Factoring: test your speed!
Hey, I won a prize. In one of the press conferences this morning, Simon Singh (of the Fermat’s Last Theorem book fame) set us a problem: what numbers are the factors of 323? (As in, what numbers multiplied together produce that number?) Prize, a green spinny thing “because I know you’re not motivated by money”.
I wrote the answer but was trying to do the maths to confirm it when he called time. But he acccepted my scribbled numbers as proof. (Generous - I might just have been a chancer who was quick at writing.)
Answer in the comment, because I know you’ll want to do it yourself. Give yourself 30 seconds.
- These posts might be related (the database thinks..):
- Speed cameras redux (1 December 2004; score: 47.92%)
- As Homer Simpson would say: woo-hoo! (15 June 2005; score: 44.02%)
- Arguments against speed cameras demolished wholesale (10 October 2004; score: 35.29%)
![[Valid RSS]](wp-includes/images/valid-rss.png "Validate my RSS feed"){kind=small}
September 6th, 2004 at 1:41 pm
Here’s the working, first. 323 isn’t divisible by 3 (the digits don’t add up to a multiple of 3) so it’s none of those sorts of numbers.
Now the only question is, what numbers multiplied together produce a 3? Not many. 3 and 1, but this isn’t a -3 x 11 sort of number. That leaves 7 and 9 - makes 63. Add 100 from 10 x 10, add 160 from 70 x 90, so we’re in the right region… in fact, just right: 17 x 19 = 323.
Singh’s point was how hard factorisation is compared to multiplication. I thought one could fake it a bit with a Newton-Raphsom method, but didn’t quite pluck up the courage to ask.
September 6th, 2004 at 2:19 pm
Numbers eh? This is more my level:
http://news.bbc.co.uk/1/hi/sci/tech/3625256.stm
Any chance of the two gentlemen making an appearance, you think?