I promised divisibility in January, and have yet to deliver. In fact, I’ve posted almost nothing about anything since the New Year. But my energy returns, and things will start rolling again. Now.

This is classic. I tried to divide my collection of rocks into two equal piles, but there was one left over. Then I tried to divide it into three equal piles, but there was one left over. When I tried to divide it into four equal piles, there were three left over. When I tried to divide it into five equal piles, there were three left over. When I tried to divide it into six equal piles, there was one left over. Then I divided it into seven equal piles. Can you imagine doing all this dividing with almost a thousand rocks? How many did I have?

This (I believe) is not classic: Consider the sum of the remainders when a number is divided by 1, 2, 3, 4, and 5. Under what conditions will the number be a multiple of this sum?

1. January 11, 2008 pm31 9:02 pm 9:02 pm

For the second puzzle, as an example, when 22 is divided by 1, 2, 3, 4, and 5, the remainders are 0, 0, 1, 2, and 2 with a sum of 5. 22 is not a multiple of 5.

January 12, 2008 am31 12:08 am 12:08 am

The first one is very classic (the “applied Chinese Remainder Theorem,” more or less). I’ve never seen the second one before — it’s certainly not a classic in the same way.

3. January 13, 2008 am31 1:44 am 1:44 am

Are you sure you have the list of remainders right?

For this set of remainders
n(i) 2 3 4 5 6 7
a(i) 1 1 3 3 1 0

I see only two solutions that are less than a thousand, and I wouldn’t have called either of them “almost a thousand”. The next solution greater than a thousand is close(-ish), but beyond a thousand isn’t traditionally “almost”, and in any case is not within ten percent of it.

Should I post them anyway?

January 13, 2008 am31 2:21 am 2:21 am

[Comment moved to solution thread, here]

5. January 13, 2008 am31 3:13 am 3:13 am

Efrique,

“almost” is in the eye of the beholder.