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 CryptArithmetic Problem

Can you solve the below alphametic riddle by replacing letters of words by a numbers so that the below equation holds true?

BASE +
BALL
---------
GAMES
----------

 
Find numbers replaced letters here! 

Source 

The CryptArithmetic Problem's Solution

What was the problem?

Let's first recall the given equation.

  BASE +
  BALL
---------
GAMES
----------

We are assuming repeating the numbers are not allowed. 

Let's first take last 2 digits operation into consideration i.e. SE + LL = ES or 1ES (carry in 2 digit operation can't exceed 1). For a moment, let's assume no carry generated.

10S + E + 10L + L = 10E + S .....(1)

9 (E - S) = 11L

To satisfy this equation L must be 9 and (E - S) must be equal to 11. But difference between 2 digits can't exceed 9. Hence, SE + LL must have generated carry.So rewriting (1),

10S + E + 10L + L = 100 + 10E + S

9 (E - S) + 100 = 11L

Now if [9 (E - S)] exceeds 99 then L must be greater than 9. But L must be digit from 0 to 9. Hence, [9 (E - S)] must be negative bringing down LHS below 100. Only value of E - S to satisfy the given condition is -5 with L = 5. Or we can say, S - E = 5.

Now, possible pairs for SE are (9,4), (8,3), (7,2), (6,1), (5,0). Out of these only (8,3) is pair that makes equation SE + LL = SE + 55 = 1ES i.e. 83 + 55 = 138. Hence, S = 8 and E = 3.

Replacing letters with numbers that we have got so far.

    1---------
  BA83 +
  BA55
---------
GAM38
----------

Now, M = 2A + 1. Hence, M must be odd number that could be any one among 1,7,9 (since 3 and 5 already used for E and L respectively).

If M = 1, then A = 0 and B must be 5. But L = 5 hence M can't be 1

If M = 7, then A = 3 or A = 8. If A = 3 then B = 1.5 and that's not valid digit. And if A = 8 then it generates carry 1 and 2B + 1 = 8 again leaves B = 3.5 - not a perfect digit.

If M = 9, then A = 4 (A = 9 not possible as M = 9) and B must be 7 with carry G = 1.Hence for first 2 digits we have 74 + 74 + 1 = 149.

Finally, rewriting the entire equation with numbers replacing digits as -

    1
---------
  7483 +
  7455
---------
14938
----------

So numbers for letters are S = 8, E = 3, L = 5, A = 4, B = 7, M = 9 and G = 1.