For every positive integer ‘n’ the h(n) is defined to be product of all even integer from 2 to n inclusive. If ‘P’ is the smallest prime factor of h(100)+1, then P is between ?
a) 3 and 10
b) 10 and 20
c) 20 and 30
d) 30 and 40
e) 40 and above.
Solve if you can !! A good GMAT Question by Ankit Gurgaon.
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Ans : E
Ans: A
A is the ans….the smallest prime factor is 2..
2+1 = 3…
so i guessed lik that….
dheeraj can u explain ur answer so that i can understand more…
Is this question tough? its a dumb question
The question asks for smallest prime factor of any number X. And amongst the answer choices there is only one prime factor listed. i.e. A.
This is a wrong question, there can be only one smallest…. Even if you have two solutions for an equation you say “X or Y” not “X and Y”
This question is asking for the smallest prime factor of (100 x 98 x … 4 x 2 + 1) = *********0…(at least 20 zeroes here)…01. So you need to find a factor of at least 10..(twenty zeroes here)..01, since this number is huge, the prime factor in itself will be greater than the range of all the options except for E. Therefore the answer is E and there is nothing wrong with the question/answers.
E is the correct choice.
Take Gmat team..
can u provide the explanation….
how come E?
h(100) + 1 or
(2 x 4 x 6…….98 x 100) + 1 will be an odd integer since we know h(100) is even.
we also know that h(100) is divisible by 3, 5, ….49 as h(100) is product of even integers and it contains 6, 10, 14…..98.
SO h(100) + 1 will not be divisible by 3, 5, 7, 9,…….49.
(Lets take for example 6 which is div, by 3, 6+1 will not be div. by 3. 6+3 will be div. by 3.
And so 15+5 is div. by 5.
And 7+7 is div. by 7)
Infact h(100) + 1 will be a prime no. But we do not need to go that far.
From the above we know that h(100) + 1 will have a prime factor P greater than 49 i.e. E(greater than 40)
Nice question…easy but tricky.
Good explainations Keshav
Thanks for the explanation Keshav…
good keshav…..
E
isay it is E becoz summation of x/2 from 2 to 100 taking even numbers as mentioned n is even then substituting x=2,4,6,8,10………100 then we will get a sequence of 1,2,3,4,5,6………..50 then using progression formula a+(n-1)d we get 1+(50-1)1=51 where 51 is least prime then adding 1 it will become even
Sartihajennila,
You lost me there. Seriously what are doing above exactly?
Saritha, We cannot employ progression formula here since it is not a summation problem. It’s asking for the product of even intergers between 2 and 100. Anyhow, I agree with Keshav’s explanation.
too gud expaltion
keep it up
When we are multiplying all numbers from 2 to 100, the result already contains the prime numbers upto 100 as factors. If we simply add 1, we now have an odd number which can not be divided by any of the numbers in 2 to 100 (because 1 is not a multiple of any of these numbers).
The smallest prime factor will be very large, possibly the number itself.
“E”
Sorry, I agree with Keshav. My explanation is not correct.
By the way, this is a question from new GMAT Powerprep software…
The smallest prime factor may not be the number itself…
P(10) + 1 = 3841 = 23 *167. But definitely its more than 10.
good xplanation Keshav..but I could not get how were u able to assume h(100)+1 as a prime number..any clarification behind that?
Has to be E….
The answer is E. 40 and above.
This question comes straight from the GMATPrep diagnostic test software available on http://www.mba.com.
When I first saw it some years back, it stumped me for literally 2 weeks. Daily, I went to bed thinking about it, and woke up thinking about it.
Then, of course, since it is a GMAT question, there is an elegant 2 minute answer, I figured it out.
The series (2 x 4 x 6 x . . . 96 x 98 x 100) can be factored as 2 x (1 x 2 x 3 x . . . 48 x 49 x 50); in other words, 2 x 50!.
Once you see that, you realize that 2 x 50! is a multiple of every prime factor less than 50, from 2 to 47.
Likewise, none of those prime factors of 2 x 50! is a factor of (2 x 50!) + 1 (Conversely, (2 x 50!) + 1 cannot be a multiple of any prime from 2 to 47.).
Therefore, the smallest prime factor of (2 x 50!) + 1 must be greater than 47. The correct answer choice is E.
Option E
i agree with keshav;s explanation gratm man
above email is fake but the answer is not
Answer is E
Choice A to D doesnt not contain prime numbers
therefore its answer is ultimately easy.
since these two numbers are consecutive integers thier HCF is 1. We know that if the HCF of two integers is 1, the prime factors of these numbers will be different.
Prime factors of h(100) include all prime numbers from 2- 47 as h(100) is divisible by 3, 5, ….49 since h(100) is product of even integers and it contains 6, 10, 14…..98.
Now h(100)+1 will have prime factors different from h(100)’s. Therefore h(100)+1 will have prime factors greater than 47.
I hope it helps.
Ans. E
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