Ram went to a Hotel for a Wedding Reception.

Option B

IMO B

First person - 1,2,3,4,5,6...........................
Second Person- 2,4,6,8,10.................
Third- 3,6,9,12,........................

If we See there is a pattern in cases where light bulb is flipped by 3 persons
e.g- 4=2^2
9=3^2
25=5^2
there are all squares of prime number.

Req cases= squares of {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,}= 11# Count....... Cuz 37^2>1000

B. 11

1st Person will touch all the bulbs. So we need to think about all the bulbs which would be touched exactly twice by other people. The bulbs that would be touched twice should be the person representing the prime numbers and the person representing the square of the prime number. Hence Bulb # 4, 9, 121, ...., 961 would always be touched only twice except for the first answer. Hence the total bulbs touched by exactly 3 people is 11.

Use the answers to see which bulb was flipped exactly three times. Thus, a light bulb x will be flipped by all persons for whom x is a multiple of their person number.

A. 10 will be flipped by persons 1, 2, 5, 10 whose person numbers are factors of 10.
B. 11 will be flipped by persons 1, 11 whose person numbers are factors of 11.
C. 16 will be flipped by persons 1, 2, 4, 8, 16 whose person numbers are factors of 16.
D. 25 will be flipped by persons 1, 5, 25 whose person numbers are factors of 10.
E. 31 will be flipped by persons 1, 31 whose person numbers are factors of 31.

Only Option D, 25 has three factors making it our answer.

For the given conditions, we should look for the numbers, which have only three divisors.

So, all the square of the prime numbers will have exactly three divisors. i.e (1, prime number and the number itself)

The numbers 9,25,49,121,169,289,361,529,841 and 961

Answer is 10

All prime #s from 3 to 31? 10 total?

follow the prime number from 2,3,5,7,11,13,17,19,23,29,31 and then square the number you will get number you want i hope i am not missing anything

OA B.

Thats a very tricky question and it took me wayyyyyy too long.

But I did figure it out with pattern making. Would definitely guess on such question in Gmat, hail precious time :p

So, by pattern we can see that

Person 1 flips bulbs 1,2,3,....,1000

Person 2 flips bulbs 2,4,6,....,1000

Person 3 flips bulbs 3,6,9,....,999

Clearly, person N flips multiple of N bulbs..

So, bulb 1 is flipped by all since 1 is factor of all numbers.

Bulb 2 is flipped by 2 people as 2 as factors (1,2)

Thus, in short, to find bulbs with exactly 3 people flipping, we need numbers who have exactly 3 factors !!!

Example : 4 = 2^2 = factors = 2+1= 3.

Thus, all the SQUARES of primes will constitute these numbers !!

2 ^2
3^2
5^2

Uptil 31^2 which is 961 ....all will have EXACTLY 3 factors.

Beyond 961 we exceed 1000 so notnto count them.

Therefore, count all prime numbers from 2-31.

i.e. 11.

Do we really get such lengthy plus tricky questions? It would take me 6-7 minutes to do this

Posted from my mobile device

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
2 4 6 8 10 12 14
3 6 9 12
4 8 12

and so on, by now it would have become clearer that any no. which has exactly 3 factors is only going to repeat 3 times. e.g. no. 4- 1, 2, 4
now, the ask becomes to find the no.s that have only 3 factors and these no.s are less than 1000.
The above statement essentially asks to calculate the squares of PRIME no.s, since only a square of a prime no. can have 3 factors- 1, prime no., its square

Prime no.s(memorize) sets are- 4, 4, 2, 2, 3, 2, 2
i.e. first set has 4 no.s- 2,3,5,7
second set has 4 no.s- 11, 13, 17, 19
third set has 2 no.s - 23, 29
so on and so forth

Basically- the highest square of prime no. less than 1000 is \(31^2\) = 961
Total no.s = 2,3,5,7,11,13,17,19,23,29,31
count= 11

bulbs that were flipped by exactly 3 persons will have total three factors. These bulbs will be squares of the prime number between 1-1000.
4 - 1,2 and 4.
9 - 1,3, and 9 ...
32^2 = 1024, so prime numbers will have to be less than 32
number of primes between 1- 32: 2,3,5,7,11,13,17,19,23,29,31 =11
Ans: B

Answer:E
1st person flipping light bulb- 1,2,3,.......1000= (1000)( total number of bulb flipped by 1st person)
2nd person flipping light bulb - 2,4,6,.....1000= (500)
. . .... .
. . .......
50 th person flipping light bulb- 50,100,150 .....1000=(20)
51st person flipping light bulb- 51,102,153.....969=( 19)
........................
................
100th person flipping light bulb= 100,200,300.....1000=(10)
...........
500th person flipping light bulb=500,1000 =(2)
..........
1000th person flipping light bulb=1000 only =(1)
Hence number of bulbs flipped by exactly 3 persons = 19+10+2=31

Only perfect squares of prime numbers have 3 factors. Till 1000, we can use these prime numbers 2,3,5,7,11,13,17,19,23,29,31 as 31^2=961. So 11 (B).