aiemdao
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Posts posted by aiemdao
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I dont really understand the question, why there are seven red x, what is symbol ( in the middle of dividend.
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20 minutes ago, bonanova said:
=)) thanks
Dividend = abcdefgh
Divisor = Y
7Y have 3 digit
=> the first number of quotient must be 8 or 9. It's canot be 8 because:
Call Z= abcd - 8Y, Z have 2 digit
10Z +e -7Y = 3 digit
The third number of quitient( call it is n) x Y = 3 digit and the subtraction of nY get 2 digit ,ít impossible
So we got 97809.
We know 9Y >= 1000, 8Y<1000
So Y =[112-124]
I check and divinded is 12128316
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the dividend is 97089 ??
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Let number bottles from 1 to 1000
10 firt bottle are drinked by each crimals
45 bottles next art drinked by 10C2 group of criminals
...
So on..
=> actually, 10 criminals can find 1 poison bottle in 1024 bottles
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1,2,3,4,8,9,10 (miles)
Any cost >= 24 cent will can pay by two kind of stamps
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to number coins from 0 to n-1
Convert number of bazillion coin to binary with head is 1, tale is 0
So need to flip: log(2,n) or Int(log(2,n)) +1
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i think
Do not bet the first card
There is 50% chance, the first card is black. so bet all the other card , will win 1$
If the first card iss red, still bet all the card follow untill have more red than black.
Totally, >50% chance win 1%
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I think after 7 move
( O-X ) : E4 -H5 , H3-H7 , H8-D4, E6-E2, D6-G7, C6-B6, G6
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Is O win if we get XOOOOOX?
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4 light ones/48
when 14(A) = 14(B)
The remaining 20 coins = C
A = 7 (A1) + 1(A2) + 6(A3)
Compare (B+A1) and ( A2 + C )
If .... = .....
=> have 2 light coins each side, in B(1) A1(1) C(2)
=> A2 +A3 = 7 heavy coins
If .... >….
A,B have 0 light coin ,
Or B have 1 light coin, C have 2, A1 have 0
=> get 7 coins in A1
If ....<….
C have 0 light coins
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1
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Spoiler
(1) Each ant have 14C7 = 3432 ways to go
because their speed is equal sao thay will met when they go 7 step
=> the probability = ( (7C0)^4 + (7C1)^4 + ... (7C7)^4)/(3432^2) = 0.288
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1 hour ago, bonanova said:
@aiemdao Hi, and welcome to the den. This might help your thinking:
If anyone has a red hat, then the other two would say their hat is one of two colors. But only BMAD made that statement. (So there are no red hats.) As you point out, that means she saw two yellow hats. So Rocdocmac and plasmid have the yellow hats. That makes BMAD's hat blue.
So when you say R would never say "three colors" you're overlooking the case that he could see a blue hat and a yellow hat.
I mean because their comment are sequent . So Rocdocmac know he dont have Red ( Plasmid say one of three ) , and he cannot have B too because if he have blue , Plasmid will know he have Red :v
=>he already know he have Yellow
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something wrong
BMAD: My hat is one of two colors
=> P and R 's hat is : RY, RB, YY, BR, YR
=> if R has B => P know he has R
=> R never say: My hat is one of three colors
Dividend please (on steroids)
in New Logic/Math Puzzles
Posted
??? there are 3 digits before green digits , but there are 4 subtraction different from all after