Own digits power sum
- Description
For the purposes of this task, an own digits power sum is a decimal integer which is N digits long and is equal to the sum of its individual digits raised to the power N.
- Example
The three digit integer 153 is an own digits power sum because 1³ + 5³ + 3³ = 1 + 125 + 27 = 153.
- Task
Find and show here all own digits power sums for N = 3 to N = 8 inclusive.
Optionally, do the same for N = 9 which may take a while for interpreted languages.
C
<lang c>#include <stdio.h>
- include <math.h>
- define MAX_DIGITS 9
int digits[MAX_DIGITS];
void getDigits(int i) {
int ix = 0;
while (i > 0) {
digits[ix++] = i % 10;
i /= 10;
}
}
int main() {
int n, d, i, max, lastDigit, sum, dp;
int powers[10] = {0, 1, 4, 9, 16, 25, 36, 49, 64, 81};
printf("Own digits power sums for N = 3 to 9 inclusive:\n");
for (n = 3; n < 10; ++n) {
for (d = 2; d < 10; ++d) powers[d] *= d;
i = (int)pow(10, n-1);
max = i * 10;
lastDigit = 0;
while (i < max) {
if (!lastDigit) {
getDigits(i);
sum = 0;
for (d = 0; d < n; ++d) {
dp = digits[d];
sum += powers[dp];
}
} else if (lastDigit == 1) {
sum++;
} else {
sum += powers[lastDigit] - powers[lastDigit-1];
}
if (sum == i) {
printf("%d\n", i);
if (lastDigit == 0) printf("%d\n", i + 1);
i += 10 - lastDigit;
lastDigit = 0;
} else if (sum > i) {
i += 10 - lastDigit;
lastDigit = 0;
} else if (lastDigit < 9) {
i++;
lastDigit++;
} else {
i++;
lastDigit = 0;
}
}
}
return 0;
}</lang>
- Output:
Same as Wren example.
Go
<lang go>package main
import (
"fmt" "math" "rcu"
)
func main() {
powers := [10]int{0, 1, 4, 9, 16, 25, 36, 49, 64, 81}
fmt.Println("Own digits power sums for N = 3 to 9 inclusive:")
for n := 3; n < 10; n++ {
for d := 2; d < 10; d++ {
powers[d] *= d
}
i := int(math.Pow(10, float64(n-1)))
max := i * 10
lastDigit := 0
sum := 0
var digits []int
for i < max {
if lastDigit == 0 {
digits = rcu.Digits(i, 10)
sum = 0
for _, d := range digits {
sum += powers[d]
}
} else if lastDigit == 1 {
sum++
} else {
sum += powers[lastDigit] - powers[lastDigit-1]
}
if sum == i {
fmt.Println(i)
if lastDigit == 0 {
fmt.Println(i + 1)
}
i += 10 - lastDigit
lastDigit = 0
} else if sum > i {
i += 10 - lastDigit
lastDigit = 0
} else if lastDigit < 9 {
i++
lastDigit++
} else {
i++
lastDigit = 0
}
}
}
}</lang>
- Output:
Same as Wren example.
Wren
Includes some simple optimizations to try and quicken up the search. However, getting up to N = 9 still took a little over 4 minutes on my machine. <lang ecmascript>import "./math" for Int
var powers = [0, 1, 4, 9, 16, 25, 36, 49, 64, 81] System.print("Own digits power sums for N = 3 to 9 inclusive:") for (n in 3..9) {
for (d in 2..9) powers[d] = powers[d] * d
var i = 10.pow(n-1)
var max = i * 10
var lastDigit = 0
var sum = 0
var digits = null
while (i < max) {
if (lastDigit == 0) {
digits = Int.digits(i)
sum = digits.reduce(0) { |acc, d| acc + powers[d] }
} else if (lastDigit == 1) {
sum = sum + 1
} else {
sum = sum + powers[lastDigit] - powers[lastDigit-1]
}
if (sum == i) {
System.print(i)
if (lastDigit == 0) System.print(i + 1)
i = i + 10 - lastDigit
lastDigit = 0
} else if (sum > i) {
i = i + 10 - lastDigit
lastDigit = 0
} else if (lastDigit < 9) {
i = i + 1
lastDigit = lastDigit + 1
} else {
i = i + 1
lastDigit = 0
}
}
}</lang>
- Output:
Own digits power sums for N = 3 to 9 inclusive: 153 370 371 407 1634 8208 9474 54748 92727 93084 548834 1741725 4210818 9800817 9926315 24678050 24678051 88593477 146511208 472335975 534494836 912985153