Let N be an integer expressed in decimal.
Let D(N) be the sum of the digits of N expressed in decimal.
Let S(N) be the sum of the squares of the digits of N expressed in decimal.
Let C(N) be the sum of the cubes of the digits of N expressed in decimal.
Find the smallest pandigital number N (that is, containing all of the digits 0 to 9) which contains D(N), S(N) and C(N).
d=[0:9];
d1=sum(d);
d2=sum(d.^2);
d3=sum(d.^3);
fprintf(' %5d %5d %5d
',d1,d2,d3)
45 285 2025 (sums of digits, squares, cubes)
shows it can't be done with just 10 digits, as the single 2 would have to be followed by 8, 0 and 5 in different places.
Similar problems occur if one of the digits is allowed to repeat in an 11-digit number:
for a=0:9
d=[0:9 a];
d1=sum(d);
d2=sum(d.^2);
d3=sum(d.^3);
fprintf('%1d %5d %5d %5d
',a,d1,d2,d3)
end
repeat sum of
digit digs sqrs cubes
0 45 285 2025
1 46 286 2026
2 47 289 2033
3 48 294 2052
4 49 301 2089
5 50 310 2150
6 51 321 2241
7 52 334 2368
8 53 349 2537
9 54 366 2754
So, allow two of the digits to repeat or one to triplicate:
for a=0:9
for b=a:9
d=[0:9 a b];
d1=sum(d);
d2=sum(d.^2);
d3=sum(d.^3);
fprintf('%1d %1d %5d %5d %5d
',a,b,d1,d2,d3)
end
end
I went through the output manually and found one that works, marked to the right of its line below:
extra
digits D(N) S(N) C(N)
0 0 45 285 2025
0 1 46 286 2026
0 2 47 289 2033
0 3 48 294 2052
0 4 49 301 2089
0 5 50 310 2150
0 6 51 321 2241
0 7 52 334 2368
0 8 53 349 2537
0 9 54 366 2754
1 1 47 287 2027
1 2 48 290 2034
1 3 49 295 2053
1 4 50 302 2090
1 5 51 311 2151
1 6 52 322 2242
1 7 53 335 2369
1 8 54 350 2538
1 9 55 367 2755
2 2 49 293 2041
2 3 50 298 2060
2 4 51 305 2097
2 5 52 314 2158
2 6 53 325 2249
2 7 54 338 2376
2 8 55 353 2545
2 9 56 370 2762
3 3 51 303 2079
3 4 52 310 2116
3 5 53 319 2177
3 6 54 330 2268
3 7 55 343 2395
3 8 56 358 2564
3 9 57 375 2781
4 4 53 317 2153
4 5 54 326 2214
4 6 55 337 2305
4 7 56 350 2432
4 8 57 365 2601
4 9 58 382 2818
5 5 55 335 2275
5 6 56 346 2366
5 7 57 359 2493
5 8 58 374 2662
5 9 59 391 2879
6 6 57 357 2457
6 7 58 370 2584 1 2584 370 6679
6 8 59 385 2753
6 9 60 402 2970
7 7 59 383 2711
7 8 60 398 2880
7 9 61 415 3097
8 8 61 413 3049
8 9 62 430 3266
9 9 63 447 3483
I placed the unused digits from D(N), S(N) and C(N) to the left and right of the concatenated S(N) and C(N), from low to high. (The D(N), as you can see is incorporated in the C(N).)
So the pandigital in question is 125,843,706,679.
Edited on August 18, 2023, 1:20 pm
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Posted by Charlie
on 2023-08-18 13:18:23 |
computer-aided solution
