Before I left home, I was an avid pub-quizzer and was often invited to join my dad and his friends at a quiz they did in a neighbouring village. It was always a very good evening out, with 10 rounds on assorted topics, finishing with a General Knowledge round. The 100th question in the quiz was always a maths question, and it always met with the same response: the rest of the team would throw down their pencils and say “Over to you, WE can’t do maths!” And my dad and I would get our heads together and come up with the answer, and the rest of the team would always be amazed that we’d managed to get it right.

Last time I went home, my dad and I went out for a long walk together, and as we walked, he asked me the questions from last week’s quiz. The format was still the same, although it seems my general knowledge has improved over the last ten years as I knew the answers to questions on a wide range of topics. (Bringing the answers to mind seemed to be harder than it used to be however!) We got to Question 100 just as we decended the hill back into the village:

*58 sweets are shared out among 10 children. If every boy gets 5 sweets, and every girl gets 7 sweets, how many girls were there?*

Now thinking back to how I would have solved it when I was 18 and finishing A level maths, I’m sure I would have quickly set up a system of equations: let x be the number of girls, and y be the number of boys. Then x+y = 10, 7x+5y = 58, solve the simultaneous equations, done. However, this is not how I tackled it on the way down the hill, with no pencil or paper to hand. Instead, I went for the much simpler method: if they were all boys, there would only be 50 sweets. Every time I swap a girl in, the number of sweets increases by 2, so I need to swap in four girls. So there are 4 girls and 6 boys. My dad was amazed by the simplicity of this method (even after I told him it only came to mind because I’d spent most of last week thinking about Cinema Problem!)

Having a method that will always work is a great thing – my dad’s teammates are in awe at his continued success on the maths question each month. I used to take great comfort in having an arsenal of reliable strategies to tackle particular classes of problem, and indeed these are a very important part of any mathematician’s toolkit. But also important is having the creativity to play with a problem, spot a shortcut, try different things out, and see a more elegant route to a solution than using a sledgehammer to crack a nut. I only wish my dad’s teammates could be persuaded to have a go!

February 11, 2010 at 21:51 |

Thanks Alison

It is always interesting to see how different people work out problems like this. I have given the problem to a group of 11 and 12 year old children as a weekend problem. I look forward to seeing how their brains tick! Personally I worked it out as follows:

If the girls had 7 sweets each then I was looking for a multiple ending in an 8 or a 3. (Hence 28 or 4 girls.) This then left me with 30 sweets which is divisible by 5. (6 boys.)

I hit the wrong button so ignore previous message!

Thanks

February 12, 2010 at 12:52 |

Glad you liked it Jonathan, I’d be very interested to hear what the children come up with. I like the way you used the structure of the 5x table to help you to solve the problem.

February 16, 2010 at 13:21 |

Yes, that’s interesting. There’s a trade off between looking for an elegant way and using a sledgehammer, and I think mathematicians are trained in both. I think there’s all sorts of little tricks and bits of experience that can simplify a problem like that, but that if it’s any more complicated than that, we’re trained to say “well, there may be an easy way, but I’m sure XXXX’s theorem can solve it in 3 minutes so its not worth looking for a one-off elegant solution at the moment”

February 20, 2010 at 17:18 |

The simultaneous equations is the sledgehammer to use if you are determined not to use the information that the number of boys and the number of girls are both integers.

Naturally, the solutions are simpler to arrive at if you use all the information in a problem, not just some.

March 4, 2010 at 16:13 |

I looked at it in a similar way to you, but the phrasing was different. I noticed that everyone has at least 5 sweets, so imagined giving everyone 5 sweets each. The problem we are left with is now:

“8 sweets are shared out among 10 children. If every boy gets 0 sweets, and every girl gets 2 sweets, how many girls were there?”I expect I was strongly influenced by the fact I’m currently solving problems where the first step is always to try and make the problem simpler, before thinking about solving it. đŸ˜›

October 25, 2010 at 18:05 |

Thats just how I solved it!