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# If n is a positive integer and the product of all the

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VP
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If n is a positive integer and the product of all the [#permalink]

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30 Jan 2005, 18:29
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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?

A. 8
B. 9
C. 10
D. 11
E. 12

Please explain the way to find it.
Thanks.
VP
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30 Jan 2005, 19:42
"D".

convert 990 in prime factors = 11x3x3x5x2

u will see that for the product to be divisble by 990....product shud atleast contain a 11.....so n shud go upto 11 .....so least n = 11
SVP
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30 Jan 2005, 20:22
What if the number is 189 instead of 990?
VP
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30 Jan 2005, 20:30
HongHu wrote:
What if the number is 189 instead of 990?

good one buddy....shud be "9".
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30 Jan 2005, 21:17
You got it.
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31 Jan 2005, 09:06
banerjeea_98 wrote:
HongHu wrote:
What if the number is 189 instead of 990?

good one buddy....shud be "9".

I just have a quick question, why is it 9 and not 7?

I was wondring because 189=7*3*3*3 so I thought the answer was going to be 7.
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31 Jan 2005, 09:32
From 1 to seven you've only got two 3s, one from 3, another from 6, but you need three 3s. That's why I used this number 189, to set the trap.
Director
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31 Jan 2005, 10:07
HongHu wrote:
From 1 to seven you've only got two 3s, one from 3, another from 6, but you need three 3s. That's why I used this number 189, to set the trap.

Honghu, can you please explain in detail how u solved the question. thanks.
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Re: product and integer n [#permalink]

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31 Jan 2005, 11:32
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?

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?

A. 8
B. 9
C. 10
D. 11
E. 12

Please explain the way to find it.
Thanks.
VP
Joined: 25 Nov 2004
Posts: 1486
Followers: 7

Kudos [?]: 104 [0], given: 0

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31 Jan 2005, 12:06
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.
Director
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31 Jan 2005, 13:13
thank you MA. thats a very good explanation indeed.
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31 Jan 2005, 13:22
can you explain what you understood? I still have a hard time following...

vprabhala wrote:
thank you MA. thats a very good explanation indeed.
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31 Jan 2005, 13:25
MA wrote:
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.

Oh!!! Now I get it! Thanks for the explination!
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31 Jan 2005, 14:48
Current Student
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31 Jan 2005, 19:03
guys explain this to me please!

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?
VP
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31 Jan 2005, 19:43
fresinha12 wrote:
guys explain this to me please!

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]
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31 Jan 2005, 20:11
got it....thanks!

banarjee nice explanation...

banerjeea_98 wrote:
fresinha12 wrote:
guys explain this to me please!

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]
VP
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01 Feb 2005, 21:25
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
VP
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02 Feb 2005, 00:05
Thanks guys. i am truly overwhelmed.........
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04 Feb 2005, 12:51
D as only way to have 990 is to have 11 which is prime
04 Feb 2005, 12:51
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