math puzzle

[Apache]

here is a math puzzle (not made by myself)

rules:
noone who has already seen the solution
no internet
solutions must be explained

A guy has robots with 7, 8 and 9 arms. Unfortunately, the ones with 7 arms a bugged and everything they say is a lie. The others are ok, they always tell the truth. One day the guy overhears 4 robots talking but he doesn't see them. They're talking about their arms. The first robot claims "Alltogether we have 28 arms."
"No" the second one said "we have 27!"
"26" said the third one.
And the fourth said "Wrong, only 25."

How many arms do they all have? (unknown is a possible answer as well for some)

have fun

[vornksr]

Needs clarification.

Does

A guy has robots with 7, 8 and 9 arms.

mean "A guy has robots with 7, 8, and 9 arms, and no robots with any other number of arms"?

And does

"Alltogether we have 28 arms."

mean "The total number of arms possessed by this group of 4 robots is no more and no less than 28"?

[Apache]

yes and yes

[clementine]

The four robots collectively share seven firearms.

Since they collectively share seven firearms, they all lie. Therefore they neither have 28, 27, 26, or 25 firearms. Nevermind how many appendages they possess?

[Apache]

u mad?^^

[clementine]

u mad?^^

creative solutions stated confidently are my specialty. Tell me if I'm wrong so I can try to think of another.

[Cryptonic]

No, this riddle doesn't work.
The only possible number out of those 4 is 28.
So, we assume each has 7 arms.
So, each one is lying.
But one said 28, so he isn't lying, so they can't have 28.
but that means that one is telling the truth, but 28 is the only number that work.


So, this puzzle just fails.

[Apache]

creative solutions stated confidently are my specialty. Tell me if I'm wrong so I can try to think of another.

oh no you got it all figured out -.-


"So, this puzzle just fails. "
This is a lie, too^^

[Cryptonic]

oh no you got it all figured out -.-


"So, this puzzle just fails. "
This is a lie, too^^

Nah, I think you told it wrong or something.
25 = impossible between 4 robots that have 7, 8, or 9 arms
26 = impossible between 4 robots that have 7, 8, or 9 arms
27 = impossible between 4 robots that have 7, 8, or 9 arms
28 = possible if all robots have 7 arms. but all robots with 7 arms lie and one said there are 28.
For this riddle to work, either 28 shouldn't have been said by the robots, or 28 shouldn't have been the highest number.

So, either robots with 7 arms don't always lie, or robots with 8 or 9 arms can also lie.
Or both.

[Apache]

oh dafuq it's 6, 7 and 8
i double checked the sum but not this -.-
so the answer was 'this shit doesn't make sense'
now new puzzle with 6, 7 and 8 arms

[Cryptonic]

well, that means 3 have to have 7 arms, because 3 are liars.

So, they are 7 7 7 and 6 = 27

[Gingerape]

WELL I GUESS ITS 27 THEN

[Apache]

and order?

[Cryptonic]

The first robot claims "Alltogether we have 28 arms." (7 arms)
"No" the second one said "we have 27!" (6 arms)
"26" said the third one. (7 arms)
And the fourth said "Wrong, only 25." (7 arms)

[Apache]

answer is correct
that was easy. let's try some harder stuff if you want^^

n is an integer 1 <= n <=100
for how many n is n^n a square number?

no calculator + count (explanation required anyway)

[Cryptonic]

wouldn't it just be all the even numbers?
so, 50 intergers

[Apache]

i said it's harder, so don't solve all my puzzles so fast -.-
no it's really harder, it's not so easy. there are more

[Cryptonic]

Yea, you're right. I forgot about odd squares working. so 54?
All evens + 9, 25, 49, and 81?

[Apache]

still not

[Voss]

i think its all even numbers, plus 1, so 51.

[Apache]

nope

[Apache]

tip: cryptonic was very close^^

[Cryptonic]

55, i forgot 1

[Apache]

lacks explanation^^ why all even numbers and all square numbers