I'm getting the answer as B = 26

n = 2*3*5*7*9*11*13/7*11*k

n = 2*3*5*9*13/k.

Check the answer choices

A. 22 = 2*11 - will leave 11 in the denominator - not be an integer.
B. 26 = 2*13 - will be an integer as you can cancel out 2 and 13 from the numerator and denominator.

C. 35 = 5*7, this option will leave 7 in the denominator - not an integer.
D. 54 = 3*16 - will leave 8 in the denominator
E. 60 = 5*12 - will leave 2 in the denominator.

Whats the source of this question?

light_speed wrote:15. n= 2.3.5.7.9.11.13/77k

. = multiply


If n is an integer and then which of the following could be the value of k?
(A) 22
(B) 26
(C) 35
(D) 54
(E) 60

I tried breaking down 77 into its primes (7 x 11) and canceled the identical numbers from the numerator and denominator but got stuck after that. Answer is C

n= 2. . .9. .13/ k

nk = 2 x 9 x 13

k = 2 x 9 x 13 / n

as you can see if n = 9

k = 26

Hence, B

Am I missing something ?

it is B. typo

it is B. typo

..HEiiiiii..
why cant it be 54 ???

after cancelling out ..it gives ...2*3*5*9*13*/k
If K is 54 then the value is aswell an integer ...what say ???

Hey your right even 54 is possible.

I made an mathematical error in my previous post

54 = 3*18 = 3*2*9

You'll get an integer in this case as well.

So the answer is D and B I guess!!!.

mals24 wrote:Hey your right even 54 is possible.

I made an mathematical error in my previous post

54 = 3*18 = 3*2*9

You'll get an integer in this case as well.

So the answer is D and B I guess!!!.

Hi mals24,
indeed you will get an integer with 54 but still it will not be suff for our question as the LHS contains 15*n.
So, if if you remove 2*3*9 from the R.H.S, 15*n cant be equal to R.H.S
Hope it helps.
And as its a PS question, it cant have 2 answers in GMATLAND.

15. n= 2.3.5.7.9.11.13/77k

Its 15*n?? Gosh I thought 15 was the question number!!!

If that is not the case, then 54 is incorrect.

Thanks jimmiejaz for correcting me.

light_speed wrote:15. n= 2.3.5.7.9.11.13/77k

. = multiply


If n is an integer and then which of the following could be the value of k?
(A) 22
(B) 26
(C) 35
(D) 54
(E) 60

I tried breaking down 77 into its primes (7 x 11) and canceled the identical numbers from the numerator and denominator but got stuck after that. Answer is C

15n = 2*3*5*7*9*11*13/7*11*k

n = 2*3*5*7*9*11*13/3*5*7*11*k (divided both sides by 15=3*5)

n = 2*9*13/k

For n to be an integer, k must be a factor of 2*9*13. Only (b) 26 (i.e. 2*13) fills this role.

RIDICULOUS....
OOPS.....Its 15*n?? Gosh I thought 15 was the question number!!!

Sorries!!!!