kunalkulkarni wrote:For every positive integer n, the function f(n) is defined as the product of all even integers from 2 to n, inclusive. If m is the smallest prime factor of f(100) + 1, then value of m is:
A. between 2 and 10
B. between 10 and 20
C. between 20 and 30
D. between 30 and 40
E. greater than 40
Important Concept:
If k is a positive integer that’s greater than 1, and if k is a factor (divisor) of N, then k is not a factor of N+1
For example, since 7 is a factor of 350, we know that 7 is not a factor of (350+1)
Similarly, since 13 is a factor of 247, we know that 13 is not a factor of (247+1)
Now let’s examine f(100)
f(100) = (2)(4)(6)(8)….(96)(98)(100)
Factor to get: f(100) = 2[(1)(2)(3)(4)….(48)(49)(50)]
Since 2 is in the product of f(100), we know that 2 is a factor of f(100), which means that 2 is
not a factor of f(100)+1 (based on the above
rule)
Similarly, since 3 is in the product of f(100), we know that 3 is a factor of f(100), which means that 3 is
not a factor of f(100)+1 (based on the above
rule)
Similarly, since 5 is in the product of f(100), we know that 5 is a factor of f(100), which means that 5 is
not a factor of f(100)+1 (based on the above
rule)
.
.
.
.
Similarly, since 47 is in the product of f(100), we know that 47 is a factor of f(100), which means that 47 is
not a factor of f(100)+1 (based on the above
rule)
So, we can see that
none of the primes from 2 to 47 can be factors of f(100)+1, which means the smallest prime factor of f(100)+1 must be greater than 47.
Answer =
E
Cheers,
