The Race of Multi-Tasking Clowns

Many contestants entered the unicycle race, but only the best multi-tasking clown came out on top, considering that each had to juggle clubs while trying to win the race! Most of the pack were soon disqualified after dropping a club or falling off the unicycle. In the end, four of the best clowns crossed the finish line. 

From this information and the clues below, can you determine each clown's full (real) name, club color (one is silver), and placement?

Places: 1st, 2nd, 3rd, 4th
First Names: Kyle, Matt, Jake, Leon
Surnames: Turner, Pettle, Vertigo, Wheeley
Colors: Green, Orange, Silver, Red

1. Mr. Turner (who isn't Matt) finished one place ahead of the red club juggler.

2. The four are Leon, Mr. Vertigo, the one who juggled with orange clubs, and the one who came in third place.

3. Kyle finished behind (but not one place behind) Mr. Wheeley.

4. Matt and Mr. Pettle finished consecutively, in some order.

5. The one who juggled with green clubs finished two places behind Jake.

Know here final stats of the race! 

In The Race of Multi-Tasking Clowns

What was the puzzle?

Let's take a look at the given data.

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Places: 1st, 2nd, 3rd, 4th

First Names: Kyle, Matt, Jake, Leon

Surnames: Turner, Pettle, Vertigo, Wheeley

Colors: Green, Orange, Silver, Red


1. Mr. Turner (who isn't Matt) finished one place ahead of the red club juggler.

2. The four are Leon, Mr. Vertigo, the one who juggled with orange clubs, and the one who came in third place.

3. Kyle finished behind (but not one place behind) Mr. Wheeley.

4. Matt and Mr. Pettle finished consecutively, in some order.

5. The one who juggled with green clubs finished two places behind Jake.  

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We will make a table like below and fill it step by step.

 
 STEPS :

1] As per clue (2), Leon, Mr. Vertigo, the one who juggled with orange clubs, and the one who came in third place are supposed to be different. Hence, Leon or Vertigo didn't finish at 3. So, Pettle or Turner must be at 3.

2] But as per (1), Matt isn't a Turner, and as per (4) Matt and Pettle are two different persons. Hence, Matt didn't finish at 3 for sure.

3] As per (3), Mr. Wheely too can't be at 3 as in that case no placement would be left for Kyle. And Jake too can't be at 3 as in that case no placement would be left for the one who juggled with green clubs.

So, if Leon, Matt and Jake aren't at 3 then Kyle must be at 3 having surname Pettle or Turner.

4] With that, Mr. Wheely must have finished first as clue (3) suggests.

5] Now, as per (1), Turner didn't finish at no.4 since there would be no place for red club juggler. Also, as per (4), Pettle can't be at 4 since Matt can't be at 3 (STEP 3). Hence, at no.4, Vertigo is there.

6] As per (2), Vertigo and Leon are 2 different contestants. Also, as per clue (5) Jake can't be at no.4 as in that case there would be no place left for the one who juggled green clubs. Hence, the first name of Vertigo is Matt.

7] Now, clue (4) suggests Pettle is at no.3

8] The only place for the Turner is no.2 and by (1), Kyle Pettle is red club juggler.

9] Now, the only locations left for the contestants pointed by clue (5) i.e. for Jake and green club juggler are 2 and 4.

10] The only location left for Leon is no.1 and as per (2), Leon and orange club juggler are different. Hence, Leon must be a silver club juggler and Jake Turner is orange club juggler.

CONCLUSION : 

So, the final stats of race are - 

   

Ultimate Test of 3 Logic Masters

Try this. The Grand Master takes a set of 8 stamps, 4 red and 4 green, known to the logicians, and loosely affixes two to the forehead of each logician so that each logician can see all the other stamps except those 2 in the Grand Master's pocket and the two on her own forehead. He asks them in turn if they know the colors of their own stamps:

A: "No."

B: "No."

C: "No."

A: "No."

B: "Yes."

What color stamps does B have? 

'THIS' could be his color combination! 

Earlier logicians had been part of "Spot On The Forehead!" and  "Spot on the Forehead" Sequel Contest

The Wisest Logic Master!

What was the challenge in front of him?

Let's denote red by R and green by G. Then, each can have combination of RR, RG or GG.

So, there are total 27 combinations are possible.

