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Re: product and integer n [#permalink]
31 Jan 2005, 10:32

I looked at it, and broke 990 to its prime, then 11*5*3*3*2

i notice that the only prime number that shows up is 11, therefor it will be the smallest from this selection, but I think techically the smallest possible value should be 2?

Honghu can you please explain your reasoning with 189?

Antmavel wrote:

If n is a positive integer and the product of all the integers from 1 to n, inclusive, is divisible by 990, what is the least possible value of n?

another good question to be discussed and put my 2 cts views.

n to be divided by 990, n must posses all the prime factors of 990. the prime factors of 990 are 2, 3, 3, 5, and 11. the question stipulated that n is a product of all positive integers 1 to n means no single integer can repete if we need one more integer in between 1 to n. for example, here we need one more 3, we have to include 6 to fulfill the requirement of 3. similarly, to fulfill the requirement to 11, we have to stretch the n up to 11.

in case of 189, n is 9 because the prime factors of 189 are 3,3,3,and 7. we need three 3s which we can get from 1 to n=9. this n (9) also fullfill the requirement of 7 also, therefore the least possible value of n=9. n can not be 7. if n is 7, then the product of 1 to n (1x2x3x4x5x6x7) can not be divided by 189.

another good question to be discussed and put my 2 cts views.

in case of 189, n is 9 because the prime factors of 189 are 3,3,3,and 7. we need three 3s which we can get from 1 to n=9. this n (9) also fullfill the requirement of 7 also, therefore the least possible value of n=9. n can not be 7. if n is 7, then the product of 1 to n (1x2x3x4x5x6x7) can not be divided by 189.

I understand the part upto the breaking of the prime factors but I dont understand what you mean by we need 3 3s before we need 7?

ok.....for 189 you have prime factors 3x3x3x7....now for this to divide product from 1....n ....u have to have atleast go upto 7 numbers...right ?
However, in those first 7 numbers , u will only get one 3 and another three from 6 (3x2)...u need another 3 as the denomintor 3x3x3x7 has to divide the product evenly...so the product shud go all the way upto 9 to get another 3.

product = 1x2x3x4x5x6(i.e 3x2)x7x8x9 (i.e. 3x3)....u see now u have atleast three 3s and one 7...which can be divided completely by 189 (3x3x3x7)....right ? [/b]

I understand the part upto the breaking of the prime factors but I dont understand what you mean by we need 3 3s before we need 7?

ok.....for 189 you have prime factors 3x3x3x7....now for this to divide product from 1....n ....u have to have atleast go upto 7 numbers...right ? However, in those first 7 numbers , u will only get one 3 and another three from 6 (3x2)...u need another 3 as the denomintor 3x3x3x7 has to divide the product evenly...so the product shud go all the way upto 9 to get another 3.

product = 1x2x3x4x5x6(i.e 3x2)x7x8x9 (i.e. 3x3)....u see now u have atleast three 3s and one 7...which can be divided completely by 189 (3x3x3x7)....right ? [/b]

Honnestly the first explanation from MA was really great and banerjeea_98 finished the work in beauty Thanks guys, I totally understand now what I totally didn't understand 5mn ago....

Princeton and Kaplan are useless compare to your explanations