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20 Oct 2015, 12:39
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
74% (01:21) correct
26%
(01:53)
wrong
based on 1024
sessions
History
Date
Time
Result
Not Attempted Yet
What is the remainder when \(\frac{(3^{84})}{26}\)
(A) 0
(B) 1
(C) 2
(D) 24
(E) 25
(A) 0
(B) 1
(C) 2
(D) 24
(E) 25
ENGRTOMBA2018
Joined: 20 Mar 2014
Last visit: 01 Dec 2021
Posts: 2,325
Given Kudos: 816
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
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24 Oct 2015, 11:58
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BrainLab
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.
24 Oct 2015, 13:32
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BrainLab
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.