1.  RR RR GG
2.  RR GG RR
3.  GG RR RR

4.  GG GG RR
5.  RR GG GG
6.  GG RR GG

7.  RR RG GG
8.  GG RG RR
9.  RG RR GG
10.RG GG RR
11. RR GG RG
12. GG RR RG

13. RR RG RG
14. GG RG RG
15. RG RR RG
16. RG GG RG
17. RG RG RR
18. RG RG GG

19. RR RR RG
20. GG GG RG
21. RG RR RR
22. RG GG GG
23. RR RG RR
24. GG RG GG

25. RR RR RR
26 .GG GG GG

27. RG RG RG.

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1. Now, obviously (19) to (26) are invalid combinations as those have more than 4 red or green stamps.

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2. In first round, everybody said 'NO' thereby eliminating (1) to (6) combinations. That's because, for example, if C had seen all red (or all green) then he would have known color of own stamps as GG (or RR). Similarly, A and B must not have seen all red or all green.

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3. For (9 - RG RR GG), A would have responded correctly at second round as NO of B had eliminated GG and NO of C had eliminated RR for A in first round. Similarly, (10) is eliminated.

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4. For (11 -  RR GG RG ), C would would have responded correctly immediately after NO of A had eliminated GG and NO of B had eliminated RR for him in first round. With similar logic, (12) also get eliminated.

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5. Remember, B has guessed color of own stamps only in second round of questioning.

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6. For B in (13), his logic would be I can't have RR (total R>4) but GG [ No.(11) - RR GG RG] and RG can be possible. But (11) is eliminated by C's response in first round. That leaves, (13) in contention.

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7. Similarly, if it was (14 - GG RG RG) combination, then B's thought would be - I can't have GG (total G>4) but can have RR as in (12) - GG RR RG which is already eliminated by C's NO response at the end of first round. Hence, I must have RG. That means (14) also remains in contention.

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8. On similar note, (17), (18) remains in contention after A's NO at the start of second round.

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9. If it was (15), then A would have been responded with RG when C's NO in first round eliminates RR (as proved in 2 above) and GG (as proved in 4 above) both. Similarly, (16) is also eliminated after C's NO in first round as above.

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11. For (27), B's logic would be -

"If I had RR then A must had seen RR-RG and had logic - 

"Can't have RR (total R>4); if had GG then C would have answered with RG after I and B said NO in first round itself. Hence, I must tell RG in second round."

Similarly, A's response at the start of second round eliminates GG for me.

Hence, I must have RG combination."

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10. So only possible combinations left where only B can deduce color of own stamps are -

7.  RR RG GG
8.  GG RG RR

13. RR RG RG
14. GG RG RG

17. RG RG RR
18. RG RG GG

27. RG RG RG.

If observed carefully all above, we can conclude that B must have RG color combination of stamps after observing A's and C's stamps as above to correctly answer in the second round.


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Plan an Unbeatable Strategy

Two people play a game of NIM. There are 100 matches on a table, and the players take turns picking 1 to 5 sticks at a time. The person who takes the last stick wins the game. (Both players has to make sure that the winner would be picking only 1 stick at the end) 

Who has a winning strategy?

And what must be winning strategy in the person who takes the last stick looses?

This could be the winning strategy! 

Planned The Unbeatable Strategy!

What is the game?

The first person can plan an unbeatable winning strategy.

CASE 1 : The person picking last stick is winner.

All that he has to do is pick 4 sticks straightaway at the start leaving behind 96 stick. Then, he has to make sure that the count of remaining stick will be always divisible by 6 like 96, 90, 84, 78......6. 

So if the opponent takes away 2 sticks in his first turn, then first person has to take 6 - 2 = 4 sticks leaving behind 90 sticks there. That is if the opponent takes away X stick the first person need to pick 6 - X sticks.

Now, when there are 6 stick left, even if opponent takes away 5 sticks then 1 stick will be left for the first person.

And even if the opponent picks 4 sticks then first person will take 2 remaining sticks.

CASE 2 : The person picking last stick is looser. 

Now the first person need to take away 3 sticks in first turn leaving behind 97. Next, he has to make sure the count of remaining sticks reduced by 6 after each of his turn. That is, the count should be like 91,85,79,72......7.

So if the opponent takes away 4 sticks in his first turn, then first person has to take 6 - 4 = 2 sticks leaving behind 91 sticks there. That is if the opponent takes away X stick the first person need to pick 6 - X sticks.

