If the greatest common factor of positive integers n and m is 15, and

Question

If the greatest common factor of positive integers n and m is 15, and the remainder when  (n+x)^{32}  is divided by 15 is 1, which of the following CANNOT be the value of x?

A. 1
B. 4
C. 9
D. 11
E. 14

Are You Up For the Challenge: 700 Level Questions

ScottTargetTestPrep · Accepted Answer

Since the GCF of n and m is 15, n must be a multiple of 15. If we expand (n + x)^32, every term in the expansion will have a factor of n except the last term, which is x^32. The terms that have a factor of n will be divisible by 15 (thus, the remainder is 0 when they are divided by 15). Therefore, the only way the remainder could be 1 when (n + x)^32 is divided by 15 would be if x^32 divided by 15 would yield a remainder of 1.

Now, let’s look at the choices.

A) 1

If x = 1, then 1^32 = 1 and when 1 is divided by 15, the remainder is 1.

B) 4

If x = 4, then 4^32 = (4^2)^16 = 16^16. Notice that when 16 is divided by 15, the remainder is 1. Therefore, when 16^16 is divided by 15, the remainder will be same as when 1^16 is divided by 15, and therefore, that remainder will be 1.

Let’s skip C for the moment and look at D and E first.

D) 11

Since 11 - 15 = -4, then x = 11 is equivalent to x = -4. However, since (-4)^32 = 4^32, then the remainder will be equal to the remainder when x = 4, which is 1.

E) 14

Similarly, since 14 - 15 = -1, then x = 14 is equivalent to x = -1. However, since (-1)^32 = 1^32, then the remainder will be equal to the remainder when x = 1, which is 1.

We have rejected choices A, B, D, and E. Therefore, the correct answer must be C.

Answer: C

nick1816 · Answer

GCD(n,m)=15
hence, we can write n=15k

(n+x)^{32}

=  (15k+x)^{32}

= 15  k^{32}  + 32C1 ( 15  k)^{31} *x+.......+32C1*  15 k* (x)^{31} +  x^{32}

All the terms are divisible by 15 except the last one. Hence, the remainder when  (n+x)^{32}  is divided by 15 is same as when  x^{32}  is divided by 15

A. 1= (15*0+1)

hence,  1^{32}  will give 1 as a remainder when divided by 15

B.  4^2 =16= (15+1)

(4^2)^{16} =  (15+1)^{16}  will give remainder 1 when divided by 15

D. 11= (15-4)
 11^{32} =  (-4)^{32}  mod 15
 11^{32} =  (4)^{32}  mod 15
 11^{32} =  1  mod 15

E. 14= (15-1)

(15-1)^{32}  will give 1 as remainder when divided by 1.

C

Mansoor50 · Answer

I wd like to know if the following is correct

\frac{(n +11)^{32}}{15}  =  \frac{(15 +11)^{32}}{15} =  \frac{(26)^{32}}{15} =  \frac{(13^{32}	imes 2^{32})}{15} =  \frac{(13^{32}	imes 2^{32})}{3 	imes 5}

now Remainder of  \frac{13^{32}}{3 }  is 1

and the remainder of  \frac{2^{32}}{5 }  =  \frac{4^{16}}{5 }  is  (-1)^{16}

giving us the overall remainder of 1 for this answer option.

is the arithmetic correct?

Jane121393 · Answer

the remainder when (n+x) ^32 : 15 is 1 => the remainder when x^32 : 15 is 1 because n =15k
 Bunuel , is there anything wrong with my solution?
A. The remainder when (1^2) :15 is 1 => the remainder when (1^32) : 15 is 1
B. The remainder when (4^2) :15 is 1 => the remainder when (4^32) : 15 is 1
C. The remainder when (4^2) :15 is 6 => => the remainder when (4^32) : 15 is 6 => OA is C
Please correct me. Thanks.

zhanbo · Answer

Great analysis! But I think you can explain D and E better:

D: 11^2 = 121 = 15*8 + 1
E: 14^2 196 = 15 * 13 + 1

gmatapprentice · Answer

Took me more than 5 mins but my solution is as follows:

GCf=15 hence
m,n=(1*15,3*5,5*3)
To calculate n+x=15, we know n or x can take up any of the following values
1+14=15 (a and c out)
4+11=15 (b and d out)
5+10=15

Whats left is 9.

SuryaNouliGMAT · Answer

I am still unclear about the value of n=15, could you please elaborate hoe the value of N is found from GCD. Thanks & Kudos in advance

gmatapprentice · Answer

SuryaNouliGMAT

Just to clarify n is not equal to 15, idea is to figure out different values for n given that the GCF (n,m) is 15.

Hope that makes sense.

Kinshook · Answer

Asked: If the greatest common factor of positive integers n and m is 15, and the remainder when  (n+x)^{32}  is divided by 15 is 1, which of the following CANNOT be the value of x?
gcd (n,m) = 15 = 3*5; n = 15k; where k is not a multiple of 3 or 5.
(n+x)^32mod15 = 1
A. 1
(15k+1)^32mod15 = 1
B. 4
(15k+4)^32mod15 = 4^32mod15=1^16mod15 = 1
C. 9
(15k+9)^32mod15 = 9^32mod15= 6^16mod15 = 6^8mod15 = 6^4mod15 = 6^2mod15 = 6mod15 = 6
D. 11
(15k+11)^32mod15 = 11^32mod15=1^16mod15 = 1
E. 14
(15k+14)^32mod15 = 14^32mod15=1^16mod15 = 1

IMO C

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

If the greatest common factor of positive integers n and m is 15, and

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

If the greatest common factor of positive integers n and m is 15, and

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards