**What was the problem?**

**L**et's first recall the given equation.

BASE +

BALL

---------

GAMES

----------

We are assuming repeating the numbers are not allowed.

Let's first take

**last**

**2 di**

**gits**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---------

BA

**83**+

BA

**55**

---------

GAM

**38**

----------

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.