Power digit sum
2014-09-11
Problem 016: Power digit sum
2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
What is the sum of the digits of the number 2^1000?
Solution:
v00000000000000000000000000000000000000000000000000000000000
v00000000000000000000000000000000000000000000000000000000000
v00000000000000000000000000000000000000000000000000000000000
v00000000000000000000000000000000000000000000000000000000000
v00000000000000000000000000000000000000000000000000000000000
v00000000000000000000000000000000000000000000000000000000001
> "0":::::00p01p02p03p04p05p v // INIT
v p64*8"}"p60*p62:6p61:"<" <v p66-1g66 <
>06g1-66p 0> 66g16g%66g16g/g"0"-+ 66g|
v $# < .
>46g:!#^_ 1-46p 06g1-66p 076p > 66g!#^_ v @
v61g66+"0"%+55:+g67*2-"0"g/g61 g66%g61g66<
>g%66g16g/p55+/76p 66g1-66p ^
v00000000000000000000000000000000000000000000000000000000000
v00000000000000000000000000000000000000000000000000000000000
v00000000000000000000000000000000000000000000000000000000000
v00000000000000000000000000000000000000000000000000000000000
v00000000000000000000000000000000000000000000000000000000001
> "0":::::00p01p02p03p04p05p v // INIT
v p64*8"}"p60*p62:6p61:"<" <v p66-1g66 <
>06g1-66p 0> 66g16g%66g16g/g"0"-+ 66g|
v $# < .
>46g:!#^_ 1-46p 06g1-66p 076p > 66g!#^_ v @
v61g66+"0"%+55:+g67*2-"0"g/g61 g66%g61g66<
>g%66g16g/p55+/76p 66g1-66p ^
Start
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Output:
Stack: (0)
Explanation:
Here I implemented a long multiplication algorithm. Then I took the value 1 and doubled it 1000 times. The calculation of the digit sum was then easy.
| Interpreter steps: | 27 332 672 |
| Execution time (BefunExec): | 4.23s (6.46 MHz) |
| Program size: | 60 x 14 (fully conform befunge-93) |
| Solution: | 1366 |
| Solved at: | 2014-09-11 |