If k=m(m+4)(m+5) k and m are positive integers. Which of the

Question

If k=m(m+4)(m+5) k and m are positive integers. Which of the following could divide k evenly?

I.3 
II.4 
III.6

A. I only
B. II only
C. III only
D. I and III only
E. I, II and III

einstein10 · Answer

k is product of 3 numbers.
there are two possibilities when m is even and when m is odd:
1: if m is even then k will be evenly divided by 4 due to factor m and (m+4)
also it will be divided by 3 and 6 because m(m+4) is always divided by 3  and since m is even m(m+4) will always divided by 3 and 6.

2. when m is odd, it is not necessarily divided by 4. but k will be divided by 3 &6.
there will always one 2 and one 3.
e.g if m=1,3.. k is divided by 4 but if k=5 then it will not.

Bunuel · Answer

You would be correct if question were: "Which of the following  MUST  divide k evenly". Then answer is I (3) and III (6).

But question asks: "Which of the following  COULD  divide k evenly" and in this case answer is I, II and III (m can take all three options).

dimitri92 · Answer

Answer should be I,II and III because what we are asked in "COULD" not "MUST" here

craky · Answer

Simple way to solve this:
m=1, m+4=5, m+5=6

Than just write numbers:

m m+4 m+5
1 4 5
2 5 6
3 6 7
4 7 8
5 8 9
6 9 10
7 10 11
8 11 12

Numbers in each column increase by one.
Now you see that each combination is divisible by 3 and 6 but not 4.

A little rough approach, but it works.

jlgdr · Answer

Let me take a crack at this m(m+4)(m+5) are really three consecutive integers (m)(m+1)(m+2) So I is always true II is NOT always true. Cause if m is odd then m+4 is odd too, and the only even is m+5 which can or cannot be a multiple of 4 III the product of three consecutive integers is always a multiple of 3! Hence the answer here should be I and III Hope it helps Cheers! J

Bunuel · Answer

ANY number COULD divide k. The correct answer is I, II and III. Check here: if-k-m-m-4-m-5-k-and-m-are-positive-integers-which-of-the-92808.html#p714621

Sukant2010 · Answer

Hi,
According to your approach, the first combination ain't divisible by 3 and 6. Then how can the combination MUST be evenly divided by 3 and 6 as stated by all here. 
I think the ans can be all the 3 options only because the question asked is 'WHICH COULD'. But if the question asked 'WHICH MUST', then I think there would have been no solution.
Any expert if please could comment on this.

arunspanda · Answer

It is not clear why m, m+4 and m+5 are consecutive integers.

PareshGmat · Answer

Corrected the series

Alphatango · Answer

All three divides k evenly

Abhishek009 · Answer

1. Let m = 3

k=m(m+4)(m+5)

So, k=3(3+4)(3+5)

Or, k=3*7*8  ( Divisible by 3 and 4 )

2. Let m = 2

k=m(m+4)(m+5)

So, k=2(2+4)(2+5)

Or, k= 2*6*7  ( Divisible by 6 ) 
  
Thus,answer must be all three of the options....

hellosanthosh2k2 · Answer

One quick thing jumped into my mind while seeing this question. May be helpful

m (m+1) (m+2) (m+3) (m+4) (m+5)

if we observe, difference between (m+3) and m is 3
so properties with respect to divisiblity by 3 for (m)(m+4)(m+5) will be same for (m+3)(m+4)(m+5)

so we ll take a leap from m(m+4)(m+5) to (m+3)(m+4)(m+5) => we have 3 consecutive integers

Let us attack the options
(1). 3 => ofcourse, product of 3 consecutive integers is always divisible by 3
(2). 4
 if m is odd, (m+3) = (m+5) = even, we have two 2s, definitely divisible by 4
(3) 6
 as we have seen from above the product is always divisible by 3 and could be divisible by 4, the prod could be divisible by 12 => could be divisible by 6

so all three options could divide evenly => (E)

MHIKER · Answer

The question says "could be evenly" divisible. So, plugging a number one or two option may valid and plugging another number other options may be valid.

Let  m=1 , 
I.  k=1(1+4)(1+5)=5*6 , which is divisible by 3 and 6.

If  m=4 
II.  k=(4+4)(4+5)=8*9 , which is divisible by 4 and 3.

Thus, k could be evenly divisible by 2, 4, and 6

The answer is E

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If k=m(m+4)(m+5) k and m are positive integers. Which of the

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If k=m(m+4)(m+5) k and m are positive integers. Which of the

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