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product of the first twelve

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by sanju09 » Thu Jan 14, 2010 1:31 am
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
Last edited by sanju09 on Thu Jan 14, 2010 3:21 am, edited 1 time in total.
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by November Rain » Thu Jan 14, 2010 3:07 am
IMO A

52 is equal to 13*2*2

There is no 13 in the first 12 numbers
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by sanju09 » Thu Jan 14, 2010 3:22 am
November Rain wrote:IMO A

52 is equal to 13*2*2

There is no 13 in the first 12 numbers
sorry NR, try now

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by ace_gre » Thu Jan 14, 2010 4:13 pm
Same concept as NR. Factor out the choices and see which ones do not exist in the product of 12 integers...

IMO C
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by anki_jain » Fri Feb 19, 2010 10:59 am
product of the first twelve positive integers is divisible by all the given choices!!

something wrong with the options??
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by harsh.champ » Fri Feb 19, 2010 11:22 am
sanju09 wrote: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??
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by harshavardhanc » Fri Feb 19, 2010 11:22 am
anki_jain wrote:product of the first twelve positive integers is divisible by all the given choices!!

something wrong with the options??
agree!!!!
Regards,
Harsha
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by shashank.ism » Sat Feb 20, 2010 5:48 am
sanju09 wrote: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....
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by thephoenix » Sat Feb 20, 2010 8:28 am
sanju09 wrote: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
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by maaz_gmat » Fri Oct 08, 2010 3:45 am
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
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