Question

Amar, Akbar and Anthony are playing a game. Amar climbs 5 stairs and gets down 2 stairs in one turn. Akbar goes up by 7 stairs and comes down by 2 stairs every time. Anthony goes 10 stairs up and 3 stairs down each time.

Doing this they have to reach to the nearest point of 100th stairs and they will stop once they find it
impossible to go forward. They can not cross 100th stair in anyway
I) How many times can they meet in between on same stair ?
II) Who takes least number of steps to reach near hundred?

Answer 1

To find the answers, we need to calculate the number of steps each player takes before they are unable to continue:

For Amar:

5 stairs up and 2 stairs down per turn

Net gain of 3 stairs per turn

It will take 32 turns to reach 96 stairs (32 * 3 = 96)

In the 33rd turn, Amar will climb 5 stairs and reach 100 stairs.

For Akbar:

7 stairs up and 2 stairs down per turn

Net gain of 5 stairs per turn

It will take 19 turns to reach 95 stairs (19 * 5 = 95)

In the 20th turn, Akbar will climb 7 stairs and reach 100 stairs.

For Anthony:

10 stairs up and 3 stairs down per turn

Net gain of 7 stairs per turn

It will take 14 turns to reach 98 stairs (14 * 7 = 98)

In the 15th turn, Anthony will climb 10 stairs and reach 100 stairs.

I) The players will meet on the same stair whenever they end up on a stair whose number is the same for any two players. To find out the number of times they meet on the same stair, we need to calculate the lowest common multiple (LCM) of the number of turns taken by each player to reach 100 stairs.

Amar takes 33 turns

Akbar takes 20 turns

Anthony takes 15 turns

LCM(33,20,15) = 660

Therefore, they will meet on the same stair 660/33 + 660/20 + 660/15 = 20 + 33 + 44 = 97 times.

II) Anthony takes the least number of steps to reach near hundred, as he only needs 15 turns to reach 98 stairs.