Strain your brain

A collection of fairly good puzzles with detailed solution.
1.Odd wine out!

You have 1000 wine bottles, one of which is poisoned. You want to determine which bottle is poisoned by feeding the wine to rats. The poisoned wine takes 24 hours to work.

How many minimum number of rats are required to find the poisoned bottle in 24 hours?

Answer: 10

Solution:

Number the wine bottles from 1 to 1000. Write the binary equivalent of the number. Ten binary digits are required to so (2^10=1024). Let 10 rats correspond to each of the binary digits. For each wine bottles, in case its binary equivalent's digits are 1, feed wine from it to the corresponding rat. 

After 24 hrs see which all rats are dead. The wine bottle which has been fed to exactly those rats is therefore poisoned.

2. Cryptic addition

Each of the 10 letters here (m, a, r, s, v, e, n, u, t and p) represents a digit in the range 0 to 9. Same letters represent same digits and different letters represent different digits. Digits which are most frequently used are 1 and 6. What does n e p t u n e represent?

Answer: 1078610

Solution: 

Given that the most frequently used digits are 1 and 6, that would be n and u (5 times each).

n=1 , not 6 as per Column VI since sum of any two digits + carry has to be less than 30.

Therefore, u=6.  Replacing n and u in table, we get:

Columns		VI	V	IV	III	II	I
Carry		µ	w	z	y	x
				m	a	r	s
+			v	e	1	6	s
+		6	r	a	1	6	s
+		s	a	t	6	r	1
=	1	e	p	t	6	1	e

Column I: 3s+1=10.x+e

s<=9 , implies x<=2

Column II: x+2r+12=10.y+1 èodd implies x is odd, therefore x=1

2r+12=10.y, implies y<=3

If y=0 or 1, r will be negative, hence not possible

If y=2, r=8/2=4

If y=3, r=18/2=9

Therefore, y=2 or 3 and r=4 or 9.

Column III: y+a+8=10.z+6 è implies z<=1

y+a+2=10.z

If z=0, y+a will be negative, hence not possible

Therefore, z=1, y+a=8; 

If y=2, a=6 which is not possible as u=6

Therefore y=3, a=5 and r=9.

Column IV: z+m+e+a+t=10.w+t

m+e+6=10.w è implies w<=2

If w=0, m+e will be negative, hence not possible

If w=2, m+e+6=20; m+e=14 è (m=9,e=5 or m=5, e=9 or m=8, e=6 or m=6, e=8)

Since a=5 and u=6, none of these are possible.

Hence w=1, m+e+6=10; m+e=4 è (m=1,e=3 or m=3, e=1 or m=0, e=4 or m=4, e=0)

Since n=1 e or m cannot be 1. The remaining possibilities are (m=0, e=4 or m=4, e=0)

Let’s try the two possible values of e in Column I: 3s+1=10+e

If e=4, s=13/3 which is not an integer

Hence e=0, s=9/3=3, s=3, m=4

Column VI: µ+6+s=10+e, èµ=1

Column V: w+v+r+a=10.µ+p èv+5=p

Digits which are unassigned so far are 2,7,8

Clearly, v=2, p=7. Only alphabet left is t, hence t=8.

Our answer is : n e p t u n e= 1 0 7 8 6 1 0