Cryptarithmetic Questions and Answers

Clearly, W = 1, as F + N = WI (2 digit number)

Now,

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  • N + N = E => 2N = E
  
  
  
  

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  • E + E = S => (2N) + (2N) = S
  
  
  
  

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  • 4N = S
  
  
  
  

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  • N = 1, E = 2, S = 4
  
  
  
  

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  • or N = 2, E = 4, S = 8
  
  
  
  

  F I 2 4

  2 I 2 4

---------

1 I V 4 8

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  • F + 2 + carry(0/1) >=10 (as 1 carry to next step)
  
  
  
  

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  • To do this possible values of F are = {7, 8, 9}
  
  
  
  

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  • 8 is already taken by S, So, F = {7,9}
  
  
  
  

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  • F + 2 + Carry(0/1) = I
  
  
  
  

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  • 9 + Carry(0/1) = I
    
    
    
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      • 9 + 0 (carry) = I (Case 1)
      
      
      
      
    
    
    
    
    
    
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      • This is not possible as no carry to next step
      
      
      
      
    
    
    
    
  
  
  
  

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  • 9 + 1 (carry) = I (case 2)
  
  
  
  

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  • I = 0 not possible
  
  
  
  

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  • As step I + I = V should generate carry to next step i.e. Case 2
  
  
  
  

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  • What if the below equations were never valid as they were generating carries
    
    
    
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      • N + N = E => 2N = E
      
      
      
      
    
    
    
    
    
    
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      • E + E = S => (2N) + (2N) = 6
      
      
      
      
    
    
    
    
    
    
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      • 4N = S
      
      
      
      
    
    
    
    
  
  
  
  

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  • What if E + E at units digit was generating a carry to next step
  
  
  
  

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  • Possible values to do this for E are = {5, 6, 7, 8, 9}
  
  
  
  

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  • E + E = S
  
  
  
  

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  • => S = 0
  
  
  
  

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  • Also, N + N + 1(carry) = E
  
  
  
  

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  • N + N + 1 = 5
    
    
    
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      • Possible values of N to do this are N = {7, 2}
      
      
      
      
    
    
    
    
  
  
  
  

  F I 7 5

  7 I 7 5

---------

1 I V 5 0

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  • F + 7 = 1I
  
  
  
  

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  • => F + 7 + carry(0/1) >= 10
  
  
  
  

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  • => F + 7 + 1 >= 10
  
  
  
  

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  • Equation is F + 7 + 1 = I
  
  
  
  

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  • Possible values for F are ={2, 3, 4, 6, 8, 9}
  
  
  
  

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  • F = 2 not possible as it will result I = 0, S is already 0
  
  
  
  

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  • F = 3 not possible as it will result I = 1, W already 1
  
  
  
  

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  • F = 4 is possible as I will be 2
    
    
    
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      • But, step I + I + 1(Carry) = V will not generate carry as
      
      
      
      
    
    
    
    
    
    
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      • 2 + 2 = 4 and carry is needed for (Note 1)
      
      
      
      
    
    
    
    
  
  
  
  

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  • F = 6, is possible and I = 4
    
    
    
    • 
      
      
      
      • But, again I + I + 1(Carry) = V will not generate carry
      
      
      
      
    
    
    
    
  
  
  
  

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  • F = 8, is possible and I = 6
    
    
    
    • 
      
      
      
      • I + I + 1(Carry) = V will generate carry
      
      
      
      
    
    
    
    
    
    
    • 
      
      
      
      • 6 + 6 + 1 = V = > V = 3
      
      
      
      
    
    
    
    
  
  
  
  

  8 6 7 5

  7 6 7 5

---------

1 6 3 5 0

GUN+SUN= HUNT here H=1 is a carry over, U=6 as U+U=N where N = 2, also N+N=T where T= 4 this gives us 862 + 762 = 1624 i.e. 1+6+2+4= 13.

Substituting values we get and using hit and trial we get, S = 9, E = 5, N = 6, D = 7, M = 1, O = 0, R = 8, Y = 2

SEND + MORE = MONEY

9567 + 1085 = 10652

O + N + E = 0 + 6 + 5 = 11

WATC + MATC = WINNE by substituting the values with get 1278 + 9278 = 10556 where W=1 is a carryover.

Hence the value of WINNE is 10556 is the appropriate answer.

WORLD+TRADE=CENTER

  W O R L D

  T R A D E

-----------

C E N T E R

so,start from last c the value is 1 always.

Next W+T=E right.

