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Project Euler #119: Digit power sum
Project Euler #119: Digit power sum
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Any hint to why i'm getting wrong answer on all testcases except 0, 18 and 19
I found my mistake, the digital sum can be higher than 1000 with bases bigger than 10
is 10^100 bound given in decimal or base B?
decimal
I wonder how many digits would it be if it is base and the number is . One digit? How do you count ? or ?
EDIT: The answer is the latter one. This implies that the upper bound of the digit sum for bases can be beyond . I used
(100 / len(str(B-1)) + 1) * (B - 1).can any one explain me this question . not sure about the base part
This part of the question was not clear.
We know the summation and the number is on base 10 however n's base did not specified. 512 = (5+1+2)^n
Should we accept that as on base 10?
How about summation? Let (x)_b stands for number x on base b. Should we look for the following equality? (ABC)_b = ((A)_b+(B)_b+(C=_c)^n
Base of n doesn't matter. The base B only matters for the digit summation. For example, with B=2, we have 5^4=1001110001_2=625. You can express the 5 and 4 in base 2 if you want, but it doesn't change the answer.
can u post some of the output for base=2, i don't know where i am going worng