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

Morgoth wrote:

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

Well without the options its not possible to answer the question, but I can tell you the least possible value of K.

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.

stop@800 wrote:

Morgoth wrote: I think No. If I take m=2
m+5 will become 7
and 720 will not be able to handle 7

Well I was talking about the product of 6 consecutive integers i.e.

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*3

take m=3
K is 3*7*8

now this will not be divisible by 30

Please correct me if I am missing something.

Here 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.

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.

Morgoth wrote:

stop@800 wrote:

Morgoth wrote: I think No. If I take m=2
m+5 will become 7
and 720 will not be able to handle 7

Well I was talking about the product of 6 consecutive integers i.e.

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*3

take m=3
K is 3*7*8

now this will not be divisible by 30

Please correct me if I am missing something.

Here 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.

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.