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eLitmus Cryptarithmetic Problem – 2

Cryptarithmetic Questions asked in eLitmus 2

How To Solve Quickly PrepInsta
                H  O  W
             x  C  U  T
             ----------
             D  D  C  P
          D  W  P  W
       D  U  O  A   
       ----------------   
       D  C  E  P  D  P
1.Value of T + O + P ?
(a) 3(b) 8(c) 6(d) 9
2.Value of D ?
(a) 1(b) 2(c) 3(d) 4
3.Value of  2C ?
(a) 13(b) 14(c) 17(d) 11

It is highly suggested to go through these before solving

  1. Cryptarithmetic Introduction
  2. How to Solve Cryptarithmetic Problems
Row1                H  O  W
Row2             x  C  U  T
Row3             ----------
Row4             D  D  C  P
Row5          D  W  P  W
Row6       D  U  O  A   
Row7       ----------------   
Row8       D  C  E  P  D  P

Step 1

  • W x U = _W (Rule 3)
  • Case 1 :
    • When W={2, 4, 8} and U={6}
  • Case 2 :
    • When W={5} and U={3, 7, 9}

Assuming U = 6, the problem will look like – 

Row1                H  O  W
Row2             x  C  6  T
Row3             ----------
Row4             D  D  C  P
Row5          D  W  P  W
Row6       D  6  O  A   
Row7       ----------------   
Row8       D  C  E  P  D  P

Step 2

  • D + 6  + (carry) = C
    • [Carry may be 0, 1, 2 (as 3 digit addition in previous steps )]
  • Thus C = {7, 8, 9}
  • Start hit and trial C={7, 8, 9} and W={2, 4 ,8}

Hit and Trial 1

  • C = 7 and W = 2
  • Then D=9 [ As, C + W = D] If we take take D=9 then
Row1                 H  O  2
Row2              x  7  6  T
Row3              ----------
Row4 (carry)1     9  9  7  P
Row5           9  2  P  2
Row6        9  6  O  A   
Row7        ----------------   
Row8        0  5  E  P  9  P

Step 3

Assuming: C = 7 and W = 2

  • Now this causes conflict as –
    • D = 9 from last step as 7 + 2 = 9
    • D = 0 as 9 + 1 (carry) from previous step 9 + 6 = 5 (1 carry)

To values of D not possible thus thus hit and trial is wrong we must do another hit and trial

  • Let’s take C=7 and W=4
Row1                 H  O  4
Row2              x  7  6  T
Row3              ----------
Row4              1  1  7  P
Row5           1  4  P  4
Row6        1  6  O  A   
Row7        ----------------   
Row8        1  7  E  P  1  P

Step 4

Assuming C=7 and W=4

Then, D=1 as C + W = D

  • Thus, value for A = 8
  • As, 7 x 4 = _A

Rewriting the whole problem

Row1                 H  O  4
Row2              x  7  6  T
Row3              ----------
Row4              1  1  7  P
Row5           1  4  P  4
Row6        1  6  O  8   
Row7        ----------------   
Row8        1  7  E  P  1  P

Step 5

Assuming C=7 and W=4

   H  O  4 
x 7
----------
1 6 O 8
  • 7 x H = 16
  • How, is this possible?
  • This is possible when
    • 7 x 2 = 14 and 2 carry from previous step
  • Thus, H = 2
Row1                 2  O  4
Row2              x  7  6  T
Row3              ----------
Row4              1  1  7  P
Row5           1  4  P  4
Row6        1  6  O  8   
Row7        ----------------   
Row8        1  7  E  P  1  P

Step 6

  • 7 + 4 = 1 (1 carry to next step)
  • Thus, 1 (carry) + 1 + P + 8 = P
  • 10 + P = P means 1 carry to next step
  • 0 + P = P, can only be possible when, P = 0

Rewriting the problem – 

Row1                 2  O  4
Row2              x  7  6  T
Row3              ----------
Row4              1  1  7  0
Row5           1  4  0  4
Row6        1  6  O  8   
Row7        ----------------   
Row8        1  7  E  0  1  0

Step 7

  • 4 x T = _0
  • This is possible when, T = 0, T = 5
    • T = 0 not possible as P = 0 in previous step
  • Thus, T = 0

Also clearly,

  • O = 3 as,
  2 O 4 
x 6
----------
1 4 0 4
  • 6 x 4 = 4 (2 carry)
  • 6 x O + 2 (carry) = _0
    • This is only possible when O = 3
  • 6 x 3 + 2 (carry) = 20
H=2, O=3, W=4, C=7, U=6, T=5, D=1, S=0, E=9

                 2  3  4
              x  7  6  5
            ------------
              1  1  7  0
           1  4  0  4
        1  6  3  8   
        ----------------   
        1  7  9  0  1  0

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