eLitmus Cryptarithmetic Problem – 3

March 20, 2020

Cryptarithmetic Questions asked in eLitmus 3

How To Solve Quickly PrepInsta
Ques. For the following Cryptarithmetic find the answers to the below questions? 
         P  O  P
         M  U  T
        ---------
      S  O  I  U
   G  M  T  U  -
U  I  R  O  -  -
-----------------
U  U  S  M  U  U
1. Value of G + U + T ?
(a) 3 (b) 12 (c) 6 (d) 9
2. Value of S + G ?
(a) 3 (b) 9 (c) 7 (d) 8
3. Value of  2U ?
(a) 4 (b) 6 (c) 8 (d) 2

It is highly suggested to go through these before solving

  1. Cryptarithmetic Introduction
  2. How to Solve Cryptarithmetic Problems
Row 1             P  O  P
Row 2             M  U  T
Row 3           ---------
Row 4          S  O  I  U
Row 5       G  M  T  U  -
Row 6    U  I  R  O  -  -
Row 7    -----------------
Row 8    U  U  S  M  U  U

Step 1

  • As I + U = U
  • This is only possible when, I = 0
    • Also, there can not be a carry from previous step as
    • Max values of U can be 9. Thus, no carry over.

Thus I = 0

Row 1             P  O  P
Row 2             M  U  T
Row 3           ---------
Row 4          S  O  0  U
Row 5       G  M  T  U  -
Row 6    U  0  R  O  -  -
Row 7    -----------------
Row 8    U  U  S  M  U  U

Step 2

Read Rule 3 here before next step

  • P x U = _U
  • Can also be written as
  • U x P = _U (is of form X x Y = _X)

According to Rule 3, this is only possible when – 

  1. P = {6} & U = {2, 4, 8}
  2. U = {5} & P = {3, 7, 9}

Let us try for values (P,U) = (6, 2) and start hit and trial method –

Replace all the values and start hit and trial again – 

Row 1             6  O  6
Row 2             M  2  T
Row 3           ---------
Row 4          S  O  0  2
Row 5       G  M  T  2  -
Row 6    2  0  R  O  -  -
Row 7    -----------------
Row 8    2  2  S  M  2  2

Step 3

After replacing with assumed hit and trial values –

  • 6 x T = _2
    • This is only possible when T = 7
    • As 6 x 7 = 42
    • As of now, it doesn’t cause any conflict this, moving further, we create the sub problem as this –
Row 1             6  O  6
Row 2                   2 
Row 3           ---------
Row 4          G  M  7  2

Note - 0(digit) and O(alphabet) are different.
  • 6 x 2 = GM
  • We can easily predict the following
    • G is 1 as 6 x 2 results in 12
    • M is 2 + carry (from previous step)

Also,

  • O x 2 + 1(carry from previous step 6 x 2) = _7
  • Possible values for O can be 3 or 8

If, O = 3

  • 6 X 2 = GM
    • O x 2 from previous step is 3 x 2 = 6
    • This, doesn’t generate any carry from previous step, thus, M = 2
  • This is not possible as U value is already assumed to be 2

Let us try with O = 8

  • O x 2 will generate 1 carry to next step as result will be 17 (1 carry from previous step)
  • Thus, 6 X 2 + 1 carry = GM
    • M = 3 and G = 1

Rewriting the question again

Thus,I = 0 M = 3, G = 1, O = 8, P = 6, U = 2, T = 7 till now

Pending values are S R

Row 1             6  8  6
Row 2             3  2  7
Row 3           ---------
Row 4          S  8  0  2
Row 5       1  3  7  2  -
Row 6    2  0  R  8  -  -
Row 7    -----------------
Row 8    2  2  S  3  2  2

Step 4

R value can be found out as follows with this sub problem

Row 1             6  8  6
Row 2                   3
Row 3           ---------
Row 4          2  0  R  8
  • 6 x 3 from units places gives 18 and 1 carry to next step
  • 8 x 3 from tens places gives us 24 + 1 (Carry from previous step) this 25
    • Value of R = 5

Thus, I = 0, G = 1, U = 2, M = 3, R = 5,   P = 6, , T = 7, O = 8, till now

Finally, finding the value of S is also very easy

S = either 9 or 4.

We do hit and trial and find out its 4.

Next
Related Navigation

eLitmus Online Classes
Join here

2 comments on “eLitmus Cryptarithmetic Problem – 3”