Puzzle time - penalty kicks

[Axtremus]

:::

There is something special about 90%, 80% and 50% ... if the quotient is 79% or 81%, or 49% or 51%, or 60% or 70% or 40%, etc., there are many pairs of integers that satisfy (x/y) < quotient and (x+1)/(y+2) > quotient. :man-shrugging:

:::

[jon-nyc]

Lets do some arithmetic. We're looking for an 'a' and 'b' such that:

a/b<4/5
(a+1)/(b+1)>4/5

The first equation simplifies to:
5a<4b

The second simplifies to:
5a+1>4b.

So 5a+1 > 4b > 5a.

But that can't be true for positive integers a and b. So the answer is you can't skip 80%.

Ax notices some other numbers you can't skip. He noticed that 90%, 50% have the same property.

Can you generalize the property?

[taiwan_girl]

@jon-nyc said in Puzzle time - penalty kicks:

Can you generalize the property?

Yes, if you divide two integers (with the numerator smaller than the denominator) and you continually increase both by 1, you will eventually have the answer equal 0.5, 0.8, and 0.9!!

5555

[jon-nyc]

Yes you will, but what other percentages are unavoidable if you start below them and end above them?

[Axtremus]

:::

Let's see ...

From:
a/b < A/B AND (a+1)/(b+1) > A/B

We get to Ba + (B-A) > Ab > B*a

If we are talking only natural numbers, the condition boils down to (B-A) = 1.

So fractions like 1/2 (50%), 2/3 (66.666...%), 3/4 (75%), 4/5 (80%), ... 9/10 (90%) ... all fractions that can be written as X/(X+1) are "special" that way.

:::

[jon-nyc]

Ax got it!

[jon-nyc]

When I first saw this problem last week I thought it was stupid and the answer was obviously “of course you can skip 80”. I didn’t even sit down to play with it for a few days.

[Axtremus]

FWWI, @Klaus might be happy to know that this puzzle got me to install Haskell.

[Klaus]

@axtremus said in Puzzle time - penalty kicks:

FWWI, @Klaus might be happy to know that this puzzle got me to install Haskell.

There are infinitely many good reasons to install Haskell, but if all you want is list comprehensions, then you can do the same thing in many other languages, such as Python.

In Haskell you can of course write the program in a cooler and more general way using monads.

import Control.Monad
[1..500] >>= \x -> [1..500] >>= \y -> guard (x/y < 0.8 && (x+1)/(y+1) > 0.8) >> return (x,y)

[jon-nyc]

I’m a little sad nobody commented on “Diego Primadona”.

[Axtremus]

I have read @Klaus mentioning Haskell a few times in the past, and has been meaning to check it out. This puzzle is just the thing that finally got me to do. These days my primary computer programming language seems to be “go” (or “golang”), mostly because I found it quite convenient for dealing with web APIs.