GMATinsight wrote:
What is the biggest positive integer which will always divide n(n+1)(n+2) evenly for any even integer value of n?
a) 2
b) 3
c) 6
d) 12
e) 24
Since the product of m consecutive integers is divisible by m!, and since n(n + 1)(n + 2) represents a product of 3 consecutive integers, n(n + 1)(n + 2) is divisible by 3! = 6. However, we are also given that n is an even integer. In that case, we can express n as 2k where k is a integer. Therefore, we have
n(n + 1)(n + 2)
2k(2k + 1)(2k + 2)
2k(2k + 1)[2(k + 1)]
4k(k + 1)(2k + 1)
Notice that k(k + 1) is a product of 2 consecutive integers, therefore, k(k + 1) is divisible by 2! = 2 and hence n(n + 1)(n + 2) = 4k(k + 1)(2k + 1) is divisible by 4 x 2 = 8.
Recall that if a number is divisible by a and b, then it will also divisible by the LCM of a and b. Here we have n(n + 1)(n + 2) is divisible by 6 and 8; therefore, it’s also divisible by the LCM of 6 and 8, which is 24.
Alternate Solution:
Since n is even, n + 2 is also even and furthermore, since n and n + 2 are consecutive even numbers, one of them must be divisible by 4. Since either n or n + 2 is divisible by 4 and the other number is even, we have at least three factors of 2, which means the product of n and n + 2 is divisible by 8.
Since we have three consecutive integers, precisely one of these integers must be divisible by 3.
Since n(n + 1)(n + 2) is divisible by both 3 and 8, it is divisible by their LCM, which is 24.
Answer: E
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