Cryptarithmetic Problem-4

Multiplication


E Y E * M A T
--------------
         S Y I A
    G M T A +
 A I R Y +  +
--------------
 A A S M A A

Follow the steps to solve the above puzzle:
Step 1: Look for '0' or '1' in the Multiplier(M A T), sadly we do not find one.

Step 2: Now, look at the product term of E Y E * - A - = G M T A. If you have gone through the previous post, you will guess that 'E' is either 1 or 6. Since it is not 1, it should be 6. And 'A' should be an even number and hence it should be 2,4,8. Considering 'A' as 2 and rewriting the multiplication we get,

6 Y 6 * M 2 T
--------------
        S Y I 2
    G M T 2 +
  2 I R Y  + +
--------------
 2 2 S M 2 2

Step 3: Now looking at the sum in the 2nd column form the right i.e.  'I + 2 = 2', we can conclude that 'I' = 0, since there is no carry. Rewriting it would yield us,
6 Y 6 * M 2 T
--------------
        S Y 0 2
    G M T 2 +
  2 0 R Y + +
--------------
2 2 S M 2 2

Step 4: Further, looking at the sum in the 2nd column form the left i.e. 'G + 0 = 2', we can conclude that 'G' is either 1 or 0, since we already have 'I' = 0, 'G' has to be 1. Rewriting it we would have,

6 Y 6 * M 2 T
--------------
        S Y 0 2
    1 M T 2 +
 2 0 R Y + +
--------------
 2 2 S M 2 2

Step 5: Now looking at the product 6 Y 6 * M = 2 0 R Y, we can conclude that 'M' = 3. since (6 * M + carry) = 20, and the only value that seems to satisfy that equation is 3. Rewriting this would yield us,
6 Y 6 * 3 2 T
--------------
        S Y 0 2
      1 3 T 2 +
  2 0 R Y + +
--------------
 2 2 S 3 2 2
Step 6: If we look at the product term 6 Y 6 * 3 = 2 0 R Y, we can easily figure out 'Y' to be 8, so rewriting the puzzle would give us,


6 8 6 * 3 2 T
--------------
        S 8 0 2
     1 3 T 2 +
  2 0 R 8 + +
--------------
 2 2 S 3 2 2


Step 7: Now, the sum term 8 + T + 8 = 3 helps us to find out the value of 'T', yeah it is 7. How? Well there is no carry and the sum has to be 23, since no number is greater than 9 (You could aslo find out the value of 'T' from the product 686*2 as well.) We will get

6 8 6 * 3 2 7
--------------
        S 8 0 2
     1 3 7 2 +
 2 0 R 8 + +
--------------
 2 2 S 3 2 2


Step 8: This step is just a formality, since we got to know the values of the Multiplicand and the multiplier, it is a piece of cake to find out the rest of the values. And they happen to be,
6 8 6 * 3 2 7
--------------
        4 8 0 2
     1 3 7 2 +
  2 0 5 8 + +
--------------
 2 2 4 3 2 2


This is it. I know these are too many steps, but yeah practice makes man perfect, so do Practice and become a pro in solving such kind of puzzles.

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Cryptarithmetic Problem-3

Multiplication

            G  A   S 
            F   B   I
       ---------------
        F  T   B  I
    S  S  T   B -
S  A  S   F -  -
-------------------
S  R   I  S  T  I

Follow the steps to solve the above puzzle:

Step 1: Look for '0' or '1' in the Multiplier(F B I), sadly we do not find one.

Step 2: Now look at the last terms in each of the individual products of this puzzle, F T B I, S S T B and S A SF, it is clear that 'S=1' (Because, any other odd number will not give the same value when multiplied to it.) Now our cryptarithmetic problem will look something like this(I've bolded the I's just to differentiate from the 1's),

G A 1 * F B I
---------------
         F T B I
     1 1 T B +
 1 A 1 F +  +
---------------
 1 R I 1 T I

Step 3: This step is the most important step in this problem, if you recognize this, you have done a great job. In the product term, G A 1 * F - -, F*1=F and there will be no carry which is a well known fact. Now, ifyou look at the product (- A -) * F = 1 A 1 F, you will guess that, F * A leads to a product of 21(because there is 1 in the second term of the resulting product of the considered section of the puzzle, and of course no carry from the previous product term.). Now, it is not a Rocket Science to guess that F, A should be 3, 7 or 7, 3. Let us assume that F=3 and A=7, and proceed,

G 7 1 * 3 B I
---------------
        3 T B I
     1 1 T B +
  1 7 1 3 +  +
---------------
 1 R I 1 T I


Step 4: Now, looking at the term (G 7 1) * 3 = 1 7 1 3, we can easily make out that G=5. Hence our puzzle will look like,

