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

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If the greatest common factor of positive integers n and m is 15, and  [#permalink]

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02 Dec 2019, 00:35
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45% (02:32) correct 55% (02:34) wrong based on 99 sessions

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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

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Re: If the greatest common factor of positive integers n and m is 15, and  [#permalink]

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12 Jan 2020, 06:28
1
Bunuel wrote:
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

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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.

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Re: If the greatest common factor of positive integers n and m is 15, and  [#permalink]

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02 Dec 2019, 17:12
2
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

Bunuel wrote:
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

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Joined: 29 May 2017
Posts: 159
Location: Pakistan
Concentration: Social Entrepreneurship, Sustainability
Re: If the greatest common factor of positive integers n and m is 15, and  [#permalink]

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18 Dec 2019, 02:16
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?
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Re: If the greatest common factor of positive integers n and m is 15, and  [#permalink]

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24 Dec 2019, 02:21
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
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Re: If the greatest common factor of positive integers n and m is 15, and  [#permalink]

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24 Dec 2019, 03:10
nick1816 wrote:
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

Bunuel wrote:
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

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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
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Re: If the greatest common factor of positive integers n and m is 15, and  [#permalink]

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25 Dec 2019, 08:28
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.
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Joined: 22 May 2019
Posts: 11
Re: If the greatest common factor of positive integers n and m is 15, and  [#permalink]

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14 Jan 2020, 06:33
I am still unclear about the value of n=15, could you please elaborate hoe the value of N is found from GCD.

gmatapprentice wrote:
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.
Manager
Joined: 14 Nov 2018
Posts: 89
Location: United Arab Emirates
GMAT 1: 590 Q42 V30
GPA: 2.6
Re: If the greatest common factor of positive integers n and m is 15, and  [#permalink]

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17 Jan 2020, 07:01
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.
Re: If the greatest common factor of positive integers n and m is 15, and   [#permalink] 17 Jan 2020, 07:01
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