You might think that the study of maths is just for humans. You’d be wrong.

Cast your mind back to Aesop’s fable, *The Town Mouse and the Country Mouse*. The moral of the story was that tastes differ and none are any better or worse than others.

But, when reading the fable, did you ever wonder how the mice became so educated?

Hedge schools, of course.

And, much like in the brick and mortar schools we all went to, maths is a big part of the mouse curriculum.

Let’s dive into that curriculum, just a little bit, by joining a mouse classroom, right as the teacher is explaining about squares, cubes and integers.

“An integer,” the teacher is saying, “is any whole number you can think of. For example, 5, 17, 253 and –25 are all integers.

“A square is an integer multiplied by itself. So 25 is a square, because 25 is 5 multiplied by 5.

“A cube is an integer multiplied by itself three times. Therefore, 216 is a cube, as 216 is 6 multiplied by 6 multiplied by 6.

“Finally, a fourth power is an integer multiplied by itself four times. For instance, 16 is a fourth power, since 16 is 2 multiplied by 2 multiplied by 2 multiplied by 2.

“Do you all understand?”

“Yes, teacher,” chorus the little mice.

“Great,” says the teacher. “Then you should have no problem completing tonight’s homework.”

The students all groan at the trick question, as the teacher starts writing their homework on the blackboard:

#### 1. Find the smallest odd integer bigger than 1 that is both a cube and a square.

2. Prove that there is an infinite number of squares that end in 729.

3. Prove that there is an infinite number of cubes that end in 729.

4. Is there any fourth power that ends in 729?

At the end of the school day, all of the mice scurry home to make a start on their assignments. But one of our little mouse friends has lied to his teacher, and hasn’t quite gotten his head around the maths.

Can you help him do his homework?

*Scroll down for the solution to this week’s puzzle.*

**This week’s maths puzzle comes courtesy of Dr*** ***Eugene Gath, lecturer in Mathematics at the University of Limerick*** ***–*** ***who is actively involved in ***mathematics enrichment classes at UL** ***and the*** **Irish Mathematical Olympiad***.**

**Solution:**

Back in school the next day, all of the mice – including our little friend – pull out their maths homework, ready for the teacher to talk them through the solutions.

We and our friend wait nervously to see if we managed to get it all right.

“Okay,” says the teacher. “Let’s start with problem 1.”

**1. Smallest odd integer that is both a square and a cube**

A number that is both a square and a cube must be a sixth power. In order for an integer to be odd, its square or cube root must also be odd. Therefore, the answer will be the next odd number after 1 to the sixth power, ie, 3^{6}. The answer is 729.

**2. Infinite squares ending in 729**

The numbers 1,027, 10,027, 100,027, … , all have squares ending in 729.

First, we reach 27 by finding the square root of 729. From there, we use distributivity – or the ‘square of sum’ formula – to calculate the squares of other numbers ending in 27.

For instance, 1,027^{2} = (1,000 + 27)(1,000 + 27) = (1,000 x 1,000) + (1,000 x 27) + (27 x 1,000) + (27 x 27).

Of the four terms in the sum above, the first three all end in 000, while the fourth term – (27 x 27) – is in 729.

Therefore, 1,027^{2} ends in 729. (In this case, it is exactly 1,054,729.)

A similar proof works for all instances of 10…027, albeit with more zeroes in the first three terms of the sum.

If you don’t want to use distributivity, but prefer pictures, you can think of 10…027^{2} as the area of a square of side 10…027. We consider *b* as 10…000 and *a* as 27, then divide our square area into four pieces, as follows:

Here, *b*^{2}, *ab* and *ab* will end in 000, while *a*^{2} is 729.

**3. Infinite cubes ending in 729**

The numbers 1,009, 10,009, 100,009, … , all have cubes ending in 729.

This can be proved following a similar method to the last question, writing 10…09 as *p* + *q*, where *p *= 10…000 and *q* = 9, where 9 is the cube root of 729.

We then use distributivity.

#### (*p* + *q*)^{3} = (*p* + *q*) (*p* + *q*) (*p* + *q*) = *p*^{3} + 3*p*^{2}*q* + 3*pq*^{2} + *q*^{3}

Here, the first three terms – *p*^{3}, 3*p*^{2}*q *and 3*pq*^{2} – all end in 000, while *q*^{3} is 729.

The formula above can be visualised by calculating the volume of a cube of side (*p* + *q*), splitting the cube into eight separate pieces, as seen here:

**4. No fourth power ending in 729**

The last digit of a fourth power must end in 0, 1, 6, 1, 6, 5, 6, 1, 6, or 1, these being the last digits of the fourth powers of 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.

Therefore, there is no fourth power that ends in 9, and none that end with 729.

Did you and your mouse friend get them all right?

*Main image via Shutterstock*