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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
----------


BASE + BALL = GAME Solution

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

Black or White Dot on Forehead

There are five men, let's say Tarun, Harish, Lavesh, Manoj and Manish. All of them have a dot mark in their forehead. They can't see the dot on their own forehead, but can see the ones on others. The owner of WHITE dot is an honest person and will never lie, while the owner of BLACK dot always tell the lie.

This are the statement from Tarun, Harish, Lavesh, and Manish:


Tarun: 'I see 3 whites and 1 black'


Harish: 'I see 4 black'


Lavesh: 'I see 3 black and 1 white'


Manish: 'I see 4 white'


What color is the dot on each Tarun, Harish, Lavesh, Manoj, Manish forehead?


Who has which color of dot?

Know color of dot on each forehead!

Source 

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