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eLitmus Cryptarithmetic Problem – 5

Cryptarithmetic Questions asked in eLitmus 5

How To Solve Quickly PrepInsta
Ques. For the following Cryptarithmetic find the answers to the below questions?
                 T  E  A
              x  H  A  D
              ----------
              L  D  T  R
           H  R  S  A
        E  W  D  A      
       -----------------
        L  E  S  S  E  R
1. Value of  S ?
(a) 6 (b) 7 (c) 8 (d) 9
2. Which of the following follows the Pythagoras theorem ?
(a) H, A, D (b) H, A, B (c) T, E, A (d) T, E, D
3. Value of H + R + S + A ?
(a) 14 (b) 15 (c) 16 (d) 17

It is highly suggested to go through these before solving

  1. Cryptarithmetic Introduction
  2. How to Solve Cryptarithmetic Problems
Row1              T  E  A
Row2           x  H  A  D
Row3 ---------- Row4 L D T R Row5 H R S A Row6 E W D A Row7 ----------------- Row8 L E S S E R

Step 1

  • A x A = _A
  • This means by Rule2 that A = {5,6}
Row1              T  E  A
Row2           x  H  A  D
Row3 ---------- Row4 L D T R Row5 H R S A Row6 E W D A Row7 ----------------- Row8 L E S S E R

Step 2

  • Case I – when A={5} then H={3, 7, 9}
  • Case II – when A={2, 4, 8} then H={6}

Taking A = 5 from Step 1 and just replacing a values, the problem now looks like –

Row1              T  E  5 
Row2 x H  5 D
Row3 ---------- Row4 L D T R Row5 H R S 5 Row6 E W D 5 Row7 ----------------- Row8 L E S S E R

Step 3

  • Looking at the 5 x D = _R
    • Possible values for R can be either 0 or 5
    • As if you multiply 5 to number you will either get 0 or 5 in units digit
    • R value will be 0 as A is already 5
    • R = 0 

Replacing the values –

Row1              T  E  5 
Row2 x H  5 D
Row3 ---------- Row4 L D T 0 Row5 H 0 S 5 Row6 E W D 5 Row7 ----------------- Row8 L E S S E 0

Step 4

  • Let us create a subproblem –
Row1              T  E  5 
Row2           x        5
Row3           ----------
Row4           H  0  S  5

Now, in this – 

  • T x 5 = _0
    • This means there is no carry from previous step
  • This means E x 5 < 10
    • Only possible way it could happen is E = 1
  • Thus the value for S will be –
    • S = 1 x 5 + 2(carry from previous step)
    • S = 7

Values till now are –

R = 0, E = 1, A = 5, S = 7 replacing and writing it again.

Row1              T  1  5 
Row2 x H  5 D
Row3 ---------- Row4 L D T 0 Row5 H 0 7 5 Row6 1 W D 5 Row7 ----------------- Row8 L 1 7 7 1 0

Step 5

  • Now, T = 6
    • As T + 5 = _1 (no carry from previous step possible)
  • L = 2
    • L can not 1, E is already 1 
    • Thus, there must be 1 carry coming from previous step
    • L = 1 + 1(carry) = 2
  • Rewriting the problem
Row1              6  1  5 
Row2 x H  5 D
Row3 ---------- Row4 2 D 6 0 Row5 H 0 7 5 Row6 1 W D 5 Row7 ----------------- Row8 2 1 7 7 1 0

Step 5

  • D = 4
    • As D + 7 + 5 + 1(carry from previous step) = _7
    • Thus D = 4
Row1              6  1  5 
Row2 x H  5 4
Row3 ---------- Row4 2 4 6 0 Row5 H 0 7 5 Row6 1 W 4 5 Row7 ----------------- Row8 2 1 7 7 1 0

Step 6

  • H = 3
    • 6 x 5 = H0
  • H + W + 0 (no carry from previous step) = 1
  • Thus, W = 8
              6  1  5
           x  3  5  4
           ----------
           2  4  6  0
        3  0  7  5
     1  8  4  5                  
     ----------------
     2  1  7  7  1  0

[Relax – it’s going to take some time to understand the whole concept.
If you are facing any difficulty.

Please go through the Cryptarithmetic Tutorial.]

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