chicken mcnuggets

darkforce · January 27th, 2007, 11:55 pm

If chicken mcnuggets come in sizes 6, 9, 20 what is the largest no you cannot buy using a combination of these?

Traden4Alpha · January 28th, 2007, 12:17 am

I got 43 (I'll admit I used semi-ugly brute-force methods)After 43, you can construct 20 numbers in a row from that basis set. Once you can construct 20 in a row, all subsequent numbers are possible.

iwanttobelieve · January 28th, 2007, 2:42 pm

, thus, if I can go short nuggets then I should be able to buy any quantity. (selling is buying a negative quantity). Ok, that's not the spirit of this riddle.

mhughes · January 29th, 2007, 1:42 pm

You can see it pretty easily just by noticing that with 6 and 9, you can get any number 0 mod 3 (greater than 3). Since 20 is 2 mod 3, this means you can get any number 2 mod 3 greater than 23. However, you don't get any numbers that are 1 mod 3 until 40, and at that point you can get any number 1 mod 3 greater than 43. So 43 is the largest unattainable number.

MCarreira · May 2nd, 2007, 3:56 pm

Browsing Demonstrations page in Wolfram Research's site (they're promoting Mathematica 6), one finds:McNugget Problem and Frobenius Numbers