## Teaser 2923: Powerful numbers

**From The Sunday Times, 30th September 2018** [link]

George and Martha’s five daughters have been investigating large powers of numbers. Each daughter told the parents that they had worked out a number which was equal to its last digit raised to an exact power. When Andrea announced her power, the parents had a 50% chance of guessing the last digit. When Bertha announced a larger power, the same applied. When Caroline announced her power, the parents had only a 25% chance of guessing the last digit; with Dorothy’s larger power, the same applied. When Elizabeth announced hers, the parents only had a 12.5% chance of guessing the last digit. I can tell you that the five powers were consecutive two-digit numbers adding up to a perfect square.

What, in alphabetical order of the daughters, were the five powers?

[teaser2923]

## Jim Randell 12:55 pm

on17 April 2019 Permalink |We ignore the digits 0 and 1 (as

0^k = 0and1^k = 1for positivek), which leaves the digits 2 to 9.A chance of 50% is 1 out of 2, 25% is 1 out of 4, and 12.5% is 1 out of 8, so we are looking for a collection of consecutive numbers where 2 numbers have a choice from 2, another 2 have a choice from 4, and the remaining number has a choice from 8.

This Python program looks at sequences of 5 consecutive 2-digit numbers, and when it finds one that sums to a square it checks that the choices for the digits conform to this pattern.

It runs in 71ms.

Run:[ @repl.it ]Solution:The five powers are: A=44, B=46, C=43, D=47, E=45.There are only three sequences that sum to a square:

And only one of these gives the right set of choices when raising the digits 2-9 to the corresponding powers.

Analytically:

If we consider digits

dfrom 2 to 9, and look at increasing powersksuch thatd^kend in the digitdwe get:We see that there is a repeating pattern, by count of the number of digits, of: (8, 2, 4, 2)…

So for powers of: 1, 5, 9, 13, 17, … there is a choice of 8 possible digits (a 12.5% chance).

For powers of: 2, 4, 6, 8, 10, 12, … there is a choice of 2 possible digits (a 50% chance).

For powers of: 3, 7, 11, 15, … there is a choice of 4 possible digits (a 25% chance).

The five girls have chosen 5 consecutive powers with two 50% chances (2), two 25% chances (4) and one 12.5% chance (8), so the part of the sequence they have chosen is … (4, 2, 8, 2, 4).

We know the 8’s occur when

k = 4n + 1.So the powers the girls have chosen are:

These are 2-digit numbers, which limits

nto the range [3, 24].The sum of the powers is:

and this needs to be a square number.

So,

n = 11, and the girls numbers (and digits counts) are:A and B both have counts of 2, C and D both have counts of 4, E has a count of 8.

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