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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

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 [for m>0 and m is integer]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.

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

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 [for m>0 and m is integer]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.

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). _________________

Re: If k=m(m+4)(m+5) k and m are positive integers. Which of the [#permalink]

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22 Jan 2014, 12:42

utin wrote:

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

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!

Re: If k=m(m+4)(m+5) k and m are positive integers. Which of the [#permalink]

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23 Jan 2014, 02:00

Expert's post

jlgdr wrote:

utin wrote:

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

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!

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.

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.

Re: If k=m(m+4)(m+5) k and m are positive integers. Which of the [#permalink]

Show Tags

24 Jan 2014, 05:08

jlgdr wrote:

utin wrote:

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

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

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

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.

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.

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