Exponential digital sums: Difference between revisions
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my atomicint $miss = 0; |
my atomicint $miss = 0; |
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(2..$Int).map( -> $exp { |
(2..$Int).map( -> $exp { |
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if (my $sum = ($Int ** $exp).comb.sum) > $Int { last if ++⚛ |
if (my $sum = ($Int ** $exp).comb.sum) > $Int { last if ++⚛$miss > 20 } |
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$sum == $Int ?? "$Int^$exp" !! Empty; |
$sum == $Int ?? "$Int^$exp" !! Empty; |
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}) || Empty; |
}) || Empty; |
Revision as of 00:34, 11 August 2023
Some integers (greater than 1), have the property that the digital sum of that integer raised to some integer power greater than 1, is equal to the original integer.
- E.G.
92 == 81 8 + 1 == 9
Some integers have this property using more than one exponent.
183 == 5832 5 + 8 + 3 + 2 == 18 186 == 34012224 3 + 4 + 0 + 1 + 2 + 2 + 2 + 4 == 18 187 == 612220032 6 + 1 + 2 + 2 + 2 + 0 + 0 + 3 + 2 == 18
Note: every integer has an exponential digital sum equal to the original integer when using an exponent of 1. And, 0 and 1 raised to any power will have a digital sum of 0 and 1 respectively.
- Task
- Find and show the first twenty integers (with their exponents), that satisfy this condition.
- Find and show at least the first ten integers with their exponents, that satisfy this condition in three or more ways.
Raku
Implement a lazy generator. Made some assumptions about search limits. May be poor assumptions, but haven't been able to find any counterexamples. (Edit: and some were bad.)
my @expsum = lazy (2..*).hyper.map( -> $Int {
my atomicint $miss = 0;
(2..$Int).map( -> $exp {
if (my $sum = ($Int ** $exp).comb.sum) > $Int { last if ++⚛$miss > 20 }
$sum == $Int ?? "$Int^$exp" !! Empty;
}) || Empty;
});
say "First twenty-five integers that are equal to the digital sum of that integer raised to some power:";
put .join(', ') for @expsum[^25];
say "\nFirst thirty that satisfy that condition in three or more ways:";
put .join(', ') for @expsum.grep({.elems≥3})[^30];
- Output:
First twenty-five integers that are equal to the digital sum of that integer raised to some power: 7^4 8^3 9^2 17^3 18^3, 18^6, 18^7 20^13 22^4 25^4 26^3 27^3, 27^7 28^4, 28^5 31^7 34^7 35^5 36^4, 36^5 40^13 43^7 45^6 46^5, 46^8 53^7 54^6, 54^8, 54^9 58^7 63^8 64^6 68^7 First thirty that satisfy that condition in three or more ways: 18^3, 18^6, 18^7 54^6, 54^8, 54^9 90^19, 90^20, 90^21, 90^22, 90^28 107^11, 107^13, 107^15 181^16, 181^18, 181^19, 181^20 360^45, 360^46, 360^49, 360^51 370^48, 370^54, 370^57, 370^59 388^32, 388^35, 388^36 523^39, 523^42, 523^44, 523^45 603^44, 603^47, 603^54 667^48, 667^54, 667^58 793^57, 793^60, 793^64 1062^72, 1062^77, 1062^81 1134^78, 1134^80, 1134^82, 1134^86 1359^92, 1359^98, 1359^102 1827^121, 1827^126, 1827^131 1828^123, 1828^127, 1828^132 2116^140, 2116^143, 2116^147 2330^213, 2330^215, 2330^229 2430^217, 2430^222, 2430^223, 2430^229, 2430^230 2557^161, 2557^166, 2557^174 2610^228, 2610^244, 2610^246 2656^170, 2656^172, 2656^176 2700^406, 2700^414, 2700^420, 2700^427 2871^177, 2871^189, 2871^190 2934^191, 2934^193, 2934^195 3077^187, 3077^193, 3077^199 3222^189, 3222^202, 3222^210 3231^203, 3231^207, 3231^209 3448^215, 3448^221, 3448^227
Wren
Well, as the digital sums are all over the place, it's difficult to know how many powers of each number should be tested to solve the task.
This agrees with the Raku solution for the first part but finds some other qualifying numbers for the second part.
Although it would be possible to use BigInt for this, GMP has been used instead to quicken up the search but even so still takes 110 seconds to run.
Arguably both 0 and 1 should qualify as all their powers will have digit sums of 0 and 1 respectively. However, as the Raku solution hasn't included them, neither have I.
import "./gmp" for Mpz
var digitSum = Fn.new { |bi|
var sum = 0
for (d in bi.toString.bytes) {
sum = sum + d - 48
}
return sum
}
var expDigitSums = Fn.new { |max, minWays, limit|
var i = Mpz.one
var c = 0
var n = Mpz.new()
while (c < max) {
i.inc
n.set(i)
var p = 1
var res = []
while (p < limit) {
n.mul(i)
p = p + 1
var ds = digitSum.call(n)
if (i == ds) {
res.add("%(i)^%(p)")
}
}
if (res.count >= minWays) {
System.print(res.join(", "))
c = c + 1
}
}
}
System.print("First twenty-five integers that are equal to the digital sum of that integer raised to some power:")
expDigitSums.call(25, 1, 100)
System.print("\nFirst thirty that satisfy that condition in three or more ways:")
expDigitSums.call(30, 3, 500)
- Output:
First twenty-five integers that are equal to the digital sum of that integer raised to some power: 7^4 8^3 9^2 17^3 18^3, 18^6, 18^7 20^13 22^4 25^4 26^3 27^3, 27^7 28^4, 28^5 31^7 34^7 35^5 36^4, 36^5 40^13 43^7 45^6 46^5, 46^8 53^7 54^6, 54^8, 54^9 58^7 63^8 64^6 68^7 First thirty that satisfy that condition in three or more ways: 18^3, 18^6, 18^7 54^6, 54^8, 54^9 90^19, 90^20, 90^21, 90^22, 90^28 107^11, 107^13, 107^15 181^16, 181^18, 181^19, 181^20 360^45, 360^46, 360^49, 360^51 370^48, 370^54, 370^57, 370^59 388^32, 388^35, 388^36 523^39, 523^42, 523^44, 523^45 603^44, 603^47, 603^54 667^48, 667^54, 667^58 793^57, 793^60, 793^64 1062^72, 1062^77, 1062^81 1134^78, 1134^80, 1134^82, 1134^86 1359^92, 1359^98, 1359^102 1827^121, 1827^126, 1827^131 1828^123, 1828^127, 1828^132 2116^140, 2116^143, 2116^147 2330^213, 2330^215, 2330^229 2430^217, 2430^222, 2430^223, 2430^229, 2430^230 2557^161, 2557^166, 2557^174 2610^228, 2610^244, 2610^246 2656^170, 2656^172, 2656^176 2700^406, 2700^414, 2700^420, 2700^427 2871^177, 2871^189, 2871^190 2934^191, 2934^193, 2934^195 3077^187, 3077^193, 3077^199 3222^189, 3222^202, 3222^210 3231^203, 3231^207, 3231^209 3448^215, 3448^221, 3448^227