Digital root/Multiplicative digital root
The multiplicative digital root (MDR) and multiplicative persistence (MP) of a number (N) is calculated rather like the Digital root except digits are multiplied:
- Set MDR to N and MP to 0
- While MDR has more than one digit:
- Find a replacement MDR as the multiplication of the digits of the current MDR
- Increment MP
- Return MP and MDR
- Task
- Tabulate the MP and MDR of the numbers 123321, 7739, 893, 899998
- Tabulate MP versus the first five numbers having that MP, something like:
MP: [n0..n4] == ======== 0: [0, 10, 20, 25, 30] 1: [1, 11, 111, 1111, 11111] 2: [2, 12, 21, 26, 34] 3: [3, 13, 31, 113, 131] 4: [4, 14, 22, 27, 39] 5: [5, 15, 35, 51, 53] 6: [6, 16, 23, 28, 32] 7: [7, 17, 71, 117, 171] 8: [8, 18, 24, 29, 36] 9: [9, 19, 33, 91, 119]
Show all output on this page.
- References
- Multiplicative Digital Root on Wolfram Mathworld.
- Multiplicative digital root on Wikipedia.
- Multiplicative digital root on The On-Line Encyclopedia of Integer Sequences.
D
<lang d>import std.stdio, std.algorithm, std.typecons, std.range, std.conv;
/// Multiplicative digital root. auto mdRoot(in int n) pure /*nothrow*/ {
auto mdr = [n]; while (mdr.back > 9) mdr ~= reduce!q{a * b}(1, mdr.back.text.map!(d => d - '0')); //mdr ~= mdr.back.text.map!(d => d - '0').mul; //mdr ~= mdr.back.reverseDigits.mul; return tuple(mdr.length - 1, mdr.back);
}
void main() {
"Number: (MP, MDR)\n====== =========".writeln; foreach (immutable n; [123321, 7739, 893, 899998]) writefln("%6d: (%s, %s)", n, n.mdRoot[]);
auto table = 10.iota.zip((int[]).init.repeat).assocArray; auto n = 0; while (table.byValue.map!walkLength.reduce!min < 5) { table[n.mdRoot[1]] ~= n; n++; } "\nMP: [n0..n4]\n== ========".writeln; foreach (const mp; table.byKey.array.sort()) writefln("%2d: %s", mp, table[mp].take(5));
}</lang>
- Output:
Number: (MP, MDR) ====== ========= 123321: (3, 8) 7739: (3, 8) 893: (3, 2) 899998: (2, 0) MP: [n0..n4] == ======== 0: [0, 10, 20, 25, 30] 1: [1, 11, 111, 1111, 11111] 2: [2, 12, 21, 26, 34] 3: [3, 13, 31, 113, 131] 4: [4, 14, 22, 27, 39] 5: [5, 15, 35, 51, 53] 6: [6, 16, 23, 28, 32] 7: [7, 17, 71, 117, 171] 8: [8, 18, 24, 29, 36] 9: [9, 19, 33, 91, 119]
Alternative Version
<lang d>import std.stdio, std.algorithm, std.typecons, std.range;
uint digitsProduct(uint n) pure nothrow {
typeof(return) result = !!n; while (n) { result *= n % 10; n /= 10; } return result;
}
/// Multiplicative digital root. Tuple!(size_t, uint) mdRoot(uint m) pure nothrow {
auto mdr = m .recurrence!((a, n) => a[n - 1].digitsProduct) .until!q{ a <= 9 }(OpenRight.no).array; return tuple(mdr.length - 1, mdr.back);
}
void main() {
"Number: (MP, MDR)\n====== =========".writeln; foreach (immutable n; [123321, 7739, 893, 899998]) writefln("%6d: (%s, %s)", n, n.mdRoot[]);
auto table = 10.iota.zip((int[]).init.repeat).assocArray; auto n = 0; while (table.byValue.map!walkLength.reduce!min < 5) { table[n.mdRoot[1]] ~= n; n++; } "\nMP: [n0..n4]\n== ========".writeln; foreach (const mp; table.byKey.array.sort()) writefln("%2d: %s", mp, table[mp].take(5));
}</lang>
More Efficient Version
<lang d>import std.stdio, std.algorithm, std.range;
/// Multiplicative digital root. uint[2] mdRoot(in uint n) pure nothrow {
uint mdr = n; uint count = 0;
while (mdr > 9) { uint m = mdr; uint digitsMul = !!m; while (m) { digitsMul *= m % 10; m /= 10; } mdr = digitsMul; count++; }
return [count, mdr];
}
void main() {
"Number: [MP, MDR]\n====== =========".writeln; foreach (immutable n; [123321, 7739, 893, 899998]) writefln("%6d: %s", n, n.mdRoot);
auto table = 10.iota.zip((uint[]).init.repeat).assocArray; auto n = 0; while (table.byValue.map!walkLength.reduce!min < 5) { table[n.mdRoot[1]] ~= n; n++; } "\nMP: [n0..n4]\n== ========".writeln; foreach (const mp; table.byKey.array.sort()) writefln("%2d: %s", mp, table[mp].take(5));
}</lang> The output is similar.