By adding numbers we should get 1 as a carry (which is C value).

so,take W as 5 and T as 7 and add now which is 12.

  5 3 6 8 4

  7 6 0 4 2

-----------

1 2 9 7 2 6

From the given data, the value of A will be 1 because it is the only carry-over possible from the sum of 2 single-digit number. T maximum it can take only 9 and there should a carryover for T to give sum as 2 digit number. So T =9, P = 0, A = 1. T + T = 18, the value of E is 8 and 1 will be a carry over to the next column. That is 1 + A + A= L = 3. And finally H = 2. Hence, 819 + 9219 = 10038. A+P+P+L=E = 1+0+0+3+8 = 12.

 

M A C + M A A R ----------- J O C K O

Step 1

Here obviously J = 1 as max carry allowed it

(nothing) + M = O, i.e. 0 + M = O this should have had been M only

It means that there is 1 carry coming from previous step

Also this M + 1 (carry) = O, is generating 1 carry to next step

This means M + 1 > 10, this is only possible when M = 9

=> O = M + 1 = 9 + 0 = 0 (1 carry)

9 A C + 9 A A R ----------- 1 0 C K 0

Step 2

C + R = 0

Since, this is at units place there can be no carry from previous step

Possible values of (C, R) can be (2, 8), (3,7), (4, 6), (6,4), (7,3), (2,8)

Let's assume, first case C = 2 and R = 8 in that case C + R, will generate 1 carry to next step A + A = K. The new problem looks like this -

9 A 2 + 9 A A 8 ----------- 1 0 2 K 0

Step 3

Thus, A + A + 1 (Carry) = K

Also, 9 + A = 2, this is possible in two ways

If, A = 3 (no carry from previous step) 9 + 3 = 2 (1 carry to next step)

If, A = 2 (1 carry from the previous step) but C value is already assumed to be 2, so this is not possible thus, A has to be 3

Thus K = 3 + 3 + 1 = 7

All values are fetched and there is no conflict.

9 3 2 + 9 3 3 8 ----------- 1 0 2 7 0

Step 4

Thus, 3 A + 2 M + 2 C = 3(3) + 2(9) + 2(2) = 9 + 18 + 4 = 31

    H O W 

+ M U C H 

-------------

P O W E R

    H 0 W 

+ 9 U C H 

-------------

1 0 W E R

     1    

 1  7 0 5

+ 9 8 C 7 

-------------

1 0 5 E 2

     1    

 1  7 0 5

+ 9 8 3 7 

-------------

1 0 5 4 2

     N O

+  G U N

     N O

----------

 H U N T

Y O U R
+ Y O U
-----------
H E A R T

Step 1

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  • Obv H = 1
  
  
  
  

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  • Y + nothing = E (not Y)
  
  
  
  

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  • Thus there is 1 carry from previous step
  
  
  
  

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  • Y + (1 carry) is generating 1 carry resulting H
  
  
  
  

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  • Thus, Y + 1 > 10 => Y = 9
  
  
  
  

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  • So, E = 0
  
  
  
  

   9 4 U R

+    9 4 U

-----------

 1 0 A R T

Step 2

1. 
   
   
   
   1. A can be A = 4 + 9 = 3
   
   
   
   

1. 
   
   
   
   1. or A = 4 + 9 + 1 (carry) = 4
   
   
   
   

1. 
   
   
   
   1. O is already 4 thus, A = 3
   
   
   
   

Step 3

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  • Thus, U + 4 < 10
  
  
  
  

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  • U < 6
  
  
  
  

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  • 4, 5, 1, 0 are already taken by O, A, H and E respectively
  
  
  
  

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  • So, U is either 5 or 2
  
  
  
  

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  • U + 4 = R
  
  
  
  

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  • 5 + 4 = R = 9
  
  
  
  

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  • 9 is already taken by Y
  
  
  
  

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  • U + 4 + 1 = R
  
  
  
  

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  • 5 + 4 + 1 = R = 0
  
  
  
  

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  • 0 is already taken by E
  
  
  
  

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  • Thus, U = 2
  
  
  
  

Step 4

   9 4 2 6

+    9 4 2

-----------

 1 0 3 6 8

Final Step

by substituting the values using hit and trial we get , B=7, A=4, S = 8, E = 3, L = 5, G = 1, M = 9

BASE +  BALL = GAMES

7483 + 7455 = 14938

A + M + E = 4 + 9 + 3 = 16

 

 

by substituting the values with 9265 + 1487 = 10752.