Puzzles Page 3
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 Each of the five characters in the word BRAIN has a different value between 0 and 9. Using the given grid, can you find out the value of each character? B R A I N 31 B B R B A 31 N I A B B 32 N I B A I 30 I R A A A 23

37  29  25  27  29
The numbers on the extreme right represent the sum of the values represented by the characters in that row. Also, the numbers on the last raw represent the sum of the values represented by the characters in that column. e.g. B + R + A + I + N = 31 (from first row)

B=7, R=6, A=4, I=5 and N=9

Make total 10 equations - 5 for rows and 5 for columns - and sovle them.

From Row3 and Row4,
N + I + A + B + B = N + I + B + A + I + 2
B = I + 2

From Row1 and Row3,
B + R + A + I + N = N + I + A + B + B - 1
R = B - 1

From Column2,
R + B + I + I + R = 29
B + 2R + 2I = 29
B + 2(B - 1) + 2I = 29
3B + 2I = 31
3(I + 2) + 2I = 31
5I = 25
I = 5

Hence, B=7 and R=6

From Row2,
B + B + R + B + A = 31
3B + R + A = 31
3(7) + 6 + A = 31
A = 4

From Row1,
B + R + A + I + N = 31
7 + 6 + 4 + 5 + N = 31
N = 9

Thus, B=7, R=6, A=4, I=5 and N=9

Submit

There are 9 coins. Out of which one is odd one i.e weight is less or more. How many iterations of weighing are required to find odd coin?

It is always possible to find odd coin in 3 weighings and to tell whether the odd coin is heavier or lighter.
1. Take 8 coins and weigh 4 against 4.
o If both are not equal, goto step 2
o If both are equal, goto step 3

2. One of these 8 coins is the odd one. Name the coins on heavier side of the scale as H1, H2, H3 and H4. Similarly, name the coins on the lighter side of the scale as L1, L2, L3 and L4. Either one of H's is heavier or one of L's is lighter. Weigh (H1, H2, L1) against (H3, H4, X) where X is one coin remaining in intial weighing.
o If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3.
 If both are equal, L4 is the odd coin and is lighter.
 If L2 is light, L2 is the odd coin and is lighter.
 If L3 is light, L3 is the odd coin and is lighter.

o If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2
 If both are equal, there is some error.
 If H1 is heavy, H1 is the odd coin and is heavier.
 If H2 is heavy, H2 is the odd coin and is heavier.

o If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4
 If both are equal, L1 is the odd coin and is lighter.
 If H3 is heavy, H3 is the odd coin and is heavier.
 If H4 is heavy, H4 is the odd coin and is heavier.

3. The remaining coin X is the odd one. Weigh X against the anyone coin used in initial weighing.
o If both are equal, there is some error.
o If X is heavy, X is the odd coin and is heavier.
o If X is light, X is the odd coin and is lighter.
In a sports contest there were m medals awarded on n successive days (n > 1).
1. On the first day 1 medal and 1/7 of the remaining m - 1 medals were awarded.
2. On the second day 2 medals and 1/7 of the now remaining medals was awarded; and so on.
3. On the nth and last day, the remaining n medals were awarded.
How many days did the contest last, and how many medals were awarded altogether?

Total 36 medals were awarded and the contest was for 6 days.

On day 1: Medals awarded = (1 + 35/7) = 6 : Remaining 30 medals
On day 2: Medals awarded = (2 + 28/7) = 6 : Remaining 24 medals
On day 3: Medals awarded = (3 + 21/7) = 6 : Remaining 18 medals
On day 4: Medals awarded = (4 + 14/7) = 6 : Remaining 12 medals
On day 5: Medals awarded = (5 +7/7) = 6 : Remaining 6 medals
On day 6: Medals awarded 6

I got this answer by writing small program. If anyone know any other simpler method, do submit it.

A number of 9 digits has the following properties:
 The number comprising the leftmost two digits is divisible by 2, that comprising the leftmost three digits is divisible by 3, the leftmost four by 4, the leftmost five by 5, and so on for the nine digits of the number i.e. the number formed from the first n digits is divisible by n, 2<=n<=9.
 Each digit in the number is different i.e. no digits are repeated.
 The digit 0 does not occur in the number i.e. it is comprised only of the digits 1-9 in some order.
Find the number.

One way to solve it is Trial-&-Error. You can make it bit easier as odd positions will always occupy ODD numbers and even positions will always occupy EVEN numbers. Further 5th position will contain 5 as 0 does not occur.

The other way to solve this problem is by writing a computer program that systematically tries all possibilities.

1/3 rd of the contents of a container evaporated on the 1st day. 3/4th of the remaining contents of the container evaporated on the second day.

What part of the contents of the container is left at the end of the second day?

