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When positive integer k is divided by 35, the remainder is 14. Which of the following must be a divisor of k?

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by BTGModeratorVI » Sun Jul 19, 2020 1:44 pm
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When positive integer k is divided by 35, the remainder is 14. Which of the following must be a divisor of k?

A. 5
B. 7
C. 11
D. 14
E. 21

Answer: B
Source: Veritas
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Source: — Problem Solving |

BTGModeratorVI wrote: ↑
Sun Jul 19, 2020 1:44 pm
When positive integer k is divided by 35, the remainder is 14. Which of the following must be a divisor of k?

A. 5
B. 7
C. 11
D. 14
E. 21

Answer: B
Source: Veritas
Say k = 35q + 14; where q is a positive integer

=> k = 7(5q + 2)

=> k must be divisible by 7. Sufficient.

Correct answer: B

Hope this helps!

-Jay
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BTGModeratorVI wrote: ↑
Sun Jul 19, 2020 1:44 pm
When positive integer k is divided by 35, the remainder is 14. Which of the following must be a divisor of k?

A. 5
B. 7
C. 11
D. 14
E. 21

Answer: B
Source: Veritas
When it comes to remainders, we have a nice rule that says:
If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

When positive integer k is divided by 35, the remainder is 14.
So, some possible values of k are: 14, 49, 84, 119, . . . etc

Let's test some values.
If k = 14, then we can ELIMINATE A, C and E, since they are not divisors of 14, and the questions says "Which of the following MUST be a divisor of k?
We're left with answer choices B and D

Let's test another value.
If k = 49, then we can ELIMINATE D, since 14 is not a divisor of 49.

By the process of elimination, we're left with B

Answer: B

Cheers,
Brent

RELATED VIDEO FROM MY COURSE
https://www.gmatprepnow.com/module/gmat ... /video/842
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BTGModeratorVI wrote: ↑
Sun Jul 19, 2020 1:44 pm
When positive integer k is divided by 35, the remainder is 14. Which of the following must be a divisor of k?

A. 5
B. 7
C. 11
D. 14
E. 21

Answer: B
Solution:

We can create the equation where q is the quotient:

k/35 = q + 14/35

k = 35q + 14

k= 7(5q + 2)

We see that 7 must be a divisor of k since k is a multiple of 7.

Answer: B

Scott Woodbury-Stewart
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