Bluezy13:


It's an exchange.
As one person is leaving the boat, the other person is entering at the same time (simultaneously).
A and b can't stay together for even a second!

My answer is correct.
Don't worry about the airtime

bigpicture001:
..on your second paragraph..if the first child and mother cross, the mother of the first child will eat up the already crossed 2nd child

Bluezy13:


I'm faster hehehe
You are not a mathematician

Gemc3:


She's still in the boat. Only her child gets out. She can't eat the other child.

bigpicture001:
...that's y dey called her a witch...dey act at the slightest chance...your not correct dear

Gemc3:


I am not disputing the fact that they are witches. It's just a riddle.

That's my final answer tho.

When you finally get the answer you believe is correct, notify me.

Thanks.

ChuksDaVinci:
And this

ChuksDaVinci:
Find the answers at the end to check whether you’re a true genius.

another riddle for genius
=
in a room, there are two doors, a table and a chair and there is also 6inches round hole at the ceiling.
ade want to escape from that dungeon(room), one door leads to a hungry lion and the other door leads to a glass tunnel with high radiation from sun which would even kill ade if he mistakenly open that door.
so how can he escape from that room?
=
Daqueen7, vivie01, chuksDavinci oya take it

bigpicture001:
.this riddle will take you an entire week..if not months...3 witches with 3 of thier children ; each child for one mother...are about to cross to the other side of a river,with a boat that can contain just two persons.(no padler) in the absence of any of the mother,her child will sure be eaten by any mother witch present..but child don't each child..how can they all cross over intact...get it. and I owe u airtime

Bsoam:


Let mothers be A, B & C and the children be a, b & c respectively. A & a cross first, a drops, A takes back the boat. b & c cross next, any one of the them can take the boat back. let's say c. c takes the boat back, then A & B cross to the other side. Any mother and child can take the boat back, let's use Aa. A & a takes the boat back, a drops and B and A cross to the other side. A & B drop and b takes the boat to the other side. This way all the mothers are on the other side. a & b comes back, a takes the boat to bring c. And everybody has successfully crossed without anyone being there without it mother.

PrincessEstates:


Let the 3 mothers be letters A, B, C
Let their 3 children be letters a, b, c respectively
Let the side of the river they're currently on be S1
Let the other side be S2

Solution:
A&a cross to S2, leaving behind B,b,C&c on S1
A returns to S1, leaving behind a on S2
b&c cross to S2, leaving behind A,B,C on S1
c returns to S1, leaving behind a&b on S2
A&B cross to S2, leaving behind C&c on S1
B&b returns to S1, leaving behind A&a on S2
B&C cross to S2, leaving behind b&c on S1
a returns S1, leaving behind A,B&C on S2
a&b cross to S2, leaving behind c on S1
b returns to S1, leaving behind A,a,B&C on S2
Finally, b&c cross to S2 and they're all complete.

At no point is any child left with another child's mother.

bigpicture001:


You got it correctly. How long did it take uu

bigpicture001:


You got it correctly. How long did it take uu

PrincessEstates:


Not long actually, my brain dey calculate well for these kin things... Lol
.
It actually took me waaaaaay longer to type out the answer