MaplePrimes - answers and comments on Question, Which number is greater?

Solution

Kitonum — Sat, 09 Feb 2013 03:18:40 Z

Good question!

We will consider the integers from 0 to 10^23 -1. We assume that each of them is written by 23 digits (from 0 to 9). If the integer is less than 10 ^ 22, then write forward the required number of zeros.
Procedure Total(N) finds the total number of numbers in the specified range 0 .. 10^23 -1 , whose sum of the digits is equal to N .

Total:=proc(N::nonnegint)

local L, i, j, k, l, m, n, a, b, M, s, S, T;

if N>207 then return 0 else

L:=[seq([i,0], i=0..54)]:

for i from 0 to 9 do

for j from 0 to 9 do

for k from 0 to 9 do

for l from 0 to 9 do

for m from 0 to 9 do

for n from 0 to 9 do

a:=i+j+k+l+m+n:

L:=subsop(a+1=[a,L[a+1,2]+1], L):

od: od: od: od: od: od:

M:=[seq([i,0], i=0..45)]:

for i from 0 to 9 do

for j from 0 to 9 do

for k from 0 to 9 do

for l from 0 to 9 do

for m from 0 to 9 do

b:=i+j+k+l+m:

M:=subsop(b+1=[b,M[b+1,2]+1], M):

od: od: od: od: od:

s:=0:

for i from 0 to 54 do

for j from 0 to 54 do

for k from 0 to 54 do

for m from 0 to 45 do

if i+j+k+m=N then s:=s+1; S[s]:=[i, j, k, m]; fi;

od: od: od: od:

S:=[seq(S[t], t=1..s)]:

T:=0;

for i in S do

T:=T+L[i[1]+1,2]*L[i[2]+1,2]*L[i[3]+1,2]*M[i[4]+1,2];

od;

fi;

end proc:

Solution of the initial problem:

add(Total(N), N=10..99);

add(Total(N), N=100..207);

38645951372988192979800
61354048627011778971400

Want for comments

Markiyan Hirnyk — Sat, 09 Feb 2013 12:41:25 Z

I wonder the courage of your thoughts! Your answer almost surely is correct in view of
> Total(1);

                               23
> Total(2);

                              276
> Total(207);

                               1
However, I don't understand its machinery. Can you comment your code?

Code comments

Kitonum — Sat, 09 Feb 2013 20:25:03 Z

@Markiyan Hirnyk

First, I build list L . It is a list of frequencies for the sum of the digits in a sequence of six digits, ie how many times this or that sum of of digits occurs.

L = [[0, 1], [1, 6], [2, 21], [3, 56], [4, 126], [5, 252], [6, 462], [7, 792], [8, 1287], [9, 2002], [10, 2997], [11, 4332], [12, 6062], [13, 8232], [14, 10872], [15, 13992], [16, 17577], [17, 21582], [18, 25927], [19, 30492], [20, 35127], [21, 39662], [22, 43917], [23, 47712], [24, 50877], [25, 53262], [26, 54747], [27, 55252], [28, 54747], [29, 53262], [30, 50877], [31, 47712], [32, 43917], [33, 39662], [34, 35127], [35, 30492], [36, 25927], [37, 21582], [38, 17577], [39, 13992], [40, 10872], [41, 8232], [42, 6062], [43, 4332], [44, 2997], [45, 2002], [46, 1287], [47, 792], [48, 462], [49, 252], [50, 126], [51, 56], [52, 21], [53, 6], [54, 1]]

For example, the sub-list [30, 50877] means that the sum of digits equal to 30 in the 6-digit sequence occurs 50877 times. The same means the list M , only for 5 digits.

Note that 23 = 6 +6 +6 +5. Therefore I find all representations of N , which is given as argument of the procedure, as a sum of 4 numbers (the first number from 0 to 54, and so on). Get the list S .

Then I go through the list S in the cycle , and, for example for N = 100, element of the list [20, 35, 30, 15] in accordance with the well-known rule of combinatorics (rule of product) gives me 35127*30492*50877*13992 variants.