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* 15k*\((x)^{31}\)+ \(x^{32}\)

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

I wd like to know if the following is correct

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

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?

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.

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

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.

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
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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

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.

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
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