Here we have an oddly shaped board.
Since this board is rather unwieldy as it is, you've decided to cut it into two pieces and rejoin these pieces to form a square.
Assuming you aren't allowed to flip either of the pieces over once the board is cut, where should you make the cut in order to make a square out of the pieces?
tetis flipped L
Cut an l block off the top and slide it right until it slots in.
Excellent!
Easier than it looks, isn't it?
Now the, let's continue.
You have a box of four different kinds of sweets arranged in no particular order. Your job is to divide these treats into five equal portions. To do so, you need to make sure each portion contains the exact same number and variety of sweets.
Got it? Good, then get started!
Start posting the dumb ones.
There are only fice green candies, first job is to divide them.
The two top right portions also include the left adiacent square to not be isolated
As there are also only 5 "ribboned" brown candies, the second portion has to extend down, not right.
The rest comes naturally
As you wish.
A boy and girl are chatting at a local cafe. Looking up from his drink, the boy says:
"You know, I'm 20, which is twice the age you were when I was the age you are now... Isn't that interesting?"
As you can see, this boy isn't the most concise of speakers, but using what he's said, how old is the girl he's chatting to?
Also this one's a classic, when it comes to dumb puzzles.
Garlic is a wonderful seasoning when used in moderation, but the smell can be pretty potent. Someone put garlic cloves in this intricate container and they're really stinking up the room. Use the two corks below to help our friend deal with the smell.
Touch the screen in the place you want to put the corks. Your answer must use no more than corks.
She's 15
i haff tvevle metchsteek
Disgusting pedo.
But also:
Nice!
A careful read of the puzzle reveals that when the boy was the girl's current age, her age was half of the boy's current age. The boy's current age is 20, and half of that is 10, so the girl must have been 10 back then.
Up his nose
Put the corks up his nose.
Took me a while to figure it out, but it's 15, but somebody else already solved it.
Ah, sweet relief!
Each of the three holes in the container is connected to the garlic, so there's no real way to seal off the container using two corks. With no way to contain the smell, our friend had no choice but to plug up his nostrils.
Hopefully he had the good sense to stay inside where no one could see him like that!
Imagine a digital clock like the one shown below. How many times will the clock display three or more of the same number in a row over the course of one day?
In case you were wondering, the clock in this puzzle displays time on a 12-hour scale, not on military time.
LUKE, HERE'S MY ANSWER:
But wait, where did the 15 come from?
3
11:10, 11:11, 11:12, 11:13, 11:14, 11:15, 11:16, 11:17, 11:18, 11:19, 01:11, 02:22, 12:22, 03:33, 04:44, 05:55
16 in 12 hours, so 32 in 24.
(20 - 10) / 2 = 5, the number of years ago the boy was talking about
I'm getting 24.
We know the girls age has to be between 10 and 20. So either you guess 15 immediately or you can use trial and error to on all the numbers that come into question.
01 00:00
02 00:01
03 00:02
04 00:03
05 00:04
06 00:05
07 00:06
08 00:07
09 00:08
10 00:09
11 01:11
12 02:22
13 03:33
14 04:44
15 05:55
16 10:00
17 11:10
18 11:11
19 11:12
20 11:13
21 11:14
22 11:15
23 11:16
24 11:17
25 11:18
26 11:19
And that all twice a day so 52 times.
>over the course of one day
Ah piss. 48 then.
7
Give it another shot!
Are you sure you haven't forgotten a few possible combinations?
Also, don't forget that you need to double your total to account for both a.m. and p.m.
You forgot 00:00
Nevermind, I'm retarded, the 00:0 ones aren't on 12-hour scale, so that's only 12:22 extra and being 17x2 = 34
Very good!
A series of three or more of the same digit appears 34 times over the course of 24 hours. See the chart above for details.
It's easy enough to spot times like 1:11 or 2:22, but many people forget about combinations like 10:00 or 11:12.
I haff tvelve metchsteek
Make it 34, forgot 10:00
FUCK YOU
SOS! Fifteen people are trapped aboard a ship that's going to sink in exactly 20 minutes. Their only chance for survival is the five-person life raft stowed on their vessel. To make matters worse, the waters around the ship are teeming with man-eating sharks, so swimming to safety is out of the question.
A round-trip to the nearest island and back to the boat takes nine minutes on the raft. How many people will live to see dry land?
All of them
>9 minutes for the first group
>9 for the second group
>Last group leaves 2 minutes before it sinks
All of them.