5 7 1 * 3 B I
---------------
        3 T B I
     1 1 T B +
  1 7 1 3 +  +
---------------
 1 R I 1 T I
Step 5: Notice that R=8, which is pretty straightforward, since there is no carry coming through. I, will either be 5 or 6, depending on whether a carry is coming through or not. Since, A is already 5, I is 6. Hence the puzzle will look like,
5 7 1 * 3 B 6
---------------
        3 T B 6
     1 1 T B +
  1 7 1 3 +  +
---------------
 1 8 6 1 T 6

Step 6: Looking at the product term (5 7 1) * - - 6 = 3 T B 6, we find that 5 7 1 * 6 = 3 4 2 6, we can easily say that, T=4 and B=2. And hence, we have completed solving this cryptarithmetic problem.
5 7 1 * 3 2 6
---------------
        3 4 2 6
     1 1 4 2 +
  1 7 1 3 +  +
---------------
 1 8 6 1 4 6

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Cryptarithmetic Problem-2

Multiplication 

			A	B	C
			D	E	C
			F	G	H
		I	A	C
A	C	A	E
A	F	A	G	C	H

Solution:

First if you see ( D E C ) * C from this we can confirm that two possibilities:

1)      E = ( 3, 7, 9 ) and C = 5

2)      E = 6 and C = ( 2, 4, 8 )

So now you are in little bit confusion which possibility I should proceed.

If you are not choosing the correct possibility time will be killed off course time is very very precious in eLitmus exam.

So, better step forward to estimate the correct possibility–)

If you see last multiple ( D E C ) * C = E if you assume C = 5 and E = (3, 7, 9 ) the only possible values ofD = (2, 4, 8) if you keep any value for D then the value E = 0 that is contradiction.

So it is very clear that first possibilities is wrong then go with the second possibility means

E = 6 and C = (2, 4, 8) Let’s take C = 2 Then the table changes as:

			A	B	2
			D	6	2
			3	0	4
		I	A	2
A	2	A	6
A	3	A	0	2	4

Clearly from the table H = 4 and G = 0 (because G + 2 = 2 means G = 0), and F = 3 (C + carry= F).

After filling this values let’s take (A B 2) * 2 = 3 0 4. It happens only when B = 5 and A = 1

Replace the values and redraw the table.

			1	5	2
			D	6	2
			3	0	4
		I	1	2
1	2	1	6
1	3	1	0	2	4

From the table (1 5 2) * D = 1 2 1 6 It will happen only when D = 8 and from the line (I + 1 + carry = 1) so from that we can say I = 9.

Replace all the values,

			1	5	2
			8	6	2
			3	0	4
		9	1	2
1	2	1	6
1	3	1	0	2	4

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Cryptarithmetic Problem-1

Multiplication

			W	H	Y
			N	U	T
		O	O	N	P
	O	Y	P	Y
O	U	H	A
O	N	E	P	O	P

When you see second multiple (W H Y ) * U = (O Y P Y)

So, The number for U = 6 and Y is (2 , 4) Lets take the value of Y is 4 and replace it in figure.

			W	H	4
			N	6	T
		O	O	N	P
	O	4	P	4
O	6	H	A
O	N	E	P	O	P

Clearly if you see the yellow portion O + 6 = N and there is no carry to the next digit so we can confirm that ‘N’ is single digit number. If ‘N’ is single digit number ‘O’ should be less than (O < 3)

Let’s take O = 1 and redraw the table…

			W	H	4
			N	6	T
		1	1	N	P
	1	4	P	4
1	6	H	A
1	N	E	P	1	P

If you observe 1 + 6 = N & N + 4 = 1 so, we can clearly the value of N =7

So replace N = 7 and redraw the table.

			W	H	4
			7	6	T
		1	1	7	P
	1	4	P	4
1	6	H	A
1	7	E	P	1	P

Take the third multiple (W H 4) * 7 = 1 6 H A    from this we get A = 8, and if you replace the H with 3 and w = 2 then the equation satisfies.

Like (2 3 4) * 7 = 1 6 3 8 so redraw the table with particular values.

			2	3	4
			7	6	T
		1	1	7	P
	1	4	P	4
1	6	3	8
1	7	E	P	1	P

The remaining numbers are (0,5,9) It is simple to analyse we can’t get 5 in place of p because when you multiply with 4 we cannot get the last digit as 5.

So T = 5, P = 0, E = 9

After redraw the table.

			2	3	4
			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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