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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 [#permalink] New post 17 Apr 2010, 00:44
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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
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Re: Algebra question [#permalink] New post 17 Apr 2010, 05:58
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


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.
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Re: Algebra question [#permalink] New post 17 Apr 2010, 06:17
Expert's post
sandeep25398 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


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).
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Re: Algebra question [#permalink] New post 17 May 2010, 20:57
Answer should be I,II and III because what we are asked in "COULD" not "MUST" here
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Re: Algebra question [#permalink] New post 03 Feb 2011, 01:42
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.
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Re: If k=m(m+4)(m+5) k and m are positive integers. Which of the [#permalink] New post 22 Jan 2014, 11:42
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!

Hence the answer here should be I and III

Hope it helps
Cheers!
J :)
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Re: If k=m(m+4)(m+5) k and m are positive integers. Which of the [#permalink] New post 23 Jan 2014, 01:00
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!

Hence the answer here should be I and III

Hope it helps
Cheers!
J :)


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
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Algebra question [#permalink] New post 24 Jan 2014, 01:30
craky wrote:
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.

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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Re: If k=m(m+4)(m+5) k and m are positive integers. Which of the [#permalink] New post 24 Jan 2014, 04:08
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.
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Re: Algebra question [#permalink] New post 25 Feb 2014, 19:03
Sukant2010 wrote:
craky wrote:
Simple way to solve this:
m=1, m+4=5, m+5=6

Than just write numbers:

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

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.



Corrected the series
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Re: Algebra question   [#permalink] 25 Feb 2014, 19:03
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