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A certain number N is divisible by the product of the first fourteen

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A certain number N is divisible by the product of the first fourteen  [#permalink]

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New post 25 Apr 2016, 23:58
2
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A
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E

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Question Stats:

63% (02:24) correct 37% (02:03) wrong based on 95 sessions

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A certain number N is divisible by the product of the first fourteen positive integers. Which of the following must be a factor of N?

A. 1,690
B. 1,210
C. 726
D. 625
E. 616

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Re: A certain number N is divisible by the product of the first fourteen  [#permalink]

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New post 26 Apr 2016, 02:41
1
Bunuel wrote:
A certain number N is divisible by the product of the first fourteen positive integers. Which of the following must be a factor of N?

A. 1,690
B. 1,210
C. 726
D. 625
E. 616


So... n is a multiple of 14! So prime numbers till 14 have occurred only once.

a) 1690 .... two 13s; out
b) 1210 .. two 11s; out
D) 625 ... 25^2 ... four 5s... but we have got only 2 ( 5 and 10)
c) 726 = 6 * 121 = two 11s (only 1 there in 14!)

Without even calculating, I can easily mark E.

But still, for peace of mind : 616 = 8 * 77 = 8*7*11 ( we have all of them)
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Re: A certain number N is divisible by the product of the first fourteen  [#permalink]

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New post 26 Apr 2016, 07:56
Bunuel wrote:
A certain number N is divisible by the product of the first fourteen positive integers. Which of the following must be a factor of N?

A. 1,690
B. 1,210
C. 726
D. 625
E. 616


The number is 14!

We are asked to the factor of the number....

Check the options -

A. 1,690 = 2^1 x 5^1 x 13^2 ; 14! will have only one 13 , so 1690 can't be a factor of 14!

B. 1,210 = 2^1 x 5^1 x 11^2; 14! will have only one 11 , so 1210 can't be a factor of 14!

C. 726 = 2^1 x 3^1 x 11^2; 14! will have only one 11 , so 726 can't be a factor of 14!

D. 625 = 5^4

No of 5's in 14! is 2 , so 625 can't be a factor of 14!

14/5 = 2
2/5 = 0

E. 616 = 2^3 x 7^1 x 11^1

14! will have 11 two's

14/2 = 7
7/2 = 3
3/2 = 1

Hence answer will be (E)
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Re: A certain number N is divisible by the product of the first fourteen  [#permalink]

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New post 02 May 2016, 19:38
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I have solved many Questions of this kind.
A general helping tip => Start with E .
Also i read a post on the MGMAT forum that said => in elimination questions the order - B,D,E,A,C is the way to go.
though i still Dont know why that is...! (maybe its a probability thing )
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Re: A certain number N is divisible by the product of the first fourteen  [#permalink]

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New post 07 May 2016, 12:30
Basically the problem is finding the factor of 14!
1690 = 2*5*13*13 (two 13's eliminate)
1210 = 2*5*11*11 (two 11's eliminate)
726 = 2*3*11*11 (two 11's eliminate)
625 = 5*5*5*5 (only two 5's in 14!)
616 = 2*2*2*7*11(all factors are present in 14!)
correct answer - E
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A certain number N is divisible by the product of the first fourteen  [#permalink]

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New post 25 Aug 2018, 17:20
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Bunuel VeritasKarishma chetan2u gmatbusters niks18 GMATPrepNow

I understood above approaches, but almost all above methods requires prime factorization of
three digits numbers and consumed more time. Any alternate approach to solve this efficiently?
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A certain number N is divisible by the product of the first fourteen  [#permalink]

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New post 25 Aug 2018, 17:55
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adkikani wrote:
Bunuel VeritasKarishma chetan2u gmatbusters niks18 GMATPrepNow

I understood above approaches, but almost all above methods requires prime factorization of
three digits numbers and consumed more time. Any alternate approach to solve this efficiently?


Hi adkikani

N is divisible by 14!. 14! contains only one 11 & 13 and N has to be even because 14! is even

Now look at the option

1690 = 169*10, 169 is square of 13 hence this will contain two 13's. therefore not a factor of N
1210 = 121*10, 121 is square of 11 hence this will contain two 11's. therefore not a factor of N
625 = 625 is square of 25 hence this will contain two 25's. therefore not a factor of N

726, is divisible by 11 (remember the divisibility rule for 11), the resulting number will again be divisible by 11. hence this also contains two or more 11s. Reject this

So your final answer has to be Option E.

If you remember the square and divisibility rule then this question can be solved within 30s by just skimming at the options
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Re: A certain number N is divisible by the product of the first fourteen  [#permalink]

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New post 25 Aug 2018, 18:03
1
adkikani wrote:
Bunuel VeritasKarishma chetan2u gmatbusters niks18 GMATPrepNow

I understood above approaches, but almost all above methods requires prime factorization of
three digits numbers and consumed more time. Any alternate approach to solve this efficiently?


Hi

Since question involves product and then asks us for the choice that has to be a factor, it will require prime factorization and I doubt you can do this without Getting into factorization.
If you know certain squares as also suggested above, it can avoid complete factorization, but yes you have to factor the choices.
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Re: A certain number N is divisible by the product of the first fourteen &nbs [#permalink] 25 Aug 2018, 18:03
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