Posts

Showing posts with the label letters

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

A Verbal Arithmetic Puzzle

In this equation, each of the letters represents uniquely a different digit in base 10:

YE × ME = TTT.

What is E + M + T + Y?


A Verbal Arithmetic Puzzle




Here is solution!

A Verbal Arithmetic Puzzle - Solution



What was the puzzle?

Let's recall the original equation.

YE × ME = TTT.

Obviously, answer TTT is product of T x 111 = T x 3 x 37. 

So YE/ME = 37 and T x 3. Either way, E = 7.

And other number T x 3 is a 2-digit number with 7 at it's unit's place.

So that has to be 9 x 3 = 27.

Therefore,

YE (or ME ) = 37 and ME (or YE) = 27

Hence,
 
E + M + T + Y = 7 + 3 + 9 + 2 = 27.



A Verbal Arithmetic Puzzle - Solution

Numbers For The Letters?

A - B = B

B * C = A

D : B = E

C * C = E

C + E = A


For which numbers stand A, B, C, D and E

Find Correct Numbers For The Letters? - Maths Puzzles

Correct Numbers For Letters


What was the question? 

Let's first rewrite all equations and number them

1. A - B = B

2. B * C = A

3. D : B = E

4. C * C = E

5. C + E = A


For which numbers stand A, B, C, D and E? 

From 1, A = 2B. Putting this in 2 gives,

C = 2.

Putting C = 2 in 4 gives, E = 4.

So from 5, A = C + E = 2 + 4 = 6.

Equation 2 gives, B = A/C = 6/2 = 3.

And equation 3 gives D = B * E = 3 * 4 = 12.

Finding Correct Numbers For Letters - Maths Puzzle


Conclusion : A = 6, B = 3, C = 2, D = 12 and E = 4

Another CryptArithmatic Problem

Replace letters with numbers assuming numbers can't be repeated. 
 
     SEND
  + MORE
  ----------
 = MONEY
 
Replace Numbers with letters - Maths Puzzle

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.
 
Maths Puzzle -  Correct numbers for letters
 
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
----------


Replace letters with numbers!
 
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
----------


BASE + BALL = GAME Solution

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