PUZZLES 1

2 ball 8 run
(Suppose)
In a match between Australia and England
Scoreboard shows:
Eng 8/0
Over 0.2
Batsman1: 4
batsman2: 4
How is it possible??
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10-coin problem (version 2)
There are n coins lying on a table (n is greater than 10).
there are exactly 10 coins with heads on them. All the rest are tails.
You have to divide the coins into two parts such that both of them contains exactly the same number of heads.
BUT you CANNOT LOOK at the coins. You have to do this without knowing/asking whether a coin has head or tails.
You are allowed to flip the coins. You can flip/turn upside-down a coin as many times you want, and can keep a record of coins which you have flipped.


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On April 1st a typist found the hammers of the typewriter resoldered in an arbitrary order, so typing a text resulted in gibberish. The typist decided to type a document in its entirety using this
typewriter. Afterwards, if the result does not represent the original text, he will type the result, and so on. Prove that the clear text will emerge sooner or later. How many iterations are enough to
guarantee that the clear text will appear, if the typewriter has, say, 46 keys? N keys?

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amazing 4 didits
there is a 4 digit number abcd which when multiplied by 4 becomes dcba.
whats the number and please tell how u worked it out.

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An interesting math problem
This problem has been proposed recently by the french association for math puzzles ("ffjm"). Consider the number x=frac(n*sqrt(2)) where sqrt() stands for square root and frac() for fractional
part. n is any positive integer. What is the largest real number smaller than any x ?
The solution can be expressed as a simple function of sqrt(2).
Good luck... It took me approximately 30 minutes. The answer is relatively simple.


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This is right, so for 46 keys we need to find the maximum LCM for a set of loop lengths that add to 46. The best I can do is 1, 1, 1, 3, 4, 5, 7, 11, 13 => 60060 iterations.

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MissingNumber = SumOfIntegers - ArraySum
ArraySum = sum of all the integers stored in array
SumOfIntegers = Sum of 1 to 100
Now
MissingNumber = SumOfIntegers - ArraySum
Ex:
Let number 41 is missed
ArraySum = 1+2+..+100 (except 41)
= 5009
SumOfIntegers = (100 * 101)/2 = 5050
MissingNumber = 5050 - 5009
= 41

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1)13,22,39,?,137
ans:72
13
13*2 - 4 == 22
22*2-5==39
39*2-6==72
72*2-7==137
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BY:-
SKU STUDENT GUIDE