Another CryptArithmatic Problem

Replace letters with numbers assuming numbers can't be repeated. 

     SEND
  + MORE
  ----------
 = MONEY

Process of decryption is here! 

Decrypting The CryptArithmatic Problem

What was the problem? 

Let's take a look at the equation once again.

       S  E N D
  +   M O R E
  -----------------
 = M O N E Y

Now letter M must be representing the carry generated & it must be 1. 

And if M = 1 then S must be 9 or 8 if carry is generated from hundreds place. 

In any case, O can be either 0 or 1. But can't be 1 as M = 1 hence O = 0.

If O = 0 then E + 0 = N i.e. E = N if there is no carry from tens place. 

Hence, N = E + 1.

Let's turn towards tens place now. With no carry from units place N + R = 10 + E .

Putting N =  E + 1, we get, R = 9. And with carry from units place, we have,

1 + N + R = 10 + E, gives us R = 8. Hence either R = 9 and S = 8 or R = 8 or S = 9.

For a moment, let's assume R = 9 and S = 8 with no carry from units place, then, 

   8 E N D
+ 1 0 9  E
=========
1 0 N E Y 

For this to be correct, we need carry at thousands place generated from hundreds 

place. That's only possible if E = 9 and carry is generated from tens place

forwarded to hundreds place. Since 9 is already being used for R this combination

is just impossible.

Hence, R = 8 and S = 9 with carry 1 from units place.

Now it looks like,

   9 E N D
+ 1 0 8 E
========
1 0 N E Y 

This means D + E >= 10 i.e. D + E = 10 + Y.  And numbers left are 2,3,4,5,6,7.

If E = 2 then D must be 8 or 9 for D + E >= 10.

Since 8 and 9 already taken, this is not possible.

If E = 3 then D can be 7 but Y would be 0 in the case. 

Since O = 0 already taken this value of E is also not valid. 

For any other value of D, D + E < 10 .

If E = 4 then D = 7 or 6 and Y = 0 or 1.

Both are taken hence this value of E in invalid.

Also,D <= 5 in the case gives  D + E < 10 .

If E = 6, N = 7 then, D <= 5. 

With D = 5 or 4, Y = 0 or 1, both are used for O and M already.

And for D = 2 or 3,D + E < 10 .

In short, this value of E is also not valid.   

So only value of E left is 5. Hence, N = 6 and D = 7. That gives, Y = 2.

To conclude,

  9 5 6 7
+ 1 0 8 5
=========
1 0 6 5 2 

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.