• ###### titanchinzocommented on "Sum of positive integers 0 through N that are divisible by 3 or 7" kumite

okey thank you

• ###### neuronupheavalcommented on "Sum of positive integers 0 through N that are divisible by 3 or 7" kumite

This does not scale well. Have you tried any n greater than 1,000,000?
Besides, your `for` loop should start with `int i = 3` obviously because this is the first divisible number by 3.

• ###### neuronupheavalcommented on "Sum of first N positive integers that are divisible by 3 or 7" kumite

Indeed, 33, 35, and 42 are struck-through, so they are not being taken into account.

• ###### ygorbunkovcommented on "Sum of first N positive integers that are divisible by 3 or 7" kumite

There's, clearly, a mistake in the example. Considering eligible numbers (divisible by 3 or 7) must not exceed N (30 in your example), 33, 35, 42 should be excluded. Furthermore, because of that the actual sum of #1, #2, #3 must be greater, which is most obvious for #3.

• ###### B1tscommented on " Kumite " kumite

I still don't understand how [[a]] extracts the first character without using spread operator o_O

• ###### rowcasedcommented on " Kumite " kumite

ok, thanks for pointing that out

Nice!

done

• ###### B1tscommented on " Kumite " kumite

Let's golf :D

• ###### ygorbunkovcommented on " Kumite " kumite

This one will drop off words beyond 3-rd (e.g. double-barrelled surname divided by space instead of hyphen), so 3 extra dots seems to be a reasonable cost.

• ###### Chrono79commented on "What is between?" javascript solution

Use spoiler flag next time please.

• ###### ygorbunkovcommented on "What is between?" javascript solution

This comment is hidden because it contains spoiler information about the solution

• ###### ttakuru88commented on "Get iterated digit sum" kumite

fail?

solution1('123')
=> 6
solution2('123')
=> 6
solution3('123')
=> 150

• ###### pawptartcommented on "Get iterated digit sum" kumite

EmBRACE the curly braces...