# Project Euler #119: Digit power sum

# Project Euler #119: Digit power sum

lester_munoz + 1 comment Any hint to why i'm getting wrong answer on all testcases except 0, 18 and 19

lester_munoz + 0 comments I found my mistake, the digital sum can be higher than 1000 with bases bigger than 10

ajayprajapati141 + 1 comment is 10^100 bound given in decimal or base B?

at361 + 0 comments decimal

kitchent + 0 comments 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)`

.

platzmaPritzma + 2 comments 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?

platzmaPritzma + 0 comments 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

cantonios + 1 comment 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.

nitishingde + 0 comments can u post some of the output for base=2, i don't know where i am going worng

Sort 12 Discussions, By:

Please Login in order to post a comment