Did you solve it? The multi-choice maths quiz taking over the world

Earlier today I set you the following problem from the international maths Kangaroo, a multiple choice challenge taken by more than seven million children every year. (For more details about this extraordinary event see the original story.)

Here is is again with the solution.

The four powers

A four digit number abcd satisfies the equation abcd = a^a + b^b + c^c + d^d.

What is a?

• 2

• 3

• 4

• 5

• 6

[Note: abcd means that the first digit is a, the second is b, the third is c and the fourth is d. It does not mean a x b x c x d. The numbers a, b, c and d do not have to be different.]

As an extra, what are b, c and d?

Solution abcd = 3435

Workings

Let’s first list the values of x^x for x = 1 to 6.

1¹ = 1, 2² = 4, 3³ = 27, 4⁴ = 256, 5⁵ = 3125, 6⁶ > 9999.

I didn’t calculate the exact value of 6⁶, because as soon as I realised it was greater than a four digit number I knew I had an upper bound for the values of a, b, c and d. They must all be less than 6.

One of the digits must be 5, because if all the digits were 4 or below, the only way for a^a + b^b + c^c + d^d to have four digits would be 4⁴ + 4⁴ + 4⁴ + 4⁴ = 1024, which does not satisfy the equation.

So, how many of the digits are 5? Let’s go through the cases

All of them. Thus abcd = 5555 ≠ 4 x 3125. So this is not a solution.

Three. Thus abcd = (3x 3125) + (either 1, 4, 27 or 256)

= 9375 + (either 1, 4, 27 or 256)

This means that a = 9. But a ≠ 9 so this is not a solution either.

Two. Thus abcd = 6250 +(two of either 1, 4, 27 or 256). Again, this is not a solution since if it was, a = 6, which we know is false.

So, only one of a, b, c and d is 5.

We now know that abcd = 3125 + (some combinaton of three from 1, 4, 27 and 256). Thus abcd < 4000. So a = 3.

We now have the answer to the multiple choice question, but let’s try and find b, c and d. One of them is 5. So we have

abcd = 27 + 3125 + (two from 1, 4, 27 and 256)

abcd = 3152 + (two from 1, 4, 27 and 256.)

It cannot be the case that both remaining digits are 4, because if they were abcd = 3152 + 256+256 = 3664, and no 6s are allowed.

Imagine that one digit is 4, which would mean that abcd = 3408 + 1, 4 or 27, i.e abcd = 3409, 3412, or 3435.

We can eliminate 3409 because it has a 9, and 3412 because it doesn’t have a 5. The answer is 3435 = 3³ + 4⁴ + 3³ + 5⁵.

To make sure there are no other solutions, let’s see what happens when abcd = 3152 + two from 1, 4 and 27. The possibilities are 3154, 3157, 3180, 3160, 3183, 3206. Only the first one is potentially ok, since it has all digits less than 6, but the number does not satisfy the equation in the question.

We got there!

I hope you enjoyed today’s puzzle. I’ll be back in two weeks.

For more information about entering the Kangaroo in the UK, contact the UKMT. For all other countries, find your national Kangaroo association here.

You can read previous Kangaroos in the World Olympiads Bank, an archive of almost 100,000 problems from international competitions in maths, physics, chemistry, biology informatics and astronomy. In the Kangaroo section, the levels are Pre-Ecolier (lower primary), Ecolier (upper primary), Benjamin (lower secondary), Cadet (middle secondary), Junior (upper secondary), Student (sixth form). The archive was created by Sokratis Romanidis.

I’ve been setting a puzzle here on alternate Mondays since 2015. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.