# 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

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

--------------

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

--------------

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

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

--------------

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

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

--------------

6 8 6 * 3 2 7

--------------

**S 8 0 2**

1 3 7 2 +

2 0 R 8 + +

--------------

2 2 S 3 2 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

--------------

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.**

**If anyone faced any problem for solving Crypt problems comment below**

**Follow us on Facebook in Below FOLLOW US (Hit LIKE) link for latest Update**

## For Newly updated eLitmus practice sets exclusively for 2016/2017 Batch Download from below

**Category**:
Books To Follow For eLitmus Test,
Crypt Arithmatic,
Cryptarithmetic Problem,
eLitmus Syllabus And Question Paper Pattern,
Multiplication,
pH test,
Rules for solving Cryptarithmetic Problems

I love itt......

ReplyDeletehow can u say that E is 1 or 6

ReplyDeleteE*A = A

DeleteDen either E=1 or 6(4*1=4,5*1=5,6*1=6,7*1=7 or 6*2=12,6*4=24,6*6=36,6*8=48 last digit same as multiplmultipleier)

If E = 1 ,den E*T should be T,but its not,so E=6.

[Read the basic rule post]

A can also be 5 na...odd*5=5...how can you directly conclude it as 6 without crosschecking with 5?

ReplyDelete5 will never come into consideration.

Delete5xodd= 5

5xeven=0

E*T = A (lets T is even,A never be 0.because MAT (M0T)

Now assume T is odd den E*T should be E (As 5* odd=5 and we assume E is 5,T is odd).but in our case its not E.

If still not understand,u can comment here For more details explanation.

ReplyDeletehow can you say E*A ?? Why not it be A*E as per your basic rule post ??

ReplyDeleteCan't get ur question properly.

ReplyDeleteE is the number that gets multiplied and its multiplying by A times. Thats what E*A not A*E. (I think this is ur question)

How u r numbering the letter

ReplyDeleteHow u r numbering the letter

ReplyDeleteE*A = A

DeleteDen either E=1 or 6(4*1=4,5*1=5,6*1=6,7*1=7 or 6*2=12,6*4=24,6*6=36,6*8=48 last digit same as multiplmultipleier)

If E = 1 ,den E*T should be T,but its not,so E=6.

[Read the basic rule post]

A can not be 5 because if u consider the third column i.e E * T = A, (E * T = 5) to get the value of A as 5, either of E and T must be 5 which is not possible. (As 5 is already assigned to A)...hope it will help

ReplyDelete'A' should be an even number and hence it should be 2,4,8. Considering 'A' as 2

ReplyDeletewhy we dont started with 'A'=4,8 initially..?