EgmatQuantExpert wrote:
e-GMAT Question of the Week #23The 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
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.
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