Show that for all integer values of ,

. . is divisible by

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- Jan 3rd 2007, 04:34 AMSorobanQuickie #7
Show that for all integer values of ,

. . is divisible by

- Jan 3rd 2007, 02:01 PMgalactus
Soroban, am I on the right track?.

We have to show that is divisible by

Factor:

I would say because 5 is a factor of

3 is a factor of

8 is a factor of - Jan 3rd 2007, 08:41 PMSoroban
Hello, Galactus!

Of course, you meant: 5 is a factor of

Then your solution is correct.

The "Quickie" solution goes like this:

We know that the product of consecutive integers is divisible by

. . and the product of four consecutive integers is divisible by

We have: .

. . and: .

. . then is divisible by

. . then is divisible by

. . then is divisible by

Therefore, is divisible by