Fourth powers
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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. [latexpage]
[/vc_column_text][lvca_accordion][lvca_panel panel_title="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 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 $5$s or $6$s 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 $5$s or $6$s 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 $6$s 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 $6$s 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)
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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)
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