Question

There are 1000 bulbs in a row numbered 1 to 1000. First person switch on all the 1000 bulbs, next person switch off bulbs 2,4,6,8 and so on to 1000; third person changes the state( switch on bulbs which are OFF, switch off bulbs which are ON) of bulbs 3,6,9,12,15 and so on; fourth person changes the state of bulbs 4,8,12,16 and so on. This goes on till every person in a group of 1000 has had a turn. How many bulbs will be ON at the end?

• A. 31
• B. 11
• C. 49
• D. none of these

Answer 1

The correct option is A

31

Let us take some simple numbers to see if there is a pattern

If you take a number, say 6, In the first round it will be on, 2${}^{nd}$ it will be off, 3${}^{rd}$ it will be on, and 6${}^{th}$ it will be off. Hence its final state will be off.

Consider a number 4-In the first, second and fourth round it will be ON, OFF & finally ON

Consider a number 9. in the first, third and 9${}^{th}$ round it will change its state and its final state will be on

From this, you can deduce that the state of the bulb is determined by the property of the number; i.e. all numbers with odd number of factors will remain in an "on” state and all other numbers will be off. Conclusion- All the bulbs which are on are "perfect square” numbered bulbs

Thus, the bulbs which will be on are all in the positions of numbers which are perfect squares.

The question is thus, asking for the number of perfect squares till 1000= 31 (i.e. 1${}^2$,2${}^2$...........31${}^2$)