product of the first twelve

[sanju09]

The product of the first twelve positive integers is divisible by all of the following EXCEPT
(A) 50
(B) 60
(C) 75
(D) 88
(E) 210

[November Rain]

IMO A

52 is equal to 13*2*2

There is no 13 in the first 12 numbers

[sanju09]

IMO A

52 is equal to 13*2*2

There is no 13 in the first 12 numbers

sorry NR, try now

edited

[ace_gre]

Same concept as NR. Factor out the choices and see which ones do not exist in the product of 12 integers...

IMO C

[anki_jain]

product of the first twelve positive integers is divisible by all the given choices!!

something wrong with the options??

[harsh.champ]

The product of the first twelve positive integers is divisible by all of the following EXCEPT
(A) 50
(B) 60
(C) 75
(D) 88
(E) 210

Product of 1st 12 integers = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 x 12
= 1 x (2) x 3 x (2^2) x 5 x (2 x 3) x 7 x (2^3) x (3^2) x (5 x 2) x 11 x (4 x 3)
= 1 x (2^10) x (3^5) x (5^2) x 7 x 11
A:-50 = 2 x( 5^2) Thus divisible.
B:-60 = (2^2) x 3 x 5 Thus divisible.
C:-75 = 3 x (5^2) Thus divisible.
D:-88 = (2^3) x 11 .Thus divisible.
E:-210 = 2 x 3 x 5 x 7. Thus divisible.

The answer should be none of the above.
Did I go wrong somewhere??

[harshavardhanc]

product of the first twelve positive integers is divisible by all the given choices!!

something wrong with the options??

agree!!!!

[shashank.ism]

The product of the first twelve positive integers is divisible by all of the following EXCEPT
(A) 50
(B) 60
(C) 75
(D) 88
(E) 210

product = 1x2x3x4x5x6x7x8x9x10x11x12
now making it in the form of product of prime nos.
= 2x3x(2x2)x5x(2x3)x7x(2x2x2)x(3x3)x(5x2)x11x(2x2x3)
= 2^10 x 3^5 x 5^2 x 7 x11

Now 50 = 2x5x5 OK
60 = 2x2x3x5 OK
75 = 3x5x5 OK
88 = 2x2x2x11 OK
210 = 2x3x5x7 OK

I think all of them divides the product of first 12 numbers..I think some problem in the question....

[thephoenix]

The product of the first twelve positive integers is divisible by all of the following EXCEPT
(A) 50
(B) 60
(C) 75
(D) 88
(E) 210

12! CONTAINS TEN 2'S
FIVE 3'S
TWO 5'S AND
ONE 7 AND ONE 11
---> 12! WILL DIV BY ALL THE NO. FROM 1 TO 12

AFTER THIS HEY SANJU WHAT HAVE U DONE U EDITED YOUR POST AND MADE ALL POSSIBLE

[maaz_gmat]

Actually the question posted is wrong.
The actual question is as below:
The product of the first twelve positive integers is divisible by all of the following EXCEPT
(A) 210
(B) 88
(C) 75
(D) 60
(E) 34


@harsh.champ - your method is correct.
As you see, with the above choices you will be able to get solutions for all except 34.
As 34 = 17 x 2, And we don't have 17 as a factor.

OA - E