Puzzle : Ants Walk on a Stick

Twenty-five ants are placed randomly on a meter stick. Each faces east or west. At a signal they all start to march at 1 centimeter per second. Whenever two ants collide they reverse directions. How long must we wait to be sure that all the ants have left the stick?

This sounds immensely complicated, but with a simple insight the answer is immediately clear. What is it?

You need to wait for.....seconds only!
 

Analysing Ants Walk on a Stick

Read the question associated with the walk.

For a moment, let's assume that there are only 2 ants 20 cm away from either end of the stick. Now, after 30 seconds they both will collide with each other & will reverse the direction.

At 50th seconds they will be at the end of the sticks falling off the stick.

So after 80 seconds they will fall off the stick. Now, imagine if ants avoid collision & pass through (or above) each other. Still, both ants would need 80 seconds to leave the stick.

In short, 2 ants' collision & reversal in direction is equivalent to their passing through each other. The other ant continues the journey on the behalf of the first ant & vice versa.

And in case, if they were 100 cm apart, they would need 100 seconds to get off the stick. Again, after collision at halfway mark here, the other ant travels the rest of distance that other ant was supposed to travel.

On the similar note, we can say that even if there are 25 ants on the stick then each ant will cover some distance on the behalf of some other ant. And we need to wait for maximum 100 seconds if 1 of 25 ants is at the edge of the stick. 

All 25 ants together completes the journey of each others in 100 seconds. The ant which is at the edge of stick might complete journey of some other ant which might be only 10 seconds long. But the 100 seconds journey of that ant will be shared by rest of ants. 

Who is older, Joe or Smoe?

Two friends, Joe and Smoe, were born in May, one in 1932, the other a year later. Each had an antique grandfather clock of which he was extremely proud. Both of the clocks worked fairly well considering their age, but one clock gained ten seconds per hour while the other one lost ten seconds per hour. 

On a day in January, the two friends set both clocks correctly at 12:00 noon. "Do you realize," asked Joe, "that the next time both of our clocks will show exactly the same time will be on your 47th birthday?" Smoe agreed. 

Who is older, Joe or Smoe?

Know who is older in the case! 

"Smoe is older than Joe"

What was the puzzle?

Since one of the clock looses and other gains 10 seconds per hour, that means one looses 240 seconds (4 minutes) & other gains 240 seconds (4 minutes) in a day.

Both the clocks are set at 12:00 PM correctly. One has to gain 6 hours (360 minutes) and other has to loose 6 hours (360 minutes) to show the same time again. At the speed of 4 minutes per day the would need 360/4 = 90 days to show the same time again. 

On 90th day, they will come together to show 6:00. Exactly at 12 noon on 90th day one clock must be showing 6:00 PM and other must be showing 6:00 AM, if they have feature of showing AM/PM.

Now as per Joe it would be 47th birthday of Smoe on the day on which the clocks will show the same time. That means, the clocks are set correctly on the noon of 90 days prior to Smoe's birthday which is 1 May for sure but year yet to be known. 

If the year is leap year then 90th day before 1st May will be on 1st February and if it's not a leap year then it would be on January 31. Since, they have set their clocks correctly at 12:00 on some day in January, the year must not be a leap year. 

But if Smoe had been born in 1933, his 47th birthday would have been on May 1, 1980 which is leap year. Hence, Smoe must have born in 1932 and Joe in 1933.

Therefore, Smoe is older than Joe.

The story must be of 1979!

7 Additions and 1 Subtraction

Fill the boxes below with only 7 addition (+) and 1 subtraction (-) operators. 

The smartest way to find the right place 

Correct Placement of Subtraction Operator

This was the question!

For a moment. let's put '+' operator in all given boxes.

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 33.

Let 'x' be the number ahead of which '-' should be there where x can be any number from 1 to 9. 

In above addition the x is already (unknown value) added so first we need to subtract it twice from LHS; first to compensate the addition operation & second for the mandatory subtraction operator that is to be placed at right place.

So for example, if it's ahead of 4 as right place then

1 + 2 + 3 + 4 - 4 - 4 + 5 + 6 + 7 + 8 + 9 = 33.

1 + 2 + 3 + (4 - 4) - 4 + 5 + 6 + 7 + 8 + 9 = 33.

For x, which is already there & can be any from 1 to 9,

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9  - x - x = 33. 

45 - 2x = 33

2x = 12

x = 6.

Hence, the subtraction (-) sign should be ahead of 6.