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ps - divisibility 2

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ps - divisibility 2

by ccassel » Sat Apr 02, 2011 2:53 pm

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Hi,

How would you reach your answer to this question?

An even number x divided by 7 gives some quotient plus a remainder of 6. Which of the following, when added to x, gives a sum which must be divisible by 14?

(a) 1
(b) 3
(c) 7
(d) 8
(e) 13

cheers,
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by rohu27 » Sat Apr 02, 2011 9:23 pm
IMO D
i did this by pluggign in values. x shud be an even number. and give 'some quotient' so i did not take x to be y (as quotint will be zero in thi scase).
nxt value x can have is 14+6=20, so for to be divisible by 14, 8 shud be added to it.
21+6=27(but its not even)
28+6=34 - again 8 shud be added to get 42(14*3=42)

so 8 shud be added.
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by manpsingh87 » Sat Apr 02, 2011 11:03 pm
ccassel wrote:Hi,

How would you reach your answer to this question?

An even number x divided by 7 gives some quotient plus a remainder of 6. Which of the following, when added to x, gives a sum which must be divisible by 14?

(a) 1
(b) 3
(c) 7
(d) 8
(e) 13

cheers,
x=7k+6;
here k (0,2,4,6...) should be even because x is an even number, now if we add any odd integer to the x then it would becomes odd (because even+odd=odd); therefore it will not be divisible by 14; as 14 is even, hence we must add even integer to x,and the only option which is given even from the among set of options is 8 hence D.
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by force5 » Sun Apr 03, 2011 4:45 am
yes well explained D.
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by GMATGuruNY » Sun Apr 03, 2011 6:07 am
ccassel wrote:Hi,

How would you reach your answer to this question?

An even number x divided by 7 gives some quotient plus a remainder of 6. Which of the following, when added to x, gives a sum which must be divisible by 14?

(a) 1
(b) 3
(c) 7
(d) 8
(e) 13

cheers,
Plug in a value that satisfies the conditions given.

X is an even integer that is 6 more than a multiple of 7.
Let x = 20. This works because 20/7 = 2 R6.

Now add each answer choice to x = 20. If the sum is not a multiple of 14, eliminate the answer choice.

A) 20+1 = 21. Not a multiple of 14. Eliminate A.
B) 20+3 = 23. Not a multiple of 14. Eliminate B.
C) 20+7 = 27. Not a multiple of 14. Eliminate C.
D) 20+8 = 28. 28 is a multiple of 14. Hold onto D.
E) 20+13 = 33. Not a multiple of 14. Eliminate E.

The correct answer is D.
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by nicolasdryzun » Wed Jun 20, 2018 5:14 pm
Hi guys, since the x + alternatives must be divisible by 14 and x is even. The alternative must be even otherwise it wont be divisible to 14. So D. and you save 119 seconds
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