positive integers
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The options are missing 
Let me still try
m(m+4)(m+5)
will certainly be divisible by 2
bcoz
if m is odd, m+4 also odd and m+5 even
Editing for: I misinterpreted divide with divisible by.
Let me still try
m(m+4)(m+5)
will certainly be divisible by 2
bcoz
if m is odd, m+4 also odd and m+5 even
Editing for: I misinterpreted divide with divisible by.
Last edited by stop@800 on Sun Sep 28, 2008 12:38 pm, edited 1 time in total.
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Morgoth
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Well without the options its not possible to answer the question, but I can tell you the least possible value of K.acorra wrote:If k=m(m+4)(m+5) k and m are positive integers. Which of the following could divide k evenly?
We know m is a positive integer.
k = m(m+4)(m+5)
Product of 6 consecutive positive integers will be divisible by 6! or 720.
Therefore,
m(m+1)(m+2)(m+3)(m+4)(m+5) will be divisible 720 at least
The question stem is,
m(m+4)(m+5) will be at least divisible by 5*6 = 30 = 5*2*3
Now which ever option has 5,2,3 as its prime factors will divide k evenly.
Hope this helps.
I think No. If I take m=2Morgoth wrote:Well without the options its not possible to answer the question, but I can tell you the least possible value of K.acorra wrote:If k=m(m+4)(m+5) k and m are positive integers. Which of the following could divide k evenly?
We know m is a positive integer.
k = m(m+4)(m+5)
Product of 6 consecutive positive integers will be divisible by 6! or 720.
Therefore,
m(m+1)(m+2)(m+3)(m+4)(m+5) will be divisible 720 at least
m+5 will become 7
and 720 will not be able to handle 7
take m=3
The question stem is,
m(m+4)(m+5) will be at least divisible by 5*6 = 30 = 5*2*3
K is 3*7*8
now this will not be divisible by 30
Please correct me if I am missing something.
Now which ever option has 5,2,3 as its prime factors will divide k evenly.
Hope this helps.
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Morgoth
- Master | Next Rank: 500 Posts
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- Joined: Mon Sep 22, 2008 12:04 am
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stop@800 wrote:Well I was talking about the product of 6 consecutive integers i.e.Morgoth wrote: I think No. If I take m=2
m+5 will become 7
and 720 will not be able to handle 7
m(m+1)(m+2)(m+3)(m+4)(m+5)
if you take m=2
this becomes 2*3*4*5*6*7 = 5040/720 = 7
similarly you can insert m=3,4,5,6,7,.....,n.
Therefore, product of any 6 positive consecutive integers will always be divisible by 720 at least.
The question stem is,
m(m+4)(m+5) will be at least divisible by 5*6 = 30 = 5*2*3Here I made a mistake, m*(m+4)*(m+5) will always be divisible by 3 and 2 i.e. 6. I included 5 previously which is wrong.take m=3
K is 3*7*8
now this will not be divisible by 30
Please correct me if I am missing something.
m=3
3*7*8 / 6 = 28
similarly you can try for m=4,5,6,7,8,.....,n
Therefore, m*(m+4)*(m+5) will always be divisible by 6 irrespective of what positive value you choose for m.
Thanks for pointing my mistake.
Hope its clear.
yes, m(m+4)(m+5) will always be divisible by 6
Morgoth wrote:stop@800 wrote:Well I was talking about the product of 6 consecutive integers i.e.Morgoth wrote: I think No. If I take m=2
m+5 will become 7
and 720 will not be able to handle 7
m(m+1)(m+2)(m+3)(m+4)(m+5)
if you take m=2
this becomes 2*3*4*5*6*7 = 5040/720 = 7
similarly you can insert m=3,4,5,6,7,.....,n.
Therefore, product of any 6 positive consecutive integers will always be divisible by 720 at least.
The question stem is,
m(m+4)(m+5) will be at least divisible by 5*6 = 30 = 5*2*3Here I made a mistake, m*(m+4)*(m+5) will always be divisible by 3 and 2 i.e. 6. I included 5 previously which is wrong.take m=3
K is 3*7*8
now this will not be divisible by 30
Please correct me if I am missing something.
m=3
3*7*8 / 6 = 28
similarly you can try for m=4,5,6,7,8,.....,n
Therefore, m*(m+4)*(m+5) will always be divisible by 6 irrespective of what positive value you choose for m.
Thanks for pointing my mistake.
Hope its clear.
