Project Euler Forum

Find [sum]U(2,2n+1)1e11 for 2 [le] n [le]10.

What does 1e11 mean?
Do I generate each of these 9 sequences up to e^11 and add up all the terms of all the sequences?
e^11 is about 59, 874. Seems to be an odd upper limit?

1e11 = 10¹¹ = 100,000,000,000

In most programming languages either 10^11 or 10**11

1e11 means 1 times 10 to the 11'th power, which is 100,000,000,000

henk263 wrote:1e11 = 10¹¹ = 100,000,000,000

In most programming languages either 10^11 or 10**11

Or simply 1e11

In Maple, 1e11 is a double, whereas 10^11 is an integer (and thus has infinite/arbitrary precision)

That's also how I learned it in school, 1e11 means you only have 1 significant digit.

In Delphi neither 10^11 nor 10**11 exist.
One can however assign the constant 1e11 (having extended precision) to an int64 variable by truncating it.

henk263 wrote:1e11 = 10¹¹ = 100,000,000,000

In most programming languages either 10^11 or 10**11

Well, usually the 'e' form (aka scientific notation) is used in source code to denote values or by the language to display values, while (^) and (**) are functions (in C and related languages it's pow(x,y), a^b is bitwise XOR).
In Haskell, you have three exponentiation operators:
Prelude> :t (^)
(^) :: (Integral b, Num a) => a -> b -> a
Prelude> :t (^^)
(^^) :: (Integral b, Fractional a) => a -> b -> a
Prelude> :t (**)
(**) :: (Floating a) => a -> a -> a

By the way: in the problem statement 1e11 is used because otherwise we would have an index with an exponent in it.
I'm not sure this would lead to readable page rendering but I could try to change it to 10¹¹ and revert it back if it does not render well.

I changed it, what do we think?

Looks nice enough. Due to the smaller font for sub- and superscripts, it should be unambiguously parseable (hopefully).

I think it could look better using k as index and let k = 10¹¹:

...)_k for 2 [le] n [le]10, k = 10¹¹

Changed that.

Restored now

Thanks for the speedy replies.
I should have realized that "e" in this context means exponent not the Log "e".

Hi!

I'm on Problem 167 as the title suggests and I think I have found what I need to do. Can anyone confirm that for n = 2, the correct value is ??????????? That would be a great checking aid!

Thanks,
Lster

Yes, it's correct.

Thank you!

This one is one that has been giving me long-term problems!

Yup it's a tough one, but lots of fun to solve

Yup it's a tough one, but lots of fun to solve

Well, I'm pretty sure I have the answer for n=2 to 9 correct. But now need to sum for n=10 - which is causing quite some hassle!

One last thing, for n=3, is the result ????????? I have a worry that I may just be lucky on the first value, and n=3 seems a little different because...