# Euler 40

We construct a decimal fraction, by concatenating all positive integers:
0.123456789101112 etc.

We define Dx as the digit on position x. Eg D1 = 1, D2 = 2 etc.

What is the product:
D1 * D10 * D100 * D1000 * D10000 * D100000 * D1000000

Just looking at the fraction, we can see that D1 = 1 and D10 = 1.
Let’s quickly save those variables:

D1 <- 1
D10 <- 1

Adding all one-digit integers gives us 9 digits in the fraction.
Adding all two-digit integers gives us 90 * 2 digits in the fraction.

Given that we have the first 9 digits from the one-digit integers, we need to add 90 two-digit integers to get to D99.
That is 45 integers. The first is 10, the second is 11 etc. So number 45 is 54.

If you want convincing:

9 + length(10:54)*2
## [1] 99

D99 is equal to 4. 54 was the two-digit integer we had to add to get to a total of 99 digits, and the last digit in 54 is 4.
The next two-digit integer we can add is 55. Therefore D100 = 5. Lets save that:

D100 <- 5

Now we’ll find D1000.

Adding all two-digit integers brings us to a total of:

9 + 90*2
## [1] 189

The last integer we add is 99. So D189 = 9.

999-189
## [1] 810

digits to the fraction. Doing that with three-digit integers, means that we should add:

810/3
## [1] 270

Integer number 270 we add is 369. That will give us D999 = 9.
The next integer we add to the fraction will be 370. So:

D1000 <- 3

Next – D10000

Adding all three-digit integers gives at total of:

9*1 + 90*2 + 900*3
## [1] 2889

We have to at a total of:

10000-2889
## [1] 7111

digits, and do it by adding four-digit integers.

7111/4
## [1] 1777.75

of them to be precise.

Adding 1777 four digit integers brings us to:

2889 + 1777*4
## [1] 9997

The four-digit integer we add to get to that is 2776. D9997 = 6

D10000 <- 7

D100000

All four-digit integers gets us to:

9*1 + 90*2 + 900*3 + 9000*4
## [1] 38889

100000-38889
## [1] 61111

digits. We are now getting to five-digit integers, so we need to add

61111/5
## [1] 12222.2

of those.
Når vi har tilføjet 12222 tal er vi nået op på:

9 + 90*2 + 900*3 + 9000*4 + 12222*5
## [1] 99999

Integer number 12222 we add (of the five-digit variety) is 22221. D99999 = 1
The next we add is 22222 and:

D100000 <- 2

Finally D1000000

All integers up to and including five-digit integers:

9+90*2+900*3+9000*4 + 90000*5
## [1] 488889

1000000-488889
## [1] 511111

And do it by adding six-digit integers.

511111/6
## [1] 85185.17

The final six-digit integer we concatenate to the fraction before reaching the million is 185184.

When we have done that we have:

9 + 90*2 + 900*3 + 9000*4 + 90000*5 + 85185*6
## [1] 999999

digits (D999999=5).
The next six-digit integer we add is 185185. And:

D1000000 <- 1

Giving the result:

D1 * D10 * D100 * D1000 * D10000 * D100000 * D1000000
## [1] Censored

Lessons learned:
1. Sometimes you don’t have to code all that much.
2. I should practise explaining these sort of things without using quite as many words.