Help please

[navdeepbajwa]

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


Can you please provide an algebric solution

[navdeepbajwa]

Is this Idea right

m and m+5 have six numbers in between so m(m+5) must be divisible by 6 and if by 6 then must be by 3 as 3 is factor of 6

[migcua]

it sounds right...

this isn't an algebraic solution but i'd like to add that i think 4 can also divide the equation since if m were odd or even, atleast 1 of the terms would be even.
given that any number that can be divided by 2 twice is divisible by 4, and since m is a positive integer, then at least of the numerators is even and greater than 4.

ex. 3*7*8 is div by 4 bec 8=2*2*2
2*6*7 is div by 4 bec of 2 and 6=2*3

i hope that makes sense

[navdeepbajwa]

But divisibility by 4 fails if M=5

[Leon1984]

Is this official?

It seems that it COULD be divisible by all 3.
It doesn't have to, however, depends on the value of m.

In algebraic way, in order for it to be divisible by 3, one of the factors (m or m+4 or m+5) needs to be divisible by 3, and for every possible positive integer, it will be.

For 6, it needs to be divisible by 2 and 3, in other words, have a factor of 2 and a factor of 3. We can see that for every value of m, either m+4 or m+5 will be even, thus divisible by 2 and, since it is also divisible by 3, be divisible by 6.

In order to be divisible by 4, it needs 2 be divisible by 2 twice, or that 2 out of our 3 factors (m, m+4, m+5) will be even. It is possible if m is even (e*e*o=e), if m is odd we get (o*e*o=e), so in any case it will be divisible by 2 but only when m is even it will be divisible by 4.

I don't know whether it fits the answer though, how should the word "could" be approached?