Thread: I feel like an idiot

Am I missing something obvious, or is this really just exhaustion method?

"Find the smallest positive integer K which is divisible by seven and which, divided by each of the integers between two and six inclusive, leaves a remainder of one. If possible, find a larger value of K."

Hello,

Originally Posted by Hiphopopotamus

Am I missing something obvious, or is this really just exhaustion method?

"Find the smallest positive integer K which is divisible by seven and which, divided by each of the integers between two and six inclusive, leaves a remainder of one. If possible, find a larger value of K."

Hmmm do you know the Chinese remainder theorem ?

Note that if the remainder in the division by 4 is 1, then it is 1 in the division by 2 too. Because K=4k+1=2*(2k)+1. So you don't need to bother yourself working on 2.
Same goes for K=6k'+1=3*(2k)+1



Edit : never say you're an idiot.

You cannot be an idiot with the screenname 'hiphopopotamus'.
Go Flight of the Conchords!

Just thought I'd say that. Sorry I can't help with your question.

Hello, Hiphopopotamus!

Find the smallest positive integer which is divisible by 7,
and which, divided by each of the integers between 2 and 6 inclusive, leaves a remainder of one.
If possible, find a larger value of .

See if this makes sense to you . . .


The Lowest Common Multiple of {2, 3, 4, 5, 6} is 60.

Then is of the form: .

Since is divisible by 7: .


And we have: .

Since is an integer, must be divisible by 7.
. . The first time this happens is when
. . The next time this happens is when


Therefore, the smallest value is: .

And the next value is: .