My favourite Maths question of all time

This is my absolute favourite Maths question of all time. It’s hard, it’s simple, and it shows you something that all the best Maths scholarship questions do, which is this:

Knowing how to use the tools is one thing, but knowing which tool to apply to the problem at hand is something quite different.

Ready? Here we go.

Solve, without a calculator: √11⅑

Which is to say, what is the square root of eleven and one-ninth?

Some of my students are as young as nine. Most of them know what a number ‘squared’ means – that you multiply it by itself, so 22 = 2 x 2 = 4. But they don’t know what the square root of a number is. And when you explain that it’s the opposite of ‘squared’, they think it means you divide the number by itself.

Finding the square root of a number without a calculator is really hard unless it’s a square number already. 11⅑ obviously isn’t, because all square numbers are integers – whole numbers (12 = 1; 22 = 4; 32 = 9; 42 = 16). We might start with trial and error, and say that 11⅑ lies between two square numbers, 9 (= 32) and 16 (= 42). So √11⅑ must be somewhere between √9 and √16, which is to say between 3 and 4. (For what it’s worth. I do find these meanderings helpful because when you do find a more precise answer through a less familiar method, you’ve got a rough answer to compare it to.)

(My more sophisticated students like to convert fractions (which feel rather messy and arbitrary) into decimals, which feel clean and neat. 11⅑ = 11.111 recurring, which is interesting, but not much use. Actually, that’s worse than useless, because this is a question about fractions, and decimals takes us away from the right answer.)

What sort of fraction is 11⅑? It’s a mixed fraction – 11 and ⅑. A normal fraction is a ratio between two whole numbers, a/b. In fact, all rational numbers, which includes nearly all the numbers a young mathematician might encounter (pi being one example of an irrational number), can be written as a/b.

A mixed fraction looks like it’s three numbers, a + b/c, 11 + 1/9. But it’s really more useful to regard it as four numbers a/b + c/d, in this case, 11/1 + 1/9.

How would 11⅑ look if it were in the form a/b? That, by the way, is what the question meant by ‘Simplify’. How can you get 11/1 + 1/9 (a/b + c/d) into the form a/b?

By making both denominators (b and d, at the bottom of the two fractions) the same. b and d are 1 and 9, so you need to multiply 1 by 9. And you need to multiply the top of the first fraction by 9 too. 11/1 x 9/9 = 99/9. Of course, 99 divided by 9 is 11, so you haven’t actually changed the fraction at all – you’ve just put it in a different, more useful format (x 9/9 is the same as x 1, which doesn’t do anything to materially change the number).

So 11/1 x 9/9 = 99/9. What was the other half of our mixed fraction, the c/d part we needed to convert the a/b part to match denominators? Oh yes, it’s 1/9. So

11⅑ = 11/1 + 1/9
= 99/9 + 1/9
= 100/9

How does that help? We still need to find the square root of this fiddly fraction, don’t we? Well, we’ve actually got very close to solving it. We’ve turned 11⅑ into a vulgar, improper, top-heavy fraction, in the form a/b, but it turns out that was exactly what we needed. Because what do you notice about a and b, 100, and 9? What was the only kind of number we can find the square root of easily, without a calculator? Square numbers, right? And what are 100 and 9? Square numbers: 100 = 102 and 9 = 32. So:

√11⅑ = √100/9
= 10/3
= 3 1/3

QED!

Actually, I cheated a bit with the wording, which I changed from the way it appears in the exam – but I only did this to make the question harder for you, and to make a point that should hopefully make the exam you take easier. This is to say that the original question asked you to Simplify √11⅑ rather than just solve. This might have led you to the solution faster: it’s a very useful cue, the idea that you need to rearrange the format of this mixed fraction. But the point remains – sometimes you’ll need to simplify, to see things differently before you can then solve a problem.

About the author

Bio here