5 to the island, 1 back, 4 saved, 9 minutes passed.
5 more to the island, 1 back, 8 saved, 18 minutes passed.
5 in the boat, RIP the rest, so 13 total. 2 ded.
2 people didn't make it. F
Nicely done!
A moment of silence for the two who didn't make it, please...
Shit, forgot about that
These games sometimes
All of them because this is an E-rated game
It could have a rope attached to it so they can pull it back from the ship. Explain that, "Professor" Layton.
1. Fuck the rest
You are tasked with spray-painting player numbers onto your team's baseball uniforms. You've prepared 10 paper stencils, each with a number from zero to nine. With the stencils cup out, you are now ready to paint player numbers on all six jerseys.
If each jersey has room for two horizontal oriented numbers, what's the fewest number of stencils you need to number the six jerseys?
3
12:22, 01:11, 02:22, 03:33, 04:44, 05:55, 10:00, 11:10, 11:11, 11:12, 11:13, 11:14, 11:15, 11:16, 11:17, 11:18, 11:19
17 times in a 12-hour period, so 34 in one day.
Give it another shot.
You need to use as few stencils as possible to come up with six unique numbers to paint on the jerseys.
You may need to turn everything you know about numbers upside down to find the answer.
Two
0
1
00
01
10
11
Five. 6 and 9 use the same stencil
Unless they want to do "06" or "09", then they'll need six.
1 stencil, the 6
6
9
66
99
96
>69
6 is arrangements of 3 taken 2 at a time, so just 3 stencils
Do they actually need to be unique? If not, 1. If yes, 2
>1
>2
>11
>12
>21
>22
That doesn't appear on a 12-hour clock.
...
Aww no heck, just 1, just use the the 6 and make
6
9
66
69
96
9
Sharp thinking!
As illustrated above, either the 6 or 9 stencil can be flipped to produce another number. This means you get the whole job done with just one stencil
oh so it's just the 6 one, i'm fucking retarded. so 1
based
I'm being a retard and would've still outfoxed Mr Layton.
6 and 0
6
9
69
96
60
90
At the edge of town, there is a traditional private school with 10 boys, 10 girls, and a single teacher.
The school requires students to show proper respect to the teacher and other students by greeting the teacher and other students with one bow.
How many bows could you expect to see on a given morning?
Always switch
1. They do it at the same time.
210
Whichever one Waze says is the fastest.
But the answer is the one that's always busy, cause now all the cars are on the quiet one.
Switch every time. Unless you picked the car on your first try, the other door will have it.
its because the show host knows where the car is we've been over this
F
wouldn't that be 20 bows then since each student does their own bow?
>greeting the teacher and other students with one bow.
So all students have to bow only once, so just 20 bows total.
Think again...
Including the teacher and students, there are a total of 21 people on campus. Stay sharp, though, because you've probably overlooked something important when forming your answer.
Dibs on 69
u did it
it's still a stupid "puzzle" though
400
Each student has 20 people to bow to, and the teacher doesn't bow back.
Each student definetly has to bow more than once, they should've worded it better.
Quiet road, because the DMV won't open that one up to the big trucks even though the road was built for the explicit purpose of handling big truck traffic so they all use the busy road and clog it up.
Well, I mean. Do you consider this to be one or two bows? If they were shaking hands, would you think there's two handshakes occurring?
delicious brown
Forgot the text.
All students in school have to bow, but the teacher doesn't. If you remember that, you get the following results:
The boys bow to each other 90 times.
The girls bow to each other 90 times.
The girls bow to the boys and the boys bow to the girls a total of 200 times.
Lastly, the children bow to the teacher 20.
text me
1 handshake, 2 bows
a handshake explicitly requires a partner so it's a joint action. A bow does not, so they're both individually bowing at the same time.
Two. A handshake is something that can only be done with two people. A bow is an act by a single person to another. Do you consider a student bowing to his teacher to be half a bow?
Fuck you Layton, that's bullshit and you know it.
Below are three siblings: A, B, and C.
A: "I have one older brother and three younger sisters."
B: "I have two older brothers and two younger sisters."
C: "I have three older brother and one younger sister."
For all of their statements to be true, what's the smallest possible number of siblings there can be in the family?
20 because each student is only wearing one each :^)
7.
Six.
Too bad!
It's not as simple as it might seem at first glance.
