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I come in darkness

August 31, 2012 by Ankur 1 Comment

I come in darkness,
though not always at night.
To some I bring joy,
to others I bring fright
What am i?

Tried enough already?

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What king can you make if you take

August 21, 2012 by Ankur 3 Comments

What king can you make if you take,
the head of a lamb,
the middle of a pig,
the hind of a buffalo,
and the tail of a dragon?

Tried enough already?

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Palindrome dates

December 2, 2007 by Ankur 4 Comments

This year on October 2, 2001, the date in MMDDYYYY format will be a palindrome (same forwards as backwards).
10/02/2001
When was the last date that this occurred on?See Solution : Palindrome datesHide Solution

One year can have only one palindrome as the year fixes the month and date too, so the year has to be less than 2001 since we already have the palindrome for 10/02. It can’t be any year in 1900 because that would result in a day of 91, same for 1800 down to 1400. it could be a year in 1300 because that would be the 31st day. So whats the latest year in 1300 that would make a month? When i first solved it, 12th month came to my mind as we have to find the latest date, so i thought it would be 1321. But we have to keep in mind that we want the maximum year in 1300 century with a valid date, so lets think about 1390 that will give the date as 09/31, is this a valid date… ? No, because September has on 30 days, so last will be the 31st August. Which means the correct date would be 08/31/1380.

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Zeroes in 100 factorial

November 30, 2007 by Ankur 3 Comments

How many trailing zeroes are there in 100! (100 factorial) ?

See Solution : Zeroes in 100!
For every factor of 10, there will be one trailing zero similarly for every factor of 5 there will be one trailing zero (as 5*2 =10, and there are enough number of 2’s). But we have to take care of repetitions.

10, 20,…., 90 = 9 zeros

100 = 2 zeros

5, 15, 25……95 = 10 zeros

and 1 extra 5 in each of 25, 50 and 75 = 3 zeros

so total 9+2+10+3 = 24 zeros.

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Average salary

November 30, 2007 by Ankur Leave a Comment

Three coworkers would like to know their average salary. How can they do it, without disclosing their own salaries to other two?

See Solution : Average salaryHide Solution

Let the three persons are A, B and C. Now first A tells the sum of his salary and a random number to B , lets say (SA + A(random no. of A)). Now B adds the sum of his salary and a random number to the number given by A i. e. B makes the total = (SA +A) + (SB+B) and he passes this number to C. (at each step, they don’t show the no. to the third person). Now C does the same, adds his salary and a random number to the amount told by B. Now C passes this total equals to (SA+SB+SC+A+ B+C) to A. Now A subtracts his random no. (A), passes to B then B subtracts his random no. (B) and passes to C. Finally C subtracts his random no. (C) and tells everybody. They divide it by 3. Now they have the average salary.

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100 doors

November 30, 2007 by Ankur 3 Comments

There are 100 doors, all closed. In a nearby cage are 100 monkeys. The first monkey is let out, and runs along the doors opening every one. The second monkey is then let out, and runs along the doors closing the 2nd, 4th, 6th,… all the even-numbered doors. The third monkey is let out. He attends only to the 3rd, 6th, 9th,… doors (every third door, in other words), closing any that is open and opening any that is closed, and so on. After all 100 monkeys have done their work in this way, what state are the doors in after the last pass, which doors are left open and which are closed ?

See Solution : 100 doorsHide Solution

Consider door number 56, monkeys will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1 1st monkey will open the door, pass 2 2nd one will close it, pass 4 open, pass 7 close, pass 8 open, pass 14 close, pass 28 open, pass 56 close. For every pair of divisors the door will just end up back in its initial state. But there are cases in which the pair of divisor has same number for example door number 16. 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is repeated because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will be open at the end.

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