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

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). [E] was my guess as a result, but it was a complete shot in the dark.

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) [take 2 common from all the terms. There are 75 terms therefore 2^75]

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

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

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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