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lonesome307
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Gmat Prep PS 19 Sep 2007, 22:01
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Hi All,

Can someone please help me solve the following question?

Thanks



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GMATBLACKBELT
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Gmat Prep PS 19 Sep 2007, 22:47
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Im really confused with this question as well. Ok so h(n) is the product of all the even integers from 2-n.

n=100 so all the evens from 2-100. 100-2+1=99. Now 98/2=49+1 = 50 evens from 100-2.

so these (50 evens* eachother) + 1. This is going to be an odd number...

if p is the smallest prime factor... I don't know how to find the product of all the even numbers in a situation like this.

I can find sum... how do u find the product? And then from there, how do u find out whether the prime factor is greater than 40????
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Gmat Prep PS 20 Sep 2007, 10:25
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h(n)= 2*4*6*8*...*100 + 1

we can rephrase it to h(n)= 2^100*(1*2*3*4*....*50) + 1

so there is a smallest prime factor p of h(n),

if you take a look to above expression h(n)= 2^100*(1*2*3*4*....*50) + 1
you will see that there are all prime numbers from 1 to 50 included into product.

What that gives us? If you multiply a prime number to 100 (in our case it is more than 100, probably ends with more than 6 zeros) and add 1 then you will see that it will not be any more prime factor of the number (for example, 37, 3700 and 37001, 37 is not prime factor any more). So as we mentioned above there are all prime factors up to 50 in product of numbers above, so that means we don't have LEAST PRIME FACTOR up to fifty (or to be exact up to 47).

If we don't have 47 as an least prime factor than it should be larger than 47, so the asnwer is E, greater tahn 40.

Ans: E

It is easy to find a solution to the question when you have an answer, so next time pls do not provide the answer with question in the beginning :)
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Gmat Prep PS 21 Sep 2007, 10:42
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Ferihere
h(n)= 2*4*6*8*...*100 + 1

we can rephrase it to h(n)= 2^100*(1*2*3*4*....*50) + 1
so there is a smallest prime factor p of h(n),

if you take a look to above expression h(n)= 2^100*(1*2*3*4*....*50) + 1
you will see that there are all prime numbers from 1 to 50 included into product.

What that gives us? If you multiply a prime number to 100 (in our case it is more than 100, probably ends with more than 6 zeros) and add 1 then you will see that it will not be any more prime factor of the number (for example, 37, 3700 and 37001, 37 is not prime factor any more). So as we mentioned above there are all prime factors up to 50 in product of numbers above, so that means we don't have LEAST PRIME FACTOR up to fifty (or to be exact up to 47).

If we don't have 47 as an least prime factor than it should be larger than 47, so the asnwer is E, greater tahn 40.

Ans: E

It is easy to find a solution to the question when you have an answer, so next time pls do not provide the answer with question in the beginning :)
Show more


Ferihere:
I am just not getting the one marked in red . Could you elaborate.Thanks
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Gmat Prep PS 22 Sep 2007, 07:05
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ashkrs
Ferihere
h(n)= 2*4*6*8*...*100 + 1

we can rephrase it to h(n)= 2^100*(1*2*3*4*....*50) + 1
so there is a smallest prime factor p of h(n),

if you take a look to above expression h(n)= 2^100*(1*2*3*4*....*50) + 1
you will see that there are all prime numbers from 1 to 50 included into product.

What that gives us? If you multiply a prime number to 100 (in our case it is more than 100, probably ends with more than 6 zeros) and add 1 then you will see that it will not be any more prime factor of the number (for example, 37, 3700 and 37001, 37 is not prime factor any more). So as we mentioned above there are all prime factors up to 50 in product of numbers above, so that means we don't have LEAST PRIME FACTOR up to fifty (or to be exact up to 47).

If we don't have 47 as an least prime factor than it should be larger than 47, so the asnwer is E, greater tahn 40.

Ans: E

It is easy to find a solution to the question when you have an answer, so next time pls do not provide the answer with question in the beginning :)

Ferihere:
I am just not getting the one marked in red . Could you elaborate.Thanks
Show more


There are 2 to 100 even numbers, so that means all of them divisible by 2. You can rewrite them as (2*1)*(2*2)*(2*3)*(2*4)*....*(2*50) + 1
as you can see there are 50 2's, that is why 2^50 (I wrote it down as 2^100, but still it will not change the asnwer) and other numbers just 1*2*3*4*....*50 so we can rewrite it 2^50*(1*2*3*4*....*50)+1

Is it clear?
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Gmat Prep PS 24 Sep 2007, 10:35
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Ferihere
ashkrs
Ferihere
h(n)= 2*4*6*8*...*100 + 1

we can rephrase it to h(n)= 2^100*(1*2*3*4*....*50) + 1
so there is a smallest prime factor p of h(n),

if you take a look to above expression h(n)= 2^100*(1*2*3*4*....*50) + 1
you will see that there are all prime numbers from 1 to 50 included into product.

What that gives us? If you multiply a prime number to 100 (in our case it is more than 100, probably ends with more than 6 zeros) and add 1 then you will see that it will not be any more prime factor of the number (for example, 37, 3700 and 37001, 37 is not prime factor any more). So as we mentioned above there are all prime factors up to 50 in product of numbers above, so that means we don't have LEAST PRIME FACTOR up to fifty (or to be exact up to 47).

If we don't have 47 as an least prime factor than it should be larger than 47, so the asnwer is E, greater tahn 40.

Ans: E

It is easy to find a solution to the question when you have an answer, so next time pls do not provide the answer with question in the beginning :)

Ferihere:
I am just not getting the one marked in red . Could you elaborate.Thanks

There are 2 to 100 even numbers, so that means all of them divisible by 2. You can rewrite them as (2*1)*(2*2)*(2*3)*(2*4)*....*(2*50) + 1
as you can see there are 50 2's, that is why 2^50 (I wrote it down as 2^100, but still it will not change the asnwer) and other numbers just 1*2*3*4*....*50 so we can rewrite it 2^50*(1*2*3*4*....*50)+1

Is it clear?
Show more


Very nice solution, I didn't understand this approach til just now :P



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