Euler_Problem-048.b93

v   v0p01**<  >:30p::>20p\1-:  #v_$$v
5>8#<"}"* #*>:|      ^%g01*g02:\<   2
>5+:::*:*: ^^-># $# .# @#1g03%g01+g0<

[10] 10^10 (const)
[20] iPow sum
[30] total sum (temp storage)


1000      "}"8*
10^10     55+:::*:*:***

iPow (with mod [10] )

  v%g01*g02:\<
::>20p\1-:  #^_$$20g

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I like the occasionally really easy problems between the others. It's like a little break sometimes.

The "trick" here is to understand the modulo operator. If you have `(a + b) % c` you can also write `a%c + b%c`.
And also you can write `(a * b)%c` as `(a%c) * (b%c)`.

So all we do is calculate the sum kinda normally, but we do modulo `10^10` after each step (every addition and multiplication).
We guarantee this way that out numbers never exceed the range of an 64bit integer.