PASSINGGMAT wrote:
The function f(m) is defined for all positive integers m as the product of m + 4, m + 5, and m + 6. If n is a positive integer, then f(n) must be divisible by which one of the following numbers?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 11
Can you explain how to answer this please. Thank you.
Given: \(f(n)=(n+4)(n+5)(n+6)\), where \(n\) is a positive integer. Question: \(f(n)\)
must be divisible by which one of the following numbers.
Now, \((n+4)(n+5)(n+6)\) is the product of 3 consecutive integers so out of them one is definitely divisible by 3 and at least one is divisible by 2, so \(f(n)\) must be divisible by 2*3=6.
Answer: C.
Generally out of ANY \(k\) consecutive integers one is always divisible by \(k\) and at least one by \(k-1\), \(k-2\), ... For example out of ANY 5 consecutive integers there is one which is divisible by 5, and at least one which is divisible by 4, 3, and 2. That's because an integer divided by an integer \(k\) can give a remainder of: 0 (when it's divisible by \(k\)), 1, 2, ..., or \(k-1\) (total of \(k\) different remainders from 0 to \(k-1\)), so out of \(k\) consecutive integers there definitely will be one which gives a reminder of zero, so divisible by \(k\).
Which give us the following property: the product of \(k\) consecutive integers is always divisible by \(k!\), so by \(k\) too. For example: given \(k=4\) consecutive integers \(\{3,4,5,6\}\) --> the product of 3*4*5*6 is 360, which is divisible by 4!=24.
If we apply this property to the original question we'll have that the product of given 3 consecutive integers \((n+4)(n+5)(n+6)\) must be divisible by 3!=6.
Hope it's clear.