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Is it because you multiplied them all together that dividing by one of the numbers puts you in the same place you started before multiplying?
Yes.
I have no idea how this "follows".
3*5*6*7*8*9*10 = 3*5*6*7*(2*4)*9*10 => 2, 4 and 8 are factors.
If 10 = 2*5 then 2 and 5 are factors. So that's what I'm doing above.
Why didn't you remove 9 instead of 3?
If I removed 9 then 9 would no longer be a factor of what remains. There's no way to make a 9 with what remains.
And why does "6 divide 5*7*8*9"
Because we can take a 2 from 8 and a 3 from 9 to show that 6 is a factor, i.e. 5*7*8*9 = 5*7*4*(2*3)*3 = 5*7*4*6*3.
but "7 does not divide 5*8*9"
5*8*9 = 5*2*2*2*3*3, see no 7's :).
I see. This makes more sense. Thank you.
You're welcome.
Yes.
3*5*6*7*8*9*10 = 3*5*6*7*(2*4)*9*10 => 2, 4 and 8 are factors.
If 10 = 2*5 then 2 and 5 are factors. So that's what I'm doing above.
If I removed 9 then 9 would no longer be a factor of what remains. There's no way to make a 9 with what remains.
Because we can take a 2 from 8 and a 3 from 9 to show that 6 is a factor, i.e. 5*7*8*9 = 5*7*4*(2*3)*3 = 5*7*4*6*3.
5*8*9 = 5*2*2*2*3*3, see no 7's :).
I see. This makes more sense. Thank you.
You're welcome.