Assume that contents of the container is X

On the first day 1/3rd is evaporated.
(1 - 1/3) of X is remaining i.e. (2/3)X

On the Second day 3/4th is evaporated. Hence,
(1- 3/4) of (2/3)X is remaining
i.e. (1/4)(2/3)X = (1/6) X

Hence 1/6th of the contents of the container is remaining

There are four people in a room (not including you). Exactly two of these four always tell the truth. The other two always lie.

You have to figure out who is who IN ONLY 2 QUESTIONS. Your questions have to be YES or NO questions and can only be answered by one person. (If you ask the same question to two different people then that counts as two questions). Keep in mind that all four know each other's characteristics whether they lie or not.

What questions would you ask to figure out who is who? Remember that you can ask only 2 questions.
Submitted
You have 3 baskets, & each one contains exactly 4 balls, each of which is of the same size. Each ball is either red, black, white, or purple, & there is one of each color in each basket.

If you were blindfolded, & lightly shook each basket so that the balls would be randomly distributed, & then took 1 ball from each basket, what chance is there that you would have exactly 2 red balls?

There are 64 different possible outcomes, & in 9 of these, exactly 2 of the balls will be red. There is thus a slightly better than 14% chance [(9/64)*100] that exactly 2 balls will be red.

A much faster way to solve the problem is to look at it this way. There are 3 scenarios where exactly 3 balls are red:

1   2   3

-----------

R   R   X

R   X   R

X   R   R

X is any ball that is not red.
There is a 4.6875% chance that each of these situations will occur.

Take the first one, for example: 25% chance the first ball is red, multiplied by a 25% chance the second ball is red, multiplied by a 75% chance the third ball is not red.

Because there are 3 scenarios where this outcome occurs, you multiply the 4.6875% chance of any one occurring by 3, & you get 14.0625%

Consider a state lottery where you get to choose 8 numbers from 1 to 80, no repetiton allowed. The Lottery Commission chooses 11 from those 80 numbers, again no repetition. You win the lottery if atleast 7 of your numbers are there in the 11 chosen by the Lottery Commission.

What is the probablity of winning the lottery?

The probability of winning the lottery is two in one billion i.e. only two person can win from one billion !!!

Let's find out sample space first. The Lottery Commission chooses 11 numbers from the 80. Hence, the 11 numbers from the 80 can be selected in 80C11 ways which is very very high and is equal to 1.04776 * 1013

Now, you have to select 8 numbers from 80 which can be selected in 80C8 ways. But we are interested in only those numbers which are in 11 numbers selected by the Lottery Commision. There are 2 cases.
 You might select 8 numbers which all are there in 11 numbers choosen by the Lottery Commission. So there are 11C8 ways.
 Another case is you might select 7 lucky numbers and 1 non-lucky number from the remaining 69 numbers. There are ( 11C7 ) * ( 69C1 ) ways to do that.
So total lucky ways are
= ( 11C8 ) + ( 11C7 ) * ( 69C1 )
= (165) + (330) * (69)
= 165 + 22770
= 22935

Hence, the probability of the winning lottery is
= (Total lucky ways) / (Total Sample space)
= (22935) / ( 1.04776 * 1013)
= 2.1889 * 10-9
i.e. 2 in a billion.

Submit
To move a Safe, two cylindrical steel bars 7 inches in diameter are used as rollers.

How far will the safe have moved forward when the rollers have made one revolution?

The safe must have moved 22 inches forward.

If the rollers make one revolution, the safe will move the distance equal to the circumference of the roller. Hence, the distance covered by the safe is
= PI * Diameter (or 2 * PI * Radius)
= PI * 7
= 3.14159265 * 7
= 21.99115
= 22 inches approx.
SubmittIf a rook and a bishop of a standard chess set are randomly placed on a chessboard, what is the probability that one is attacking the other?

Note that both are different colored pieces.

The probability of either the Rook or the Bishop attacking the other is 0.3611

A Rook and a Bishop on a standard chess-board can be arranged in 64P2 = 64*63 = 4032 ways

Now, there are 2 cases - Rook attacking Bishop and Bishop attacking Rook. Note that the Rook and the Bishop never attack each other simultaneously. Let's consider both the cases one by one.

Case I - Rook attacking Bishop
The Rook can be placed in any of the given 64 positions and it always attacks 14 positions. Hence, total possible ways of the Rook attacking the Bishop = 64*14 = 896 ways

Case II - Bishop attacking Rook
View the chess-board as a 4 co-centric hollow squares with the outermost square with side 8 units and the innermost square with side 2 units.

If the bishop is in one of the outer 28 squares, then it can attack 7 positions. If the bishop is in one of the 20 squares at next inner-level, then it can attack 9 positions. Similarly if the bishop is in one of the 12 squares at next inner-level, then it can attack 11 positions. And if the bishop is in one of the 4 squares at next inner-level (the innermost level), then it can attack 13 positions.

Hence, total possible ways of the Bishop attacking the Rook
= 28*7 + 20*9 + 12*11 + 4*13
= 560 ways

Thus, the required probability is
= (896 + 560) / 4032
= 13/36
= 0.3611
itted ed