Divide evenly problem

IMO I and III

Could you please explain? How did you get this?

For any positive integer m, k contains a multiple that is divisible by 6, and since 3 is a factor of 6, it can also evenly divide into any value of k.

m = 1, then k = 1 * (1 + 4) * (1 + 5) = 1 * 5 * 6 --> clearly, you can see that both 3 and 6 can be divided into this.
m =2, then k = 2 * 6 * 7, again you have a factor 6 in there, so you could divide both 3 and 6
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say m = 7, then k = 7 * 11 * 12 --> 12 is a multiple of both 6 and 3.


As you can see, for any value of m, there exists a multiple in k that is divisible by 3 or 6.

But why wouldn't 4 work as well. If m = 2 that could be divisible by 4 and same if m = 3

RoadtoIVY wrote:Could you please explain? How did you get this?

i just substitute m with 1,2,3,4,5,6,7....and check for divisbilty with 3,4,6 and concluded that 3,6 is true for all cases
but 4 is not

^^ Agreed. 3 and 6 will be true for any number, whereas 4 will not.