## Saturday, 19 September 2009

Just after breakfast this morning I tweeted the following:
And very quickly I got the following replies:
1. goze01@stujohnson I have answer. Not telling yet. Get him to list the combinations of numbers when multiplied give a 0 in unit column. :-)19 minutes ago from Echofon in reply to stujohnson
2. Steepholm@stujohnson I assume there's also some way of doing it mathematically that a 9-year-old could figure out.about 1 hour ago from TweetDeck in reply to stujohnson
3. Steepholm@stujohnson I've just written a little program to do it by brute force - 32 * 3125 is the only solution.about 1 hour ago from TweetDeck in reply to stujohnson
4. traceymadden@stujohnson Suspect just being asked to find all the factors of 100,000 and identify the ones that match the criteria.about 1 hour ago from TweetDeck in reply to stujohnson
5. traceymadden@stujohnson Try looking 4 clues on how they are 'supposed' 2 solve this in their maths workbook (should lead on from last lesson).about 1 hour ago from TweetDeck in reply to stujohnson
6. pk_kent@stujohnson 32 x 3125. Do factor analysis of 100,000 - divide by 2, again by 2 etc. u end up with 2x2x2x2x2x3125about 2 hours ago from Echofon in reply to stujohnson
7. katiepiatt@stujohnson got to be 3125 x 32, I divided down 10000 by 2 until I reached a number that didn't end in 0, then found the other factorabout 2 hours ago from TweetDeck in reply to stujohnson
So my son worked it out by:
• dividing 100,000 by 2 until he got to a number that didn't end in 0 (3,125)
• then dividing 100,000 by 3,125 which gave an answer that also didn't end in 0 - 32
So thank you to @goze01, @steepholm, @traceymadden, @pk_kent and @katiepiatt!!

1. 2. What are you using to display the Twitter post on your blog Stu?

AS maths homework now moving beyond my capacity to "help", but an hour with Wolframalpha helped us figure out some tricky quadratics last night ;-)

3. I'm just dragging the mouse across the screen and copying and pasting. Pretty technical, eh?! Also, if I delete some of the irrelevant @ replies - wordpress re-numbers them because it automatically displays them as a numbered list.

I know I should know more about the maths, but heh hoh.

4. One of my A-level teachers used to make a clear distinction between maths and arithmetic, and I think my brute-force method would have definitely been classed as arithmetic. Here's the code (tidied up slightly - originally the range was from 2 to 100000 but of course you can stop when a is the square root of x):

from math import sqrt
x = 100000
max = int(sqrt(x)) + 1

for a in range(2,max):
b = x // a
if (a * b == x):
if (a % 10 != 0 and b % 10 != 0):
print "%10d %10d is a solution" % (a, b)
else:
print "%10d %10d" % (a, b)

Thinking about it later on, if you recognise that the powers of 2 never end in zero, repeatedly dividing by 2 until you get an answer which doesn't end in zero is an elegant (and therefore mathematical) way of doing it - but it isn't generalisable, it wouldn't work if one of the two factors wasn't a power of two. I wonder if there's a properly mathematical way of doing it? (Elsewhere, I know a couple of people who used to compete with each other for top place in maths during their time at Oxford, and they would know - I'll ask them).

5. The responses from an older and richer form of social media:

Prize for elegance goes to this solution though it may not be understandable by a 9-year-old (I understood powers instinctively when I was introducted to them, but I think I was 12 or 13).

100000 = 10 * 10 * 10 * 10 * 10 = 10^5 = (2 * 5)^5 = 2^5 * 5^5 = 32 * 3125

The first step is intuitive, the separation of (2 * 5)^5 into 2^5 * 5^5 may not be (but you can also go via 10*10*10*10*10 = 2*5*2*5*2*5*2*5*2*5 which makes it more obvious). You don't need to know that 2^5 is 32 and 5^5 is 3125 to recognise that powers of 2 and powers of 5 never end in zero. Someone else put this another way - factorise into primes, and then separate into two groups, one of which has all the twos and the other all the fives (the other numbers don't matter, but you can't have 2*5 in either group). I think this is over-intellectualising it though - for any number ending in zero, it's divisible by 2, and the answer either still ends in zero (in which case you can divide it by 2 again), or it doesn't (in which case you've found the answer). Less easy to say if 392849201 has two factors, neither of which ends in 9.

Someone else again (a maths PhD I believe) reminded me of what I already knew - there is no known general method for factorising large numbers (because if there was, various encryption algorithms we all rely on would not be secure), so you have to do it by a brute force method.

6. Hi Andrew. Thanks for all your help with this - v elegant indeed. He handed it in today (I made sure he noted the bits where he got help!) and I'll let you know how he gets on. Thank you :)

7. 