Parallelogram 30 • Level 3 • 28 Mar 2024Wombat Geometry

The factors of 6 are 1, 2, 3 and 6; the factors of 8 are 1, 2, 4 and 8; the factors of 13 are 1 and 13; the factors of 19 are 1 and 19; and the factors of 25 are 1, 5 and 25. We see from this that, of the numbers we are given as options, only 25 has exactly one factor other than 1 and itself. It is worth noticing that 25 is the only square number.

Wombat poo is roughly cubic, and the BBC explains why this is the case.

Removing tile A or tile B or tile D has the effect of reducing the perimeter by a distance equal to twice the side of one tile, whilst removing tile C increases the perimeter by that same distance.

Removing tile E, however, leaves the length of the perimeter unchanged.

Unlike kangaroos, wombats have their pouches upside-down, with the opening at the bottom, nearer to their back legs.

This is because wombats dig with their front legs, and having the pouch opening at the bottom avoids soil been thrown into the pouch.

Koalas are close relatives of wombats, and they also the opening to the pouch at the bottom.

This sounds risky, because koalas spend their time upright, high in trees, but the opening to the pouch is probably quite tight, so there is no risk of the baby koala falling out.

Let the input factors be p,q,r,s and t along the top row, and v,w,x,y and z in the first column, as shown in the grid on the right.

We see that wp=15 and xp=18, so that p is a divisor of both 15 and 18, and hence is either 1 or 3. However, if p=1, then it would follow that w=15. But this is not possible as wr=40. We deduce that p=3. It follows that w=5 and x=6. Since w=5 and wr=40, we have that r=8.

Therefore, as zr=56, z=7. Also, as x=6 and xs=60, s=10. Since vs=20, it follows that v=2. Then, as vq=10, q=5. Hence, as yq=20, we have that y=4 . Finally, as yt=24, we deduce that t=6.

This enables us to complete the table, as shown on the left. (Though, really, we need only to calculate the diagonal entries that we have shown in bold.)

Therefore, we have A+B+C+D+E=6+25+48+40+42=161.

Let ∠QSR =x°. Since PS is parallel to QR, the alternate angles ∠PQS and ∠QSR are equal. So ∠PQS =x°. Therefore, as ∠PSQ:∠PQS =1:5, ∠PSQ =15x°. Therefore, from the right angled triangle PQS we deduce that x+15x+90=180 and so 65x=90. Therefore x=56×90=75.

So ∠QSR is 75°.

We suppose that x, a, b, c, d, e, and f are the numbers in the circles, as shown in the diagram. These numbers are the digits 1, 3, 4, 6, 7, 8, and 9, in some order.

Let the common total of the digits in the three lines be k.

Then: x+a+b+5+x+e+f+2+5+c+d+2=3k,

that is, x+a+b+c+d+e+f+x+14=3k.

Now x+a+b+c+d+e+f=1+3+4+6+7+8+9=38

and therefore

52+x=3k.

It follows that 52+x is divisible by 3. Now x is one of 1, 3, 4, 6, 7, 8, 9, and the only one of these possible values for which 52 + x is divisible by 3 is x = 8. This gives 52 + x = 60.

Hence 60=3k, and so k=20.

In the context of the JMC, we can stop here. But to give a full mathematical solution we should satisfy ourselves that it is possible to place the digits in the circles so that each line adds up to 20.

One way this can be done is shown in the diagram opposite.