When there are 7 sticks are left then even if the opponent takes away 5 sticks then first person can force him to pick the last stick by picking only 1 stick of remaining 2. 

And if the opponent takes away 4 sticks at this stage, the first person still can force him to pick last stick by picking 2 of remaining 3 sticks.

Conclusion : The first person always has a chance to plan a winning strategy.

The Ping Pong Puzzle

Three friends (A, B and C) are playing ping pong. They play the usual way: the winner stays on, and the loser waits his/her turn again. At the end of the day, they summarize the number of games that each of them played:

A played 10
B played 15
C played 17.

Who lost the second game? 

This person played & lost the second game! 

Participant of the Second Game!

How games were played?

A played 10, B played 15 and C played 17 games. So total number of presences are 10 + 15 + 17 = 42. Every 2 presences form a game. Hence, the number of games played are 42/2 = 21.

Let's take into consideration the minimum number of games that a player can play. For that, he need to loose every game that he has played. That is, if he has played first game then he must have out in second but replaced looser of second in third game. In short, he must have played odd numbered of games like 1,3,5,7,9,11,13,15,17,19,21.That's 11 games in total.

And if he had made debut in second game then he must had played even numbered games like 2,4,6,8,10,12,14,16,18,20. That's 10 games in total.

Since, in the case only A has played 10 games, he must have made debut in second game where he lost that game to make comeback in 4th game thereby replacing looser of third game.

The Number Game!

Let’s play a game. You name an integer from 1 to 10. Then we’ll take turns adding an integer from 1 to 10 to the number our opponent has just named, giving the resulting sum as our answer. Whoever reaches 100 first is the winner.

You go first. What number should you choose?

This is how you can be winner!

Winning The Number Game!

What was the game?

Here the player whose number 'forces' sum to fall in range of 90-99 will be ending on losing side. 

That means, somehow if you 'force' the total at some point to 89 then opponent has to fall in the range of 90-99 with his number. 

To get 'door' to total 89 you have to force the previous sum to 78 so that opponent is forced to open a 'door' for total 89 for you.

And so on backward you have to make stops at 67, 56, 45, 34, 23, 12, 1.

So you have to start with 1 & achieve all above milestones.

Let's verify our conclusion. Suppose you started with 1.

You       Sum      Opponent    Sum

1            -             8              9
3           12            5              17
6           23            7              30
4           34            9              43
2           45            4              49
7           56            10            66
1           67            3              70
8           78            2              80
9           89            4              93
7           100           

YOU WIN!

100m Running Race

Lavesh, Bolt, and Lewis race each other in a 100 meters race. All of them run at a constant speed throughout the race.

Lavesh beats Bolt by 20 meters.
Bolt beats Lewis by 20 meters.

How many meters does Lavesh beat Lewis by ? 

Know here the answer! 

Source 

Winner Beats Second Runner Up by...

What was the question? 

Let Lavesh's speed be 10 m/s. Then he must have taken 10 seconds to finish the race. Since Bolt was beaten by Lavesh by 20 m he must have run 80m when Lavesh finished race in 10 seconds (t=10). So his speed would be 8 m/s.

Now Bolt requires 100/0.8 = 12.5 seconds to finish the race. When he finished, Lewis was 20m behind i.e. 80m from starting point at t = 12.5. So Lewis speed is 80/12.5 = 6.4 m/s

At t = 10 seconds, when Lavesh finished his race Lewis must be at 6.4 x 10 = 64 m from starting point. Hence Lavesh beats Lewis by 100 - 64 = 36 m.

Another method.

Let L be the speed of Lavesh, B be the speed of Bolt & W be that of Lewis. Then,

L/B = 100/80 = 5/4

L = (5/4) B  .......(1)

Similarly,

B/W = 100/80 = 5/4

B = (5/4) W .......(2)

Putting (2) into (1),

L = (5/4) x (5/4) W

L/W = 25/16

L = (25/16) W

For given time t, when Lavesh finished the race,

Distance by L/ Distance by W = 25 / 16

100 m/ Distance by W = 25 / 16

Distance by W = (16 x 100) / 25 = 64.

  
So when Lavesh finished cross line at 100 m, Lewis was at 64m i.e. 36m behind. In other words, Lavesh beats Lewis by 100 - 64 = 36 m.