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There is a basket full of hats. 3 of them are white and 2 of them are
black. There are 3 men Tom, Tim, and Jim. They each take a hat out of
the basket and put it on their heads without seeing the hat they
selected or the hats the other men selected.
The men arrange themselves
so Tom can see Tim and Jim’s hats, Tim can see Jim’s hat, and Jim can’t
see anyone’s hat.
Tom is asked what color his hat is and he says he doesn’t know.
Tim is asked the same question, and he also doesn’t know.
Finally, Jim is asked the question, and he does know.
What color is his hat?
Know color of his hat!
What was the challenge?
Since there are only 2 black hats if Tom had saw 2 black hats (on heads of Tim & Jim) then he would have realized that he must be wearing white hat. Since, he says he doesn't know color of his hat that mean he can see either 2 white hats or 1 black & 1 white hat on heads of other 2.
Now when Tim is saying he doesn't know color of his hat that means Jim must not be wearing black hat. If Jim had black hat then Tim would know that his color of hat is white but can't be black again as in that case Tom would have identified color of his hat in first attempt.
Hence, Jim must be wearing white hat!
Two people play a game of NIM. There are 100 matches on a table, and the
players take turns picking 1 to 5 sticks at a time. The person who
takes the last stick wins the game. (Both players has to make sure that the winner would be picking only 1 stick at the end)
Who has a winning strategy?
And what must be winning strategy in the person who takes the last stick looses?
This could be the winning strategy!
What is the game?
The first person can plan an unbeatable winning strategy.
CASE 1 : The person picking last stick is winner.
All that he has to do is pick 4 sticks straightaway at the start leaving behind 96 stick. Then, he has to make sure that the count of remaining stick will be always divisible by 6 like 96, 90, 84, 78......6.
So if the opponent takes away 2 sticks in his first turn, then first person has to take 6 - 2 = 4 sticks leaving behind 90 sticks there. That is if the opponent takes away X stick the first person need to pick 6 - X sticks.
Now, when there are 6 stick left, even if opponent takes away 5 sticks then 1 stick will be left for the first person.
And even if the opponent picks 4 sticks then first person will take 2 remaining sticks.
CASE 2 : The person picking last stick is looser.
Now the first person need to take away 3 sticks in first turn leaving behind 97. Next, he has to make sure the count of remaining sticks reduced by 6 after each of his turn. That is, the count should be like 91,85,79,72......7.
So if
the opponent takes away 4 sticks in his first turn, then first person
has to take 6 - 4 = 2 sticks leaving behind 91 sticks there. That is if
the opponent takes away X stick the first person need to pick 6 - X
sticks.
When
there are 7 sticks are left then even if the opponent takes away 5
sticks then first person can force him to pick the last stick by picking
only 1 stick of remaining 2.
And if the opponent takes away 4 sticks at this stage, the first person still can force him to pick last stick by picking 2 of remaining 3 sticks.
Conclusion : The first person always has a chance to plan a winning strategy.
Is it possible to fill each box in with an arithmetic operation so that this becomes a true equation?
Did you too find it true?
What wasn't looking possible?
Yes, it's possible. All you need to do is recall BODMAS (Brackets, Of, Division, Multiplication, Addition, Subtraction) rule in mathematics that we learned in school.
