) as I have not seen a single official/GMAT question that can only be solved by binomial theorem application. Almost always there were other ways to solve such a question. It is not about how many "difficult" questions you solve, there can be hundreds of them but the more important part of GMAT quant is to understand the tactics to solve 'problem groups'.
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I completely agree. I just like having many different methods in my arsenal.

Such Questions Usually have a sort of Repetition in them :

3^1 = 3/26 gives remainder 3
3^2 = 9/26 gives remainder 9
3^3 = 27/26 gives remainder 1
3^4 = 81/26 again gives remainder 3
3^5 = 243/26 again gives remainder 9 and so on.

Its following a repetion of 3...9...1 or we can say they are repeating in multiples of 3. So, we can simply divide 84 by 3 which is completely divisible.

=> As per the repetition the remainder should be 1.

Bunuel
As per cyclicity
it is 4m+4
mean last 2 digits are 81
so 81/28=78
Answer should be 3 isnot it?
like this question
https://gmatclub.com/forum/what-is-the- ... 34109.html
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No, it's 1: https://www.wolframalpha.com/input/?i=t ... of+(3%5E84)%2F7

Please show your work.
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Hi Bunuel,

Will it be possible for you to provide an approach to this type of questions? or could you guide me to a post where you have already posted an explanation to such questions? I read the entire thread and I was unable to understand the approaches and frankly it seemed like a really time consuming approaches.
Kindly help,

thanks in advance!

Check below:

REMAINDERS ON THE GMAT

Divisibility and Remainders on the GMAT
Divisibility Tips

Theory on remainders problems
Tips on remainders

Units digits, exponents, remainders problems

DS remainders problems
PS remainders problems

DS divisibility problems
PS divisibility problems

Hope it helps.
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\(\frac{(3^{84})}{26}\)

= \(\frac{3^{3*28}}{26}\)

= \(\frac{27^{28}}{26}\)

= \(\frac{(26 + 1)^{28}}{26}\)

= \(\frac{(26^{28} + 1^{28})}{26}\)

\frac{26^{Any \ Number }}{26} , will have remainder 0
\frac{1^{Any \ Number }}{26} , will have remainder 1

So, The answer will be (B) 1
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