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THE TEACHER'S QUESTION,
CLUE AND ANSWER

Brian Grainger

brian@grainger1.freeserve.co.uk

I like maths puzzles that seem to expect a solution from what appears to be very little information. The following is an example.

The maths teacher came into the staff room all of a lather.

"I have lost the notes to my lesson on Diophantine Eqautions", he exclaimed. "Has anyone found them?"

"Calm down", said the Head of Maths. "Do you remember anything about the content?"

"I remember I had created an equation where I had summed the 5^th powers of 5 different numbers."

"Do you remember anything about the solution?", asked the Head of Maths.

"I remember the answer had 4 digits, all different", said the teacher.

The Head of Maths tapped his calculator for a couple of minutes and then wrote something on a piece of paper. Passing it to the teacher he said, "Is that what you were looking for?"

"Yes! Amazing!", the teacher exclaimed. Thanking his colleague profusely he left to teach his class.

What was the equation the Head of Maths had written on the piece of paper?

Although it can easily be done with a calculator, the right spreadsheet will get the answer very quickly. You are trying to solve:

a⁵ + b⁵ + c⁵ + d⁵ + e⁵ = mnop

where:

• a, b, c, d, e are all different
• m, n, o, p are all different, (although could be equal to a, b, c, d, e)
• all numbers are integers

Have a look at the fifth powers of the first few integers to get you started.

If you look at the values of the n⁵ for n = 1, 2, 3, … you find that for n=7 or more n⁵ has more than 4 digits.

This means the 5 numbers whose 5^th powers are added must be 5 from the numbers 1 to 6.

Put that another way, it means only 1 of the six numbers 1-6 is missing.

If we calculate 1⁵ + 2⁵ + 3⁵ + 4⁵ + 5⁵ + 6⁵ we can look at the 6 possibilities by subtracting n⁵ from this sum for each value of n= 1 through 6 in turn.

The following can be produced very quickly with a spreadsheet:

From this we can see that only 2 possibilities have an answer with 4 digits and only one of these, 9076, has all digits different.

The equation the Head of Maths had written was therefore:

1⁵ + 2⁵ + 3⁵ + 4⁵ + 6⁵ = 9076

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