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

Mathematical approach

\(k=m*(m+4)*(m+5) = (m+3-3)*(m+4)*(m+5)\)

or \(k = (m+3)*(m+4)*(m+5)-3(m+4)*(m+5)\)

Now, \((m+3)\), \((m+4)\) & \((m+5)\) are three consecutive integers, hence MUST be divisible by \(3\) & \(6\)

\(-3(m+4)*(m+5)\), has \(3\) as a factor and \((m+4)\) & \((m+5)\) are consecutive integers, hence either of the two will be a multiple of \(2\). Hence the product MUST be multiple of \(3\) & \(6\)

Therefore \(k\) MUST be divisible by \(3\) & \(6\)

Option B

Plug in m = 1, m = 2, m = -7

(1)(5)(6) = 30
divisible by 3 and 6

(2)(6)(7) = 84
divisible by 3 and 6

(-7)(-3)(-2) = -42
divisible by 3 and 6

Answer B
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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
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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.
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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.
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