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This can't be right (# C4A 7LK)
Posted: 14 February 2025, 17:50
by alexxvg
Hi,
maybe I am mistaken or don't understand the rules completely.
This is the setup Checks:
A 10: b+y+p=even | b+y+p=odd
B 25: number of acending or decending values in a row
C 26: b<3 | y<3 | p<3
D 48: b<y | b<p | y<p
b=y | b=p | y=p
b>y | b>p | y>p
My first guess: b=2, y=4, p=5 --> A=true (sum is odd), C=false, D=true (b<y & b<p & y<p)
Second guess: b=4, y=2, p=5 --> B=true (2 numbers in a row)
This lead me to b=1, y=2, P=5 but was incorrect as the correct code should be b=1, y=4, p=2 but this would lead to D being wrong in my first guess.
What am I missing?
Thanks for a hint.
Re: This can't be right (# C4A 7LK)
Posted: 14 February 2025, 18:08
by buu441
D is a true/false verifier for only one of those 9 conditions, so D=true doesn't imply all 3 conditions from your code are true, but only one of those 3. You need to test further to figure out which condition D is testing for. Hope that explanation helps?
Re: This can't be right (# C4A 7LK)
Posted: 14 February 2025, 23:02
by alexxvg
Sorry buu441, but Check D (48) has all 9 conditions
b<y | b<p | y<p
b=y | b=p | y=p
b>y | b>p | y>p
and would always be true, right?
Re: This can't be right (# C4A 7LK)
Posted: 15 February 2025, 00:11
by buu441
What I meant to say is D=true for code b=2 y=4 p=5 means (b<y or b<p or y<p), not (b<y & b<p & y<p) as you had deduced.
Re: This can't be right (# C4A 7LK)
Posted: 15 February 2025, 09:22
by Jellby
The card checks one condition, and only one, but always the same (for a single game). You don't get to choose which one, and the card doesn't get to choose which one, it's already defined by the id C4A7LK. You can think it as your task to figure out which one it is in this game.
It's the same for all other cards, they check one of the conditions printed, the others don't matter. It's just that for "easy" cards the conditions are all mutually exclusive, so once you get a "yes" it can only match one of them, but for the "advanced" cards a "yes" does not pinpoint it, it just narrows it down.
Re: This can't be right (# C4A 7LK)
Posted: 03 March 2025, 00:44
by elamate
alexxvg wrote: ↑14 February 2025, 17:50
Hi,
maybe I am mistaken or don't understand the rules completely.
This is the setup Checks:
A 10: b+y+p=even | b+y+p=odd
B 25: number of acending or decending values in a row
C 26: b<3 | y<3 | p<3
D 48: b<y | b<p | y<p
b=y | b=p | y=p
b>y | b>p | y>p
My first guess: b=2, y=4, p=5 --> A=true (sum is odd), C=false, D=true (b<y & b<p & y<p)
Second guess: b=4, y=2, p=5 --> B=true (2 numbers in a row)
This lead me to b=1, y=2, P=5 but was incorrect as the correct code should be b=1, y=4, p=2 but this would lead to D being wrong in my first guess.
What am I missing?
Thanks for a hint.
First can't be 125 beacuse its a even sum but you know the sum is odd.
Your second guess means that no sequence of number in ascending order.
so you know A= sum odd, and B no sequencei ( so cant be 12, 23, 34, 45, 54,43, 32, 21)
Either y<3 or p<3. Can't be yellow because that not lead any unique solution. so p<3.
That give D: either b<y or b<p.
b<p give you answer: 142
b<y dont give any unique solution. forexample can be: 351,241,252
so the answer is: 142