Own digits power sum: Difference between revisions
(Created a new draft task, Own digits power sum, and added some solutions.) |
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<pre> |
<pre> |
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Same as Wren example. |
Same as Wren example. |
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</pre> |
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=={{header|Julia}}== |
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<lang julia>function isowndigitspowersum(n::Integer, base=10) |
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dig = digits(n, base=base) |
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exponent = length(dig) |
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return mapreduce(x -> x^exponent, +, dig) == n |
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end |
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for i in 10^2:10^9-1 |
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isowndigitspowersum(i) && println(i) |
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end |
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</lang>{{out}} |
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<pre> |
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153 |
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370 |
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371 |
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407 |
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1634 |
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8208 |
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9474 |
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54748 |
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92727 |
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93084 |
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548834 |
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1741725 |
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4210818 |
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9800817 |
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9926315 |
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24678050 |
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24678051 |
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88593477 |
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472335975 |
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534494836 |
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912985153 |
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</pre> |
</pre> |
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Revision as of 15:49, 25 October 2021
- 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.
Julia
<lang julia>function isowndigitspowersum(n::Integer, base=10)
dig = digits(n, base=base) exponent = length(dig) return mapreduce(x -> x^exponent, +, dig) == n
end
for i in 10^2:10^9-1
isowndigitspowersum(i) && println(i)
end
</lang>
- Output:
153 370 371 407 1634 8208 9474 54748 92727 93084 548834 1741725 4210818 9800817 9926315 24678050 24678051 88593477 472335975 534494836 912985153
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