If k and m are integers and k = m*(m+4)*(m+5), k must be divisible by

if m is divided by 3, so the expression is divided by 3
if m is not dividec by 3, m+1 or m+2 is divided by 3, and m+4 or m+5 is divided by 3.
in conclusion, the expression is divided by 3

m+4 and m+5 are consecutive numbers , so, one of them must be divided by 2
so, the expression is divided by 3

for 8
if m is divided by 2, but not divided by 4, m+4 is divided by 2. and m+7 is odd. so the expression is not divided by 8

answer is b.

if m is divided by 3, so the expression is divided by 3
if m is not dividec by 3, m+1 or m+2 is divided by 3, and m+4 or m+5 is divided by 3.
in conclusion, the expression is divided by 3

m+4 and m+5 are consecutive numbers , so, one of them must be divided by 2
so, the expression is divided by 3

for 8
if m is divided by 2, but not divided by 4, m+4 is divided by 2. and m+7 is odd. so the expression is not divided by 8

answer is b.

Roman Numeral I:

If m is 1, then k = 1 x 5 x 6.

If m is 2, k = 2 x 6 x 7.

If m is 3, k = 3 x 7 x 9.

In all three cases, we see that m*(m+4)*(m+5) must be divisible by 3. A more dynamic proof to show that k is divisible by 3 is to let m = 3q + r where q is an integer and r = 0, 1 or 2.

If m = 3q, obviously k is divisible by 3 since m is a multiple of 3.

If m = 3q + 1, then k is divisible by 3 since m + 5 = 3q + 6 is a multiple of 3.

If m = 3q + 2, then k is divisible by 3 since m + 4 = 3q + 6 is a multiple of 3.

Hence, k is always divisible by 3, regardless of the value of integer m.

Roman Numeral II:

We observe that m + 4 and m + 5 are two consecutive integers, hence, one of them is necessarily even. We know that k is even and from Roman Numeral I, we also know that k is divisible by 3. Thus, k must be divisible by 6.

Roman Numeral III:

We notice that m and m + 4 have the same parity (they are both odd or both even) and the parity of m + 5 is opposite to that of m (if m is even, m + 5 is odd and if m is odd, m + 5 is even). When we have a product of three integers, the product can be divisible by 8 if a) one of the integers is divisible by 8, b) one of the integers is divisible by 4 and one of the integers is divisible by 2 or c) each of the integers is divisible by 2.Since the parity of m + 5 is opposite to the remaining integers, it is impossible that each of m, m + 4 and m + 5 are divisible by 2. So if we pick a value for m such that neither of m, m + 4 and m + 5 are divisible by 4, the product cannot be divisible by 8. Such a value is m = 1, in which case k = 1*5*6 = 30. As we can see, k is not necessarily divisible by 8.

Answer: B

i understand that k is even and that k is divisible by 3 but dont follow why is this div by 6.

Also, any particular reason why you chose m as 1 2 3?

thanks

i understand that k is even and that k is divisible by 3 but dont follow why is this div by 6.

Also, any particular reason why you chose m as 1 2 3?

thanks[/quote]

For the first one, we are using the rule that if x is divisible by n and m, then x is divisible by LCM(n, m). If k is even (i.e. divisible by 2) and k is divisible by 3, then k has to be divisible by LCM(2, 3) = 6.

For the second question, the values m = 1, 2 and 3 correspond to the simplest special cases of the more general argument to follow. It's sometimes helpful to look at a few special cases before attacking the general case.

CrackverbalGMAT, avigutman could you please share your approach for this question? How can we look at it more logically instead of plugging random numbers?

Hi CrushTHYGMAT, I think you'll find exactly what you're looking for here:
https://gmatclub.com/forum/if-n-is-an-i ... l#p2662420
This will enable you to solve such problems in under 30 seconds, with no pen or paper.

Is there any reason why we can't select m as zero? So k=0(4)(5) which would not be divisible by 6.

Every integer is a divisor of 0, except, by convention, 0 itself. In this case \(\frac{0}{6} = 0\)

Here's a useful post on Number Theory