Haskell
Note that in the function mdrNums
we don't know in advance how many numbers we'll need to examine to find the first 5 associated with all the MDRs. Using a lazy array to accumulate these numbers allows us to keep the function simple.
<lang haskell>import Control.Arrow
import Data.Array
import Data.LazyArray
import Data.List (unfoldr)
import Data.Tuple
import Text.Printf
-- The multiplicative persistence (MP) and multiplicative digital root (MDR) of -- the argument. mpmdr :: Integer -> (Int, Integer) mpmdr = (length *** head) . span (> 9) . iterate (product . digits)
-- Pairs (mdr, ns) where mdr is a multiplicative digital root and ns are the -- first k numbers having that root. mdrNums :: Int -> [(Integer, [Integer])] mdrNums k = assocs $ lArrayMap (take k) (0,9) [(snd $ mpmdr n, n) | n <- [0..]]
digits :: Integral t => t -> [t] digits 0 = [0] digits n = unfoldr step n
where step k = if k == 0 then Nothing else Just (swap $ quotRem k 10)
printMpMdrs :: [Integer] -> IO () printMpMdrs ns = do
putStrLn "Number MP MDR" putStrLn "====== == ===" sequence_ [printf "%6d %2d %2d\n" n p r | n <- ns, let (p,r) = mpmdr n]
printMdrNums:: Int -> IO () printMdrNums k = do
putStrLn "MDR Numbers" putStrLn "=== =======" let showNums = unwords . map show sequence_ [printf "%2d %s\n" mdr $ showNums ns | (mdr,ns) <- mdrNums k]
main :: IO () main = do
printMpMdrs [123321, 7739, 893, 899998] putStrLn "" printMdrNums 5</lang>
- Output:
Note that the values in the first column of the table are MDRs, as shown in the task's sample output, not MP as incorrectly stated in the task statement and column header.
Number MP MDR ====== == === 123321 3 8 7739 3 8 893 3 2 899998 2 0 MDR Numbers === ======= 0 0 10 20 25 30 1 1 11 111 1111 11111 2 2 12 21 26 34 3 3 13 31 113 131 4 4 14 22 27 39 5 5 15 35 51 53 6 6 16 23 28 32 7 7 17 71 117 171 8 8 18 24 29 36 9 9 19 33 91 119
Python
Python: Inspired by the solution to the Digital root task
<lang python>try:
from functools import reduce
except:
pass
def mdroot(n):
'Multiplicative digital root' mdr = [n] while mdr[-1] > 9: mdr.append(reduce(int.__mul__, (int(dig) for dig in str(mdr[-1])), 1)) return len(mdr) - 1, mdr[-1]
if __name__ == '__main__':
print('Number: (MP, MDR)\n====== =========') for n in (123321, 7739, 893, 899998): print('%6i: %r' % (n, mdroot(n))) table, n = {i: [] for i in range(10)}, 0 while min(len(row) for row in table.values()) < 5: mpersistence, mdr = mdroot(n) table[mdr].append(n) n += 1 print('\nMP: [n0..n4]\n== ========') for mp, val in sorted(table.items()): print('%2i: %r' % (mp, val[:5]))</lang>
- Output:
Number: (MP, MDR) ====== ========= 123321: (3, 8) 7739: (3, 8) 893: (3, 2) 899998: (2, 0) MP: [n0..n4] == ======== 0: [0, 10, 20, 25, 30] 1: [1, 11, 111, 1111, 11111] 2: [2, 12, 21, 26, 34] 3: [3, 13, 31, 113, 131] 4: [4, 14, 22, 27, 39] 5: [5, 15, 35, 51, 53] 6: [6, 16, 23, 28, 32] 7: [7, 17, 71, 117, 171] 8: [8, 18, 24, 29, 36] 9: [9, 19, 33, 91, 119]
Python: Inspired by the more efficient version of D.
Substitute the following function to run twice as fast when calculating mdroot(n) with n in range(1000000). <lang python>def mdroot(n):
count, mdr = 0, n while mdr > 9: m, digitsMul = mdr, 1 while m: m, md = divmod(m, 10) digitsMul *= md mdr = digitsMul count += 1 return count, mdr</lang>
- Output:
(Exactly the same as before).