If n is the remainder when 250 is divided by 3 and m is the remainder

Question

If n is the remainder when 2^50 is divided by 3 and m is the remainder when 2^15 is divided by 5, what is m + n?

A. 6
B. 5
C. 4
D. 3
E. 2

KatariaP · Accepted Answer

There is repetitive pattern for remainders.
(2^1)/3 - Remainder is 2
(2^2)/3 - Remainder is 1
(2^3)/3 - Remainder is 2, and so on.
So for even powers, remainder is 1. Therefore remainder for (2^50)/3 is 1.

Similarly, we can observe a pattern for remainders when divided by 5.
(2^1)/5 - Remainder is 2
(2^2)/5 - Remainder is 4
(2^3)/5 - Remainder is 3
(2^4)/5 - Remainder is 1
(2^5)/5 - Remainder is 2
(2^6)/5 - Remainder is 4
(2^7)/5 - Remainder is 3
We can see that there is a pattern of 2,4,3,1. So remainder for (2^15)/5 is 3.

Therefore m+n = 1+3 = 4.

Option C is the right answer.

This is my first answer on a forum. If it helped you out, please leave a kudos.

vitaliyGMAT · Answer

2 = -1 (mod 3)

\frac{2^{50}}{3} = \frac{(-1)^{50}}{3} = \frac{1}{3}  remainder 1

2^4   = 1 (mod 5)

\frac{(2^4)^3*2^3}{5} = \frac{1*8}{5}  remainder 3

1 + 3 = 4

Answer C

Abhishek009 · Answer

\frac{2^4}{3}  = Remainder 1
So,  2^{50} = 2^{4*12}*2^2

Now,  \frac{2^2}{3}  = Remainder 1

So, m = 1

\frac{2^4}{5}  = Remainder 1
So,  2^{15} = 2^{4*3}*2^3

\frac{2^3}{5}  = Remainder 3

So, n = 3

Hence,  m + n = 1 + 3 = 4

Hence, Correct answer will be (C) 4

lauramo · Answer

Hi,

I don´t quite understand the following

243243 = Remainder 1
So, 250=24∗12∗22250=24∗12∗22

Now, 223223 = Remainder 1

So, m = 1

245245 = Remainder 1
So, 215=24∗3∗23215=24∗3∗23

235235 = Remainder 3

So, n = 3

Hence, m+n=1+3=4m+n=1+3=4

Can anyone expand the explanation?

Thanks

LeonidK · Answer

Great solution! Thanks!

ScottTargetTestPrep · Answer

Let’s determine the remainder pattern when 2 raised to an exponent is divided by 3.

(2^1)/3 = 0 remainder 2

(2^2)/3 = 4/3 = 1 remainder 1

(2^3)/3 = 8/3 = 2 remainder 2

(2^4)/3 = 16/3 = 5 remainder 1

When 2 is raised to an even exponent and is divided by 3, a remainder of 1 results. Thus, 2^50 divided by 3 has a remainder of 1.

Next we can determine the remainder when 2^15 is divided by 5. To do so, we recall that the units digit of any number divided by 5 will produce the same remainder as when the actual number is divided by 5. Thus, let’s determine the units digit of 2^15. The pattern of units digits of the base of 2 when raised to an exponent is:

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 6

2^5 = 2

We see the pattern is 2-4-8-6, and furthermore that 2^4k, in which k is a positive integer, will always have a units digit of 6.

Thus, the units digit of 2^16 = 6 and so the units digit of 2^15 = 8. Dividing 8 by 5 yields a remainder of 3; thus, dividing 2^15 by 5 also yields a remainder of 3.

Therefore, m + n = 3 + 1 = 4.

Answer: C

EgmatQuantExpert · Answer

Solution

• When  2^{50}  is divided by  3  the remainder is n. Let us find this “ n ”.

o  \frac{2^{50}}{3} = \frac{(2^2)^{25}}{30} = \frac{4^{25}}{3}

o We know when 4 is divided by 3, the remainder is 1.

o Thus  n =  _R = (1^{25}) = 1

• When  2^{15}  is divided by  5 , the remainder is  m .

o  \frac{2^{15}}{5} = \frac{ }{5}

o We know that when 16 is divided by 5 the remainder 1 and 8 divided by 5, the remainder is 3

o Thus,  m =  _R *  _R = 1 * 3 = 3

• Therefore,  m+n = 1 + 3 = 4

• Correct answer is   Option C  .

