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 $5$ 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)