dude is the answer ael_torero wrote:this question comes from the GMAT practice test software from mba.com. good luck.
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div and primes question
Source: Beat The GMAT — Problem Solving |
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albatross86
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h(n) = Product of all even integers from 2 to n, inclusive.
p = smallest prime factor of h(100) + 1. p is?
h(100) = 2*4*6*8*....*100
How many even numbers are there here? = 100/2 = 50 even numbers
Take a factor of 2 out from each of these even numbers:
h(100) = 2^50 * (1*2*3*4*....*50) = 2^50*50!
=> h(100) + 1 = 2^50*50! + 1
Now, dividing this expression by any number from 1 to 50 will always yield a remainder of 1, since 2^50*50! is divisible by every number from 1 to 50 (since it contains 50! )
Thus, no number from 1 to 50 can be a factor of h(100) + 1.
Thus, the smallest prime factor must also be greater than 50. Pick E.
p = smallest prime factor of h(100) + 1. p is?
h(100) = 2*4*6*8*....*100
How many even numbers are there here? = 100/2 = 50 even numbers
Take a factor of 2 out from each of these even numbers:
h(100) = 2^50 * (1*2*3*4*....*50) = 2^50*50!
=> h(100) + 1 = 2^50*50! + 1
Now, dividing this expression by any number from 1 to 50 will always yield a remainder of 1, since 2^50*50! is divisible by every number from 1 to 50 (since it contains 50! )
Thus, no number from 1 to 50 can be a factor of h(100) + 1.
Thus, the smallest prime factor must also be greater than 50. Pick E.
~Abhay
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Believe those who are seeking the truth. Doubt those who find it. -- Andre Gide
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GMATGuruNY
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Here's the number property rule that's being tested with this problem:
If x is a positive integer, the only factor common to x and x+1 is 1; they share no other factors. Any factor of x (other than 1) will NOT be a factor of x+1.
Let's think about why this rule holds true.
If x is a multiple of 2, how much do we need to add to get to the next largest multiple of 2? 2. So the next largest multiple of 2 will be x+2.
If x is a multiple of 3, how much do we need to add to get to the next largest multiple of 3? 3. So the next largest multiple of 3 will be x+3.
If x is a multiple of 4, how much do we need to add to get to the next largest multiple of 4? 4. So the next largest multiple of 4 will be x+4.
Using this logic, if we add 1 to x, we get only to the next largest multiple of 1. So 1 is the only factor common to both x and x+1.
Thus, in the problem above, we know that 1 is the only factor common to h(100) and h(100) + 1. They share no other factors.
h(100) = 2 * 4 * 6 *....* 94 * 96 * 98 * 100
If from each of the 50 factors listed above we factor out 2, we get:
h(100) = 2^50 (1 * 2 * 3 *... * 47 * 48 * 49 * 50)
Looking at the set of parentheses on the right, we can see that every prime number between 1 and 50 is a factor of h(100). This means that none of the prime numbers between 1 and 50 can be a factor of h(100) + 1, because h(100) and h(100) + 1 share no factors other than 1.
So the smallest prime factor of h(100) + 1 must be greater than 50.
The correct answer is E.
If x is a positive integer, the only factor common to x and x+1 is 1; they share no other factors. Any factor of x (other than 1) will NOT be a factor of x+1.
Let's think about why this rule holds true.
If x is a multiple of 2, how much do we need to add to get to the next largest multiple of 2? 2. So the next largest multiple of 2 will be x+2.
If x is a multiple of 3, how much do we need to add to get to the next largest multiple of 3? 3. So the next largest multiple of 3 will be x+3.
If x is a multiple of 4, how much do we need to add to get to the next largest multiple of 4? 4. So the next largest multiple of 4 will be x+4.
Using this logic, if we add 1 to x, we get only to the next largest multiple of 1. So 1 is the only factor common to both x and x+1.
Thus, in the problem above, we know that 1 is the only factor common to h(100) and h(100) + 1. They share no other factors.
h(100) = 2 * 4 * 6 *....* 94 * 96 * 98 * 100
If from each of the 50 factors listed above we factor out 2, we get:
h(100) = 2^50 (1 * 2 * 3 *... * 47 * 48 * 49 * 50)
Looking at the set of parentheses on the right, we can see that every prime number between 1 and 50 is a factor of h(100). This means that none of the prime numbers between 1 and 50 can be a factor of h(100) + 1, because h(100) and h(100) + 1 share no factors other than 1.
So the smallest prime factor of h(100) + 1 must be greater than 50.
The correct answer is E.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
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Stuart@KaplanGMAT
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Seriously, this is the #1 question posted on this site.
Try a search on h(100), you'll find dozens of threads, almost all of which have solutions posted by different experts.
Learn to love the search function!
Try a search on h(100), you'll find dozens of threads, almost all of which have solutions posted by different experts.
Learn to love the search function!
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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