Solution of Mathemagician's Puzzle

What was the MAGIC?

Let us name ten cards as C1, C2, C3.....C10 and initially they are in order like below.

C1 C2 C3 C4 C5 C6 C7 C8 C9 C10

1. Mavis the 'mathemagician' moved the top card to the bottom of the pack, counting '1', and turned up the next card, placing it on the table. It was the Ace. 

Therefore, card C2 must be the Ace. C2 = ACE. 

So, the order has to be as -

C1 ACE C3 C4 C5 C6 C7 C8 C9 C10.

She moved top card to the bottom & kept Ace card on the table.

C3 C4 C5 C6 C7 C8 C9 C10 C1 

2. She counted two more cards to the bottom of the pack, showed the next card - the '2' - and placed it on the table. 

Therefore, the card C5 must be '2'. C5 = 2.

C3 C4 2 C6 C7 C8 C9 C10 C1 

Moved C3, C4 to the bottom while keeping C5 = 2 on the table.

C6 C7 C8 C9 C10 C1 C3 C4

3. Counted 3 more cards to the bottom of the pack, found '3' as next card. So, the card C9 must be '3'. C9 = 3.

C6 C7 C8 3 C10 C1 C3 C4

Moved C6, C7 and C8 to the bottom while keeping C9 = 3 on table.

C10 C1 C3 C4 C6 C7 C8

4. Counted 4 more cards to the bottom of the pack, found '4' as next card. So, the card C6 must be '4'. C6 = 4.

C10 C1 C3 C4 4 C7 C8

Moved C10, C1, C3 and C4 to the bottom while keeping C6 = 4 on table.

C7 C8 C10 C1 C3 C4

5. Counted 5 more cards to the bottom of the pack, found '5' as next card. So, the card C4 must be '5'. C4 = 5.

C7 C8 C10 C1 C3 5

Moved C7, C8, C10, C1, and C3 to the bottom while keeping C4 = 5 on table.

C7 C8 C10 C1 C3

6. Counted 6 more cards to the bottom of the pack where count goes to the top of the pack after 5, found '6' as next card. So, the card C8 must be '6'. C8 = 6.

C7 6 C10 C1 C3 

Moved C7 to the bottom while keeping C8 = 6 on table.

C10 C1 C3 C7

7. Counted 7 more cards to the bottom of the pack where count goes back to the top of the pack after 4, found '7' as next card. So, the card C7 must be '7'. .C7 = 7

C10 C1 C3 7

Moved C10, C1 and C3 to the bottom while keeping C8 = 6 on table. 

C10 C1 C3.

8.  Counted 8 more cards to the bottom of the pack where count goes back to the top of the pack after 3 and 6, found '8' as next card. So, the card C3 must be '8'. .C3 = 8

C10 C1 8

Moved C10, C1 to the bottom of the pack while keeping C3 = 8 on table.

C10 C1 

9. Counted 9 more cards to the bottom of the pack where count goes back to the top of the pack after 2, 4, 6 and 8, found '9' as next card. So, the card C1 must be '9'. .C1 = 9

C10 9. 

Keeping C1 = 9 on the table leaves only 1 card in the deck.

C10

10. The final card was - ta-daah! - the '10' of Hearts. Hence, .C10 = 10

So the initial order of

C1 C2 C3 C4 C5 C6 C7 C8 C9 C10

must be 

9 A 8 5 2 4 7 6 3 10

What's The Right Answer?

On a multiple-choice test, one of the questions is illegible, but the choice of answers is listed clearly below. 

What’s the right answer?

(a) All of the below.
(b) None of the below.
(c) All of the above.
(d) One of the above.
(e) None of the above.
(f) None of the above.

'THIS' should be the right answer! 

Source
 

Well, 'THIS' is the Right Answer!

What was the question?

Let's recall the question once again.


What’s the right answer?

(a) All of the below.
(b) None of the below.
(c) All of the above.
(d) One of the above.
(e) None of the above.
(f) None of the above.


If (a) is the right answer then as per (a), all below including (b) and (f) are right answers. But (b) & (f) are contradicting each other. So, (a) can't be right.

Now if (c) is right, then as per (c), (a) also must be right. But as concluded above, (a) can't be right. So, (c) also can't be right.

Next, if (b) is correct, then (d) is also correct making 2 options right. Hence, (b) is also false.

If all (a), (b), (c) are false, then (d) has to be false.

If (f) is correct, then what (e) states also is correct making both (e) & (f) correct. Hence, (f) must be false.

That makes (e) is correct answer. 

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!