Solve if you can !! A good GMAT Question by Ankit Gurgaon.

Keshav Malhotra
#1. January 15th, 2008, at 5:06 PM.

Ans : E

Dheeraj
#2. January 16th, 2008, at 4:27 PM.

Ans: A

john
#3. January 17th, 2008, at 11:55 AM.

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…

simplyjat
#4. January 17th, 2008, at 1:16 PM.

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”

Emleung
#5. January 18th, 2008, at 11:54 PM.

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.

Dheeraj
#7. January 19th, 2008, at 10:00 AM.

Take Gmat team..
can u provide the explanation….
how come E?

Keshav
#8. January 19th, 2008, at 2:04 PM.

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)

ABP
#9. January 19th, 2008, at 11:27 PM.

Nice question…easy but tricky.

Meenakshi
#10. January 20th, 2008, at 3:19 PM.

Good explainations Keshav

Yogi
#11. January 21st, 2008, at 9:36 AM.

Thanks for the explanation Keshav…

john
#12. January 21st, 2008, at 4:37 PM.

good keshav…..

sarithajennila
#13. January 22nd, 2008, at 3:06 PM.

E

sarithajennila
#14. January 22nd, 2008, at 3:26 PM.

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

Johnkenendy10
#15. March 12th, 2008, at 11:30 AM.

Sartihajennila,

You lost me there. Seriously what are doing above exactly?

Deeks
#16. March 21st, 2008, at 2:26 AM.

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.

shailza
#17. April 3rd, 2008, at 9:26 AM.

too gud expaltion
keep it up

RD
#18. April 3rd, 2008, at 10:58 AM.

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”

RD
#19. April 3rd, 2008, at 11:00 AM.

Sorry, I agree with Keshav. My explanation is not correct.

sumanth toom
#20. April 14th, 2008, at 12:19 PM.

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.