Problem # 1

We have a bunch of rectangular blocks and we also have a chessboard. Each rectangular block is exactly the size of two adjacent squares on the chessboard. So the chessboard can be fully covered with 32 of these blocks.

Now suppose we cut off two corner squares of the chessboard (as shown in the picture). Is it possible to cover this new board (now consisting of just 62 squares) with 31 blocks?

To the Reader:
Lokvani puzzles are chosen from a multitude of sources, some of them online. Often, solutions can be found with a few minutes of searching. BUT ... that would entirely miss the point.

The idea is to have some fun and exercise your noggin. Stop slinking away to Google and try to crack it yourself!

Problem # 2

A guy walks into a 7-11 store and selects four items to buy. The clerk at the counter informs the gentleman that the total cost of the four items is $7.11. He was completely surprised that the cost was the same as the name of the store. The clerk informed the man that he simply multiplied the cost of each item and arrived at the total. The customer calmly informed the clerk that the items should be added and not multiplied. The clerk then added the items together and informed the customer that the total was still exactly $7.11.

What are the exact costs of each item?

----------------------------------------------------------------------------------------------------------------------

Please send your solutions to anil@lokvani.com.

Use "Problem Solution M-031303" as the subject line. Please include your full name in the text of the main message. The first one to submit the right answer will be profiled in the next issue of Lokvani.

Please do not post your solution in "Post Comments". No credit will be give for solutions not sent to anil@lokvani.com.

---------------------------------------------------------------------------------------------------------------------

SOLUTIONS TO PROBLEMS ON 2/27/03

Problem # 1:
To encourage Sanjeev's promising tennis career, his father offers him a prize if he wins (at least) two tennis sets in a row in a three-set series to be played with his father and the club champion alternately: father-champion-father or champion-father-champion, according to Sanjeev's choice. The champion is a better player than Sanjeev's father. Which series should Sanjeev choose?"

Solution:
Most people would intuitively think that Sanjeev is better off playing the Champ as few times as possible, and they pick Father-Champ-Father as the answer. This is the right thing to do if winning at least 2 of 3 games leads to an overall win. Unfortunately, that is the wrong thing to do if you have to win two in a row. The key insight here is that if Sanjeev loses the MIDDLE game, he will lose the match.

There are many ways to use that insight to solve the puzzle. Here is one way:
Let f be the probability of Sanjeev beating his father in any one game and c the probability of beating the champ. It is reasonable to assume that f > c.

To win two in a row, Sanjeev has to win the MIDDLE game and at least one of the other two games. In other words he has to win the middle game AND not lose both the other two games. Thus,
Prob {win} = Prob {winning middle game} x [1- Prob {lose both the other two games}] = Prob {winning middle game} x [1- (Prob {losing first game} x Prob {losing third game} )]

For the Father-Champ-Father series, Prob {win} = c [1-(1-f)(1-f)] = c f (2-f)
For the Champ-Father-Champ series, f [1-(1-c)(1-c)] = f c (2-c)
Since f > c, (2-f) < (2-c). Therefore, the probability of winning is higher with the Champ-Father-Champ series.

Problem#2:
AC = 10 (right triangle)
arcCD = arcDA (angle is bisected) = 90 degrees each
angleCAD = 45 degrees
sin(BCA) = 8/10 = 4/5
angleCDB = angleBAC
sin(CDB) = 6/10 = 3/5
sin(CDB)/BC = 1/10 = sin(BCD)/BD
BD = 10sin(BCD)
sin(BCD) = sin(BCA)cos(45)+cos(BCA)sin(45) =7sqrt(2)/10
BD = 7sqrt(2) = ~9.89

|