5
I'm not even going to check if that's right but it's the lowest number possible and that's usually right half of the time
Right on!
The smallest possible number of siblings is six.
At first glance, it might seem like there are only five. But the fact that A says he has three younger sisters means there must be at least three female siblings. Likewise, the fact that C says she has three older brothers means there must be at least three male siblings. So five siblings isn't quite enough.
Holy shit my sides
No it isn't. Three sisters, three brothers. Can't be five. It's six.
From oldest to youngest. M is male, F female. Letters the ones making the statemenst.
M A(M) B(F) M C(F) F
we don't speak of that here good sir
Oh, A is a guy.
There are 15 cookies to share amongst a large group. The first person eats one cookie then passes two equal portions of the leftover cookies to two other people. They both eat a cookie then each pass two equal portions of their remaining cookies to two other people, and so on until all of the cookies are gone.
It takes one minute to eat a cookie. Ignoring the time that it takes to pass them, what's the shortest amount of time it could take for all of the cookies to be eaten?
My nigga
Four?
Correct!
The cookies will be gone in four minutes.
The diagram above shows how you can visualize the puzzle to help you solve it. This seems like the kind of thing that might be useful in real life, but it's not a situation that arises very often, is it?
4 unless it's some bullshit retarded Layton gimmick.
5 mintues
>15 - 1/ 14-2 / 12-3 / 9-4 / 5-5/
>15 - 1/ 14-2 / 12-3 / 9-4 / 5-5/
what
It takes 15 minutes to travel from station A, the first stop on the line, to station B.
It takes five minutes to travel from station B to station C.
It takes 10 minutes to travel from station C to the last station, station D.
However, it doesn't take 30 minutes to travel from station A to station D. Given that this is a straight railway line with no branches, how many minutes does it take to travel from A to D?
FUCK YOU!
It's 10 you motherfucker, Layton.
I'm retarded, forget about my existence.
20 minutes?
A-10-C-5-B-5-D
Depends on how much time they spend on the stations? How the fuck would I know?
Choo choo!
Twenty minutes is correct!
To meet the conditions of this puzzle, station C must be located between station A and station B.
If C is before B, then it's 20, right?
>A-10-C-5-B-5-D
What the fuck does this even mean
retarded as D can be between A and B and still satisfy, as in a d b c
>Can't A->D
>Nothing stops me from going A->C->D
So 25 minutes?
Some people meet for a round-robin card tournament, where every person plays every other person once. Wallace has to leave after only a few hands, sitting out the remainder of the tournament. A total of 59 hands are played at the tournament. How many hands did Wallace play before leaving?
The card game in question is a two-player game, and no person played with the same opponent more than once. No one sat out any hands besides Wallace.
>the last station, station D.
nah
D wouldn't be the last station then.
This one only 200+ IQs will solve
Or people who remember how to calculate shit like this. There's probably 64 hands in total if Wallace had stayed but how do you calculate how many players and how many hands needed so you could deduce Wallace's hands played.
Alright, I'll give you the hints:
1. Mock up an equation with the number of people and the number of matches played as variables. That would be a good place to start.
2. In order to create the equation for the number of hands, you would need to multiply the number of players by the number of players minus one, then divide this product by two. If there were three people, there would have been three hands. If there were four people, there would have been six hands. You can assume from the conditions given that if no one left, the minimum number of hands would still have to be at least 60.
3. There were 12 people at the tournament. Now think through!
64 doesn't work. If everyone plays each opponent once, then we're not squaring the number of players.
Minimum participants is 9, so it was supposed to be 72 hands (9*8). That's as far as I've gotten.
I actually started testing how many games are played with different players and it didn't seem to make sense. The division makes it easier but I don't understand why it is there.
Answer is 4.
He played 4 matches. 11 players would play 55.
Ok, this basically give the answer. 4
Wallace played four hands. First, you need to find the total number of hands if everyone stayed. You can find it with an equation like the one here. If there were 11 people, there would've been 55 matches, and if there were 12 people, there would've been 66. Since we know 59 hands were played, there must have been 12 people at the start. When we subtract the 59 hands played from the 66 ideal, we learn that Wallace missed seven hands. Since everyone would've played 11 hands ideally, that means Wallace only played four hands.
if we count it so for 2 students it's 2 bows and for 3 students it's 6 bows, then for 20 students + 1 teacher there would be a blazing 420 bows - 20 that the teacher would do because the teacher's an asshole
so answer is 400