Fourth powers


There are only three numbers (>1)that can be written as the sum of fourth powers of their digits:

1634={{1}^{4}}+{{6}^{4}}+{{3}^{4}}+{{4}^{4}}

8208={{8}^{4}}+{{2}^{4}}+{{0}^{4}}+{{8}^{4}}

9474={{9}^{4}}+{{4}^{4}}+{{7}^{4}}+{{4}^{4}}

Find the smallest number (>1) that can be written as the sum of fifth powers of its digits.

Reveal Solution
It helps to know the fifth powers of each possible digit so here they are:

n n^5 n n^5
0 0 5 3125
1 1 6 7776
2 32 7 16807
3 243 8 32768
4 1024 9 59049

There are a few simple conclusions that can be made from this:

  • There are no one digit numbers (as we don’t include 0 or 1).
  • Two digit numbers can only use 2, 1, 0 since 3^5 is a three digit number and therefore there are none of these.
  • Three digit numbers would have to contain at least one 3 since 222 would only give a total of 96 and therefore there are none of these.

For four digit numbers 6, 5, 4, 3, 2, 1, 0 can be used

1000 \leq n < 2000 The first digit is 1 and you can’t have 5s or 6s since both give totals > 2000. One of the digits must be 4 and none of these combinations work.

2000 \leq n < 3000 The first digit is 2 and you can’t have 5s or 6s since both give totals > 3000. One of the digits must be 4 and none of these combinations work.

3000 \leq n < 4000 The first digit is 3 and you can’t have 6s since this gives totals > 4000. One of the digits must be 4 or a 5 and none of these combinations work.

4000 \leq n < 5000 The first digit is 4 and you can’t have 6s since this gives totals > 5000. One of the other digits must be 4 or a 5.

None of the combinations with a second 4 work.

Examining numbers of the form 4 _ _ _, with one 5 gives:

4150 = 45 + 15 +55 + 05

The next smallest solution is:

4151 = 45 + 15 +55 + 15

 

The following short Python script can be used to show that these are the only 4-digit solutions:

def conway(n):
for a in range(0,10):
for b in range(0,10):
for c in range(0,10):
for d in range(0,10):
m=1000*a+100*b+10*c+d
if a**n+b**n+c**n+d**n==m and m>1:
print(m)

Problem Page Coordinator: Claire Baldwin – Mathematics in Education and Industry
Acknowledgement: The IMA are indebted to MEI for sourcing and supplying Mathematics Today with these well-known puzzles.
First published in Mathematics Today (December 2017)
Image credit: Four by Bart Heird / Flickr / CC BY-NC-ND 2.0
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