These function questions are killing me. Usually only see one on the prep test but it's always early
Function h(n) is defined to be the sum of even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is:
a) between 2 and 10
b) between 10 and 20
c) between 20 and 30
d) between 30 and 40
e) greater than 40
The answer is E but I have no idea why.....
I thought it would be a smaller number just because we are looking for the smallest prime number but I guess that shows how much I know....lol
another function question
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Still trying to work this one...
The part I can't seem to figure out is how to find the sum consecutive even integers. I know the sum of consecutive integers is the average times the number of terms but the even part is messing me up. Here is where I am...
The average should be 100+2 / 2 = 51
The number of terms from A to B is B - A + 1 so B - A is 100-2 = 98, then divide by 2 since I'm only concerned with even numbers and add 1 which gives me 50 (49 + 1).
So now the sum would be 50 * 51 = 2550. But the function said F(100) + 1 so that equals 2551.
Now I think the question is asking for the smallest prime factor of 2551 and nothing seems to go into it so to me it seems like a prime number which makes the answer E.
Am I close??
The part I can't seem to figure out is how to find the sum consecutive even integers. I know the sum of consecutive integers is the average times the number of terms but the even part is messing me up. Here is where I am...
The average should be 100+2 / 2 = 51
The number of terms from A to B is B - A + 1 so B - A is 100-2 = 98, then divide by 2 since I'm only concerned with even numbers and add 1 which gives me 50 (49 + 1).
So now the sum would be 50 * 51 = 2550. But the function said F(100) + 1 so that equals 2551.
Now I think the question is asking for the smallest prime factor of 2551 and nothing seems to go into it so to me it seems like a prime number which makes the answer E.
Am I close??
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The only way to solve this question is first find out the sum of the all the consecutive even integers by using the arithmetic progression formula
Sum = 2551 as rightly pointed out...You have to divide 2551 by all the prime numbers less than 40 (3,5,7,9,11,13,17,19,23,29,31,37)....
Another way to find out if a number x is prime or not....find out sqr. rt x....if x is not divisible by any number between 1 and sqr rt. x...then x is prime (although even this is a little time consuming!!)
Sum = 2551 as rightly pointed out...You have to divide 2551 by all the prime numbers less than 40 (3,5,7,9,11,13,17,19,23,29,31,37)....
Another way to find out if a number x is prime or not....find out sqr. rt x....if x is not divisible by any number between 1 and sqr rt. x...then x is prime (although even this is a little time consuming!!)
- gabriel
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.. allen r u sure u have the q right ... not that i have a poblem with this q but i remeber to have answered a identical q .. but in that the fun h(n) was equal the multiplication of all numbers from 2 to n .. and the answer to that one was also E (that is greater than 40) ... moreover i dont think gmat wuld just giv u a number and ask to find the smallest prime number ...