integers

[Kyle Concannon]

Find the smallest positive integer that is divisible by 13, and that when divided by any of the integers from 2 to 12 (inclusive) leaves a remainder of 1.

 

[stapel]

What have you tried? (If you're thinking the answer is "371,293", please review this other post of this exercise.)

Eliz.

 

[Denis]

Find the smallest positive integer that is divisible by 13, and that when divided by any of the integers from 2 to 12 (inclusive) leaves a remainder of 1.

Sir Concannon, you are becoming a pain in the Royal Butt...
you're flooding the place with questions, showing absolutely nothing else;
I for one am disregarding your posts from now on...

 

[happy]

Then you agree with what I posted on the "other board", don't you, Denis?

 

[stapel]

There's only been eleven posts in more than a month -- here, anyway. The lack of effort displayed is a bit disheartening and the multi-posting when answers have already been provided elsewhere is somewhat disrespectful, but -- just my opinion -- it hardly seems a "flood", at least not compared to some other floods we've experienced.

Eliz.

P.S. "Other board"? "Posted"?

 

[happy]

Here- http://www.sosmath.com/CBB/viewtopic.php?t=19622

 

[pka]

Kyle Concannon, I had not forgotten about this problem.
In fact, I was interested in it, but got nowhere using number theory.
So I gave up and programmed it. Here is some pseudo code.

For K=1 to 100000
        For J=2 to 12
	        If mod(K,J)≠1 then next K
        Next J
        If mod(K,13)=0 then output K
 Next K

Low and behold, the output is 83161.
I double checked and it works.

The interesting part is that the prime factorization of 83161=(13)(6397).
I cannot see any mathematical way that we should have used to find that.
Therefore, I concluded that this must have been a programming exercise

 

[Denis]

For K=1 to 100000
        For J=2 to 12
	        If mod(K,J)≠1 then next K
        Next J
        If mod(K,13)=0 then output K
 Next K

Low and behold, the output is 83161.

Agree; next 2 are 443521 and 803881 :shock:

By the way, you can code also this way (faster):
For K = 13 to 100000 step 13
For J = 2 to 12
If mod(K,J)<>1 then next K
Next J
Output K
Next K

And to you Stapel: I saw 6 posts on Nov 29th, so looked like flooding to me.
Anyhoo, Kyle emailed me and we're buddy-buddy now :wink: