From one of the the Prep materials: I am unable to solve this. Can someone please explain? I will reveal the solution once I have a few answers, else it will kill the thrill.
For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, the p is
Options:
A: Between 2 & 10
B: Between 10 & 20
C: Between 20 & 30
D: Between 30 & 40
E: Greater than 40
Quant - PS - Arithmetic
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h(100) = 2 * 4 * 6 * ... * 100polter wrote:From one of the the Prep materials: I am unable to solve this. Can someone please explain? I will reveal the solution once I have a few answers, else it will kill the thrill.
For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, the p is
Options:
A: Between 2 & 10
B: Between 10 & 20
C: Between 20 & 30
D: Between 30 & 40
E: Greater than 40
= (2 * 1) * (2 * 2) * (2 * 3) * ... * (2 * 50)
= 2^(50) * (1 * 2 * 3 ... * 50)
Then h(100) + 1 = 2^(50) * (1 * 2 * 3 ... * 50) + 1
Now, h(100) + 1 cannot have any prime factors 50 or below, because dividing this value by any of these prime numbers will give a remainder of 1.
So, the correct answer is E.
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h(100) = 2 * 4 * 6 .... * 100
h(100) = 2*1 * 2*2 2*3 ...2 * 50.
This means h(100) has all numbers from 1 to 50 as a factor. So h(100) + 1 cannot have any of these numbers as its factor. So any factor of h(100)+1 is greater than 50. hence greater than 40.
So my answer is E.
h(100) = 2*1 * 2*2 2*3 ...2 * 50.
This means h(100) has all numbers from 1 to 50 as a factor. So h(100) + 1 cannot have any of these numbers as its factor. So any factor of h(100)+1 is greater than 50. hence greater than 40.
So my answer is E.
Vineesh,
Just telling you what I know and think. I am not the expert.
Just telling you what I know and think. I am not the expert.
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Since the difference between them is 1, h(100) and h(100)+1 are consecutive integers. Consecutive integers are co-primes: they share no factors other than 1.polter wrote:From one of the the Prep materials: I am unable to solve this. Can someone please explain? I will reveal the solution once I have a few answers, else it will kill the thrill.
For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, the p is
Options:
A: Between 2 & 10
B: Between 10 & 20
C: Between 20 & 30
D: Between 30 & 40
E: Greater than 40
Let's examine why:
If x is a multiple of 2, the next largest multiple of 2 is x+2.
If x is a multiple of 3, the next largest multiple of 3 is x+3.
Using this logic, if we go from x to x+1, we get only to the next largest multiple of 1. So 1 is the only factor common both to x and to x+1. Integers that share no factors other than 1 are called coprimes.
Thus, in the problem above, h(100) and h(100)+1 are coprimes. They share no factors other than 1.
h(100) = 2 * 4 * 6 *....* 94 * 96 * 98 * 100
Factoring 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). Since h(100) and h(100)+1 are coprimes, none of the prime numbers between 1 and 50 can be a factor of h(100)+1.
Thus, 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].
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