1000 doors problem world famous puzzle

This is my second post today I am going to discuss 1000 doors problem it is world famous puzzle asked in many aptitude exams like CAT/E LITMUS/GRE/GMAT and interviews the puzzle is

There are 1000 doors in a high school with 1000 students. The problem begins with the first student opening all 1000 doors next the second student closes doors 2,4,6,8,10 and so on to door 1000 the third student changes the state (opens doors closed, closes doors s open) on doors 3,6,9,12,15 and so on the fourth student Changes the state of doors 4, 8, 12, 16 and so on. This goes on until every student has had a turn. How many doors will be open at the end?

before  discussing this puzzle I want to clear your basic concepts to understand this puzzle

if you list the factors of 24, you get (1, 24), (2, 12), (3, 8), (4, 6).  thing to notice here are Since the factors are listed in pairs, there are an even number of them.

if you list the factors of 36, you get (1, 36), (2, 18), (3, 12), (4, 9), and (6, 6). The last pair of factors is just the square root of 36 listed twice.

Since 6 only counts once as a factor of 36, 36 has nine  factors, an odd number.

so core funda to solve this puzzle is perfect square hace odd number of factors

initially all the doors are closed

person  1 came  opens all of the doors, they're all open. Now person 2 goes, and all of the even numbered doors means multiples of 2  (see in  below table)  are shut. Now it's 3's turn: door 3 is shut, door  6 is open again, and 9 is shut. but if we continue to solve in this type of manner it will take lot of time to work on 1000 doors and 1000 persons

doors	Door 1(close)	Door 2(close)	Door 3 (close)	…………………	Door1000(close)
Person 1	Open	Open	Open	……………….	open
Person 2	Open	Close	Open	……………….	Close
Person 3	Open	Open	Close	……………………	close
…………
Person 1000	………………………	………………………..		……………………………

now think in this manner suppose think door number 36 who can change the state of door number 36 yes those persons number that are factor of 36 so see in table person 1 came he will open the closed door person 2 will close person 3 will open person 4 close person 6 open and in last…….person 36 open .as door 36 is a perfect square having odd number of factors the last condition is open so every perfect square door will have open state  in the end .up to 1000 there are 31 perfect squares doors number last one is

31² = 961

DOOR NUMBER 36

FACTORS               

1	OPEN
2	CLOSE
3	OPEN
4	CLOSE
6	OPEN
9	CLOSE
12	OPEN
18	CLOSE
36	OPEN

                          

now think about door number 24 which is not a perfect square number it have even number of factors so will have closed position in the end

DOOR NUMBER 24 FACTORS

1	OPEN
2	CLOSE
3	OPEN
4	CLOSE
6	OPEN
8	CLOSE
12	OPEN
24	CLOSE

so answer of this puzzle is 31 doors will be open in the end 

see video solution of this puzzle 

https://www.youtube.com/watch?v=cQhaaictt7M

Number of trailing zeros in a product important for CAT,E LITMUS,PLACEMENTS,GMAT,GRE/SSC CGL tier 2

In this post I would like to discuss some of the  fundamental ideas that can be used to solve number of trailing zeros in a product important for CAT,E LITMUS,PLACEMENTS,GMAT,GRE/SSC CGL tier 2. If you have just started your preparation you might find this article helpful.first of all we should have idea about number of trailing zeros trailing zeros are a sequence of 0s  after which no other digits follow.

think number 210 have one trailing zeros but have you ever thought why in this number only one trailing zero because 210 can be written as in form of prime factors

2 × 5 × 3 × 7 = 210  only one pair of 2 and 5 so  there is only there is only one zero at the end of the product  so number of trailing zeros depends upon number of pairs of 2’s and 5’s or We can say that for n number of zero’s at the end of the product we need exactly n combinations of “5 × 2 “. For example

Again 2 × 3 × 5 × 6 × 7 × 15 = 2 × 3 × 5 × 2 × 3 × 7 × 3 × 5

= 2 × 5× 2 × 5 × 3 × 3 × 3  × 7

= 100 × 189 = 18900

Thus there are two zeros because there are two combinations of “5 × 2 “.

Now, 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9

= 1 × 2 × 3 × 2 × 2 × 5 × 2 × 3 × 7 × 2 × 2 × 2 × 3 × 3

= 1 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5 × 3 × 3 × 3 × 3 ×  7

= 1 × 2⁷ × 5 × 3⁴ × 7

= 1 × 2⁶ × 2 × 5 × 3⁴ × 7

= 10 × 2⁶ × 3⁴ × 7

= 10 × 64 × 81 × 7

= 10 × 36288 = 362880

Thus there is only one zero at the end of the product since there is only one combination of 5 × 2.

Again 4 × 125 × 3 = 2 × 2 × 5 × 5 × 5 × 3

= 2² × 5³ × 3

=  2 × 5× 2 × 5    × 5 × 3

= 100 × 15 = 1500  Thus there are only two zeros at the end of the product since there is only one combination of “5 × 2”.

some CAT/ELITMUS /GRE /GMAT/SSC CGL MAINS  RELATED QUESTIONS

Find the number of trailing zeros in 5⁵×10¹⁰×15¹⁵×20²⁰×25²⁵×30³⁰

(a) 130         (b) 80                          (c) 100                         (d) None 

think why answer is  80  since there is only 80 combinations  of “5 × 2”.

Find the number of trailing zeros in 1¹×2²×3³×4⁴------------100¹⁰⁰

 (a) 1500                       (b) 1350                  (c) 1300                (d) 1050

click on this video for solution this question