eLitmus Cryptarithmetic Problem – 2

Cryptarithmetic Questions asked in eLitmus 2 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

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

2 comments on “eLitmus Cryptarithmetic Problem – 2”

• Vijayant

These questions were really helpful for eLitmus exam got placed via eLitmus in 2 companies with an offer of 6 LPA. 0
• Atulya PrepInsta

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