i am wondering whether i clearly state my question?
"zero reply" indicates cannot solve the problem or cannot understand the question?

hope that someone can reply me whether there is one solution or not.

thanks everybody.

seanceserene wrote:is there a formular to have the product of n consecutive integers?

when n is big, for instance, 1*2*3*...*121, it is time-consuming to calculate the product.

ask a shortcut to have product

I do not think there is a formula or a shortcut to calculate the product of n Consecutive numbers. I am confident that GMAT will not make you do things you can't do within 2-3 minutes. So, if you can post the specific question, I may be of help.

ajith wrote:

seanceserene wrote:is there a formular to have the product of n consecutive integers?

when n is big, for instance, 1*2*3*...*121, it is time-consuming to calculate the product.

ask a shortcut to have product

I do not think there is a formula or a shortcut to calculate the product of n Consecutive numbers. I am confident that GMAT will not make you do things you can't do within 2-3 minutes. So, if you can post the specific question, I may be of help.

If n is the product of the integers from1 to 20 inclusive, what is the greatest integer k for which 2k is a factor of n?

A. 10
B. 12
C. 15
D. 18
E. 20

seanceserene wrote:
If n is the product of the integers from1 to 20 inclusive, what is the greatest integer k for which 2k is a factor of n?

A. 10
B. 12
C. 15
D. 18
E. 20

If n is the product of the integers from1 to 20 inclusive, what is the greatest integer k for which 2^k is a factor of n?

16 has 4, two's as it's factor
8 has 3
4,12 and 20 has 2

Rest of the even numbers 5 of them has 1
so k = 4+3+3*2 +5 =18

Another way to solve this will be

floor(20/2) + floor(20/4) + floor(20/8)+ floor(20/16)
10+5+2+1 =18

seanceserene wrote: If n is the product of the integers from1 to 20 inclusive, what is the greatest integer k for which 2k is a factor of n?

A. 10
B. 12
C. 15
D. 18
E. 20

there is no short cut for finding the value of n!
but there is a one for this type of question

divide n by 2 note the quotient
for 20 divided by 2 quotient is 10
then div 10 by 2 and note down the q=5
5/2 q=2
div 2/2 q=1
div 1/2 q=0

summ all the quotient you get the ans
here uts 10+5+2+1=18

I am clearly missing something here. If k is 20 then 2k = 40. 40 is definitely a factor of 20!. So IMO answer should be E.

@ajith can you please explain why the answer will be 18 ? It did not ask to calculate H.C.F. or L.C.M which I think you did.

Please explain.

sarthak wrote:I am clearly missing something here. If k is 20 then 2k = 40. 40 is definitely a factor of 20!. So IMO answer should be E.

@ajith can you please explain why the answer will be 18 ? It did not ask to calculate H.C.F. or L.C.M which I think you did.

Please explain.

i apologize for giving 2K. it should be 2^K.

thephoenix wrote:

seanceserene wrote: If n is the product of the integers from1 to 20 inclusive, what is the greatest integer k for which 2k is a factor of n?

A. 10
B. 12
C. 15
D. 18
E. 20

there is no short cut for finding the value of n!
but there is a one for this type of question

divide n by 2 note the quotient
for 20 divided by 2 quotient is 10
then div 10 by 2 and note down the q=5
5/2 q=2
div 2/2 q=1
div 1/2 q=0

summ all the quotient you get the ans
here uts 10+5+2+1=18

great. but is it applicable when the consecutive sequence is not starting with 1 ?
and, since you only use the last number "20", does rest of the numbers are irrelavant in this calculation?

just wondering the theorem behind the trick.