Posts

Showing posts with the label cryptarithms

One More Alphamatic Problem?

In the following  puzzles, replace the same characters by the same numerals
so that the mathematical operations are correct.
               
Note - Each letter represents a unique digit and vice-versa.
 
ABCB - DEFC = GAFB
     :          +      -
  DH  x     AB =    IEI
---------------------------
 GGE + DEBB = DHDG
 
One More Alphanumeric Problem?
 
 
 
 
Here is the SOLUTION 
 
 

One More Alphamatic Solution!


Look at the problem first!

Rewriting the problem once again,

ABCB - DEFC = GAFB
   :         +       -
  DH x   AB    =    IEI
-------------------------
 GGE + DEBB = DHDG
 
We have 6 equations from above -
 
(1) A B C B - D E F C = G A F B  

(2) G G E + D E B B = D H D G  

(3) G A F B - I E I = D H D G 

(4) D E F C +  A B = D E B B 

(5) A B C B : D H  = G G E  

(6) D H x A B = I E I  

Steps :

1. From (1), we have B - C = B. That's possible only when C = 0.

2. If C = 0 then in (1), for tens' place subtraction i.e. C - F = F the carry need to 
    be taken from B. And that subtraction looks like 10 - F = F. Obviously, F = 5.

3. From (3), we see D in result seems to be carry and carry never exceeds 1 
    even if those numbers are 999 + 9999. So, D = 1. 
 
4. From (1), since C = 0, at hundreds' place (B - 1) - E = A and from (4),
    we have F + A = B (since first 2 digit of first numberremain same in result
    indicating no carry forwarded in addition of FC + AB = BB.
 
    So placing F = B - A in (B - 1) - E = A gives, F = E + 1. Since, F = 5, then E = 4. 
 
5. In (3), G at the thousands' place converted to D without actually subtraction 
    of digit from IEI. Since, G and D are different numbers some carry must have been
    taken from G.



    As D = 1 then G = 2.

6. From (1), A - D = G and D = 1 and G = 2 then A = 3 since if carry had been taken 
    from A then A = 4 which is impossible as we already have E = 4. 
 
7. From (2), E + B = G i.e. 4 + B = 2 only possible with B = 8.

8. With that, in (2), carry forwarded to  G + B = D making it 
    1 + G + B = 1 + 2 + 8 = 11 = 1D  i.e. carry 1 forwarded to G + E = H making it 
    1 + G + E = H = 1 + 2 + 4 = 7.
    Therefore, H = 7 and no carry forwarded as digit D in second number remains
    unchanged in result. 

9. Now (6) looks like - 17 x 38 = 646 = IEI = I4I. Hence, I = 6. 

 
To sum up,
 
A = 3, B = 8, C = 0, D = 1, E = 2, F = 5, G = 3, H = 7 and I = 6
 
One More Alphanumeric Solution!

 
Eventually, all above 6 equations after replacing digits in place of letters look - 

1. 3808 - 1450 = 2358  ✅
 
2. 224 + 1488 = 1712   ✅
 
3. 2358 - 646 = 1712   ✅
 
4. 1450 + 38 = 1488    ✅
 
5. 3808 : 17 = 224      ✅
 
6. 17 x 38 = 646         ✅  
 
Rewriting in the given format,

3808 - 1450 = 2358
      :         +       -
    17 x     38 =  646
-----------------------------
  224 + 1488 = 1712

Follow me on Blogarama