Thanks,
Saquib
Quant Expert
e-GMAT

Aiming to score Q50 or higher in GMAT Quant ? Attend this webinar on   2nd April   to learn a structured approach to  solve 700+ Number Properties question in less than 2 minutes .  Register

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saraheja · Answer

243243 = Remainder 1
So, 250=24∗12∗22250=24∗12∗22

Now, 223223 = Remainder 1

So, m = 1

245245 = Remainder 1
So, 215=24∗3∗23215=24∗3∗23

235235 = Remainder 3

So, n = 3

SidKan · Answer

2 = -1 (mod 3)

2503=(−1)503=132503=(−1)503=13 remainder 1

2424 = 1 (mod 5)

(24)3∗235=1∗85(24)3∗235=1∗85 remainder 3

1 + 3 = 4

Answer C

afa13 · Answer

Hello  VeritasPrepKarishma  and  mikemcgarry ,
I solved this problem by identifying the pattern however my error was that i started from 2^0 thus resulting in a different numbering of the sequence. When to consider starting from the zero pattern and when not to?
Thank you.

niks18 · Answer

Hi  afa13

if you are solving this question using the remainder pattern method and you are starting with  2^0 , then note that  2^0=1  and  \frac{1}{3} , the remainder will be  1  and next  2^1=2  when divided by 3 will have a remainder of 2 and so on

can you explain what sequence you were getting with your method

EgmatQuantExpert · Answer

Hey afa13, Ideally in a remainder question, one should never start with 0 as power. Because any number raised to 0 will be equal to 1 and any number which divides 1 (other than 1 itself) always gives the remainder as 1. For example:- \frac{1}{20} = 1 , \frac{1}{21} = 1 So, whenever you want to solve a reminder question to find the pattern, always start from 1 as x^0 changes the number. We want the pattern with 2 in this case... 2^0 is not 2..hence we won't be able to see the pattern. Hope this helps.

Regards

afa13 · Answer

Hi  niks18 ,
I was getting the same pattern1-2-1-2 and 1-2-4-3 but since i was starting with 2^0 instead of 2^1 m and n changed. I should have gotten 2-1-2-1 and 2-4-3-1 patterns. But as  EgmatQuantExpert , not to start with 0.
Thanks  niks18  and  EgmatQuantExpert  for answering.

gracie · Answer

there are 4 units digit values for the exponent cycle of 2: 2,4,8,6
50/4 gives remainder of 2➡2nd value in cycle is 4
remainder of 4/3=1=n
15/4 gives a remainder of 3➡3rd value in cycle is 8
remainder of 8/5=3=m
3+1=4
C

dracobook · Answer

EgmatQuantExpert  - thanks for sharing this solution.
Looks like you did the following:
1) Reduce the problem by converting the base to numbers greater than the divisor
2) Find the remainder when these base(s) are divided by the divisor
3) multiplied the remainders together

Is there somewhere I can learn more about this method of solving remainder problems?

saraheja · Answer

apply rule of cyclicity. m+n would be 4.

EgmatQuantExpert · Answer

Hey  dracobook

We teach these concepts in our live sessions and concept files...which our paid students can access.
However, in short, you can follow these steps to solve any remainder question:
 1- Look for a power p, such that N^p when divided by D, gives remainder 1 or -1
2- Once you have that..see much power is still left to resolve.

For example: In this case, if we have  2^{51}  instead of  2^{50} , we would have had  (2^2)^{25}  *  2^1 , which means we will get 1 *  2^1 
Finally, we multiply remainders to get our answer.

Note: There can be some slight variations also..which I have not mentioned here. 
Give us some time..we will try to write a short article on this and you can also buy our quant course to learn more about Number Properties and other topics.

Regards,
Ashutosh
e-GMAT

KanishkM · Answer

2^n follows the unit digit cyclicity as 2 4 8 16

Now 2^15 will have 3 as remainder, m

2^50 will have 1 as remainder,n

C

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If n is the remainder when 250 is divided by 3 and m is the remainder

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If n is the remainder when 250 is divided by 3 and m is the remainder

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