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If k=m(m+4)(m+5) k and m are positive integers. Which of the
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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 A. I only B. II only C. III only D. I and III only E. I, II and III
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Re: If k=m(m+4)(m+5) k and m are positive integers. Which of the
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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: If k=m(m+4)(m+5) k and m are positive integers. Which of the
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17 Apr 2010, 06:17
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: If k=m(m+4)(m+5) k and m are positive integers. Which of the
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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: If k=m(m+4)(m+5) k and m are positive integers. Which of the
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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
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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
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23 Jan 2014, 01:00



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Re: If k=m(m+4)(m+5) k and m are positive integers. Which of the
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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
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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: If k=m(m+4)(m+5) k and m are positive integers. Which of the
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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: If k=m(m+4)(m+5) k and m are positive integers. Which of the
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22 May 2016, 02:59
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 All three divides k evenly
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Re: If k=m(m+4)(m+5) k and m are positive integers. Which of the
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24 Sep 2017, 06:06
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 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....
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If k=m(m+4)(m+5) k and m are positive integers. Which of the
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10 Jan 2018, 11:07
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)




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