The function f(n) is defined as the product of all integers from 1 to

Question

The function f(n) is defined as the product of all integers from 1 to n, inclusive, and the function g(n) is defined as the product of all odd integers from 1 to n, inclusive, where n is a positive integer. If p is a prime factor of \(\frac{f(150)}{g(150)} + 1\), then which of the following must be true

A. p < 10
B. 10 < p < 25
C. 25 < p < 50
D. 50 < p < 75
E. p > 75

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e-GMAT Question of the Week #23

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Official Answer

E

A. p < 10
B. 10 < p < 25
C. 25 < p < 50
D. 50 < p < 75
E. p > 75

e-GMAT          Question of the Week #23

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pandeyashwin · Accepted Answer

Let k = f(n)/g(n) + 1
product of all integers / product of odd integers = product of even integers

Question is asking us to find the product of all even integers between 1 and 150.

k = 2x4x6x...150.
k = 2^75(1*2*3*...75)

Now k+1 and k are co prime (two consecutive no's are co prime) . So k+1 won't have any factors between 2-75 inclusive

So p > 75

pandeyashwin · Answer

i hope this will be clearer

f(150) = 1x2x3x4x5x....150 (product of all integers between 1-150 inclusive)
g(150) = 1x3x5x7x9...149 (product of all odd integers between 1-150)

\frac{f(150)}{g(150)}  = 2x4x6x8x10....150

=> 2^75 (1*2*3*4*...75)

Prime Factor for  \frac{f(150)}{g(150)}  = all prime factors from 1 to 75

Since two consecutive terms have no prime factors common , Prime Factor for  \frac{f(150)}{g(150)}+ 1  will not have any term between 1 and 75

Therefore p>75

keatross · Answer

Going to need some help on this one for sure (and I do not understand the solution posted)....

I tried to find a pattern between the sizes of f(n) and g(n) given their properties. Didn't really result in much I could use.

I did, however, notice that g(n) will have to skip half of the #s from 1-150, which is 75. Using this I kind of theorized that f(n) will be 75(x all even numbers) times larger than g(n).   was my guess as a result, but it was a complete shot in the dark.

EgmatQuantExpert · Answer

Solution 
  Given:  
 • f(n) = 1 * 2 * 3 * 4 * … * (n - 1) * n
• g(n) = 1 * 3 * 5 * 7 * … * (n - 1), if n is even, or g(n) = 1 * 3 * 5 * 7 * … * n, if n is odd
• p is a prime factor of  \frac{f(150)}{g(150)} + 1

To find:  
 • Which of the given options is always true?

Approach and Working:   
 • f(150) = 1 * 2 * 3 * 4 * … * 149 * 150
• g(150) = 1 * 3 * 5 * 7 * … * 149
• Implies,  \frac{f(150)}{g(150)} = 2 * 4 * 6 * 8 * … * 148 * 150 = 2^{75} * (1 * 2 * 3 * 4 * … * 75) = 2^{75} * 75! 
 o Let us assume that  2^{75} * 75! = N

• So,  \frac{f(150)}{g(150)} + 1 = 2^{75} * 75! + 1 = N + 1 
• Now, we know that any two consecutive integers are co-primes
 o Thus, N and N + 1 are co-primes
o If we observe,  N = 2^{75} * 75! , which is a multiple of all the primes from 1 to 75,
o So, all the prime factors from 1 to 75 are not the factors of N + 1, as N and N + 1 are co-primes

• Therefore, we can say that the prime factor of N + 1 must be greater than 75

Hence the correct answer is Option E.

Answer: E

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aarushisingla · Answer

Hi, 
Could you please explain the highlighted part as I am unable to understand the same

GMATGuruNY · Answer

Since the difference between them is 1,  \frac{f(150)}{g(150)}  and  \frac{f(150)}{g(150)} + 1  are consecutive integers.
Consecutive integers are COPRIMES: they share no factors other than 1.

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.
Thus, 1 is the ONLY factor that x and x+1 have in common.
In other words:
x and x+1 are COPRIMES.

Implication:
 \frac{f(150)}{g(150)}  and  \frac{f(150)}{g(150)} + 1  are COPRIMES.
They share no factors other than 1.

\frac{f(150)}{g(150)} =

\frac{1*2*3*4*5..146*147*148*149*150}{1*3*5...145*147*149}

= 2*4*6...146*148*150

= (2*1)(2*2)(2*3)...(2*73)(2*74)(2*75)

= 2^{75}(1 * 2 * 3 *... * 73 * 74 * 75)

Looking at the set of parentheses on the right, we can see that every prime number between 1 and 75 is a factor of  \frac{f(150)}{g(150)} .
Since  \frac{f(150)}{g(150)}  and  \frac{f(150)}{g(150)}+1  are coprimes, NONE of the prime numbers between 1 and 75 can be a factor of  \frac{f(150)}{g(150)}+1 .
Thus:
For p to be a prime factor of  \frac{f(150)}{g(150)}+1 , it must be GREATER THAN 75.

E

Prince1890Sharma · Answer

Can someone explain, when we are saying two numbers are coprime why are we considering the answer choice as E....How it will be greater than 75. @e-GMAT

Bunuel · Answer

Since f(150) is the product of all integers from 1 to 150, inclusive, then f(150) = 1*2*3*...*150. Since g(150) is the product of all odd integers from 1 to 150, inclusive, then g(150) = 1*3*5*...*149. Thus, f(150)/g(150) will be the product of all even integers from 1 to 150, inclusive: 2*4*...*150 = 2^75(1*2*...*75) Now, 2^75(1*2*...*75) and 2^75(1*2*...*75) + 1 are two consecutive integers, which means that they do not share any factor other than 1. For example, 20 and 21, being consecutive integers, do not share any factor other than 1 (they are co-prime). Since 2^75(1*2*...*75) will have all primes less than 75 in it, then 2^75(1*2*...*75) + 1 won't have any primes less than 75. Therefore, the smallest prime of 2^75(1*2*...*75) + 1, whatever it is must be greater than 75. Answer: E. Similar questions to practice: https://gmatclub.com/forum/for-every-po ... 26691.html https://gmatclub.com/forum/for-every-po ... 49722.html https://gmatclub.com/forum/x-is-the-pro ... 56545.html https://gmatclub.com/forum/if-n-is-a-po ... 44553.html https://gmatclub.com/forum/for-every-ev ... 68636.html https://gmatclub.com/forum/for-any-inte ... 12494.html https://gmatclub.com/forum/if-a-and-b-a ... 44714.html https://gmatclub.com/forum/the-function ... 08309.html https://gmatclub.com/forum/for-every-po ... 81815.html https://gmatclub.com/forum/for-any-inte ... 31701.html https://gmatclub.com/forum/let-p-be-the ... 32329.html https://gmatclub.com/forum/does-the-int ... 26735.html https://gmatclub.com/forum/if-x-is-an-i ... 00670.html https://gmatclub.com/forum/if-n-20-17-t ... 43638.html Hope it helps.

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The function f(n) is defined as the product of all integers from 1 to

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