A G E                 *O A T          ---------------                 S O A R                 H O G         ...

*O A T
---------------
S O A R
H O G
G O T O
----------------
G E C O I R

Step 1: Figure out the characters which can not take value 0
Now you will ask the question "how do I figure out which character can be or can not be 0 ?" Follow these rules and you will be able to find them very easily:
• No leading digit of a number can be 0. What does that mean ? That simply means the left most digit in a number can't be 0(ever saw a number like 023 or 007 ?). Following this rule in our case, none of these can be 0: A, O, S, H, G

• None of the characters in multiplier(O, A and T in our case) will be 0. You can argue "Hey are you crazy ? You are wrong !". OK wait a moment.

Can you tell me what is the value of A in the question below ?
A G E
*O A T
---------------
S O A R
A A A
G O T O
----------------
G E C O I R

Yes, you are thinking right. A will be 0. I think you know the reason. After all you are an engineer.

Do you really think eLitmus is gonna ask you such an easy question ? Don't even think of that.

So simple rule of thumb is that just proceed assuming no digit in multiplier will be 0 and you can land on the solution faster

So where did we reach ? None of these will be 0: A, O, S, H, G, T
• No trailing digit in multiplicand will be 0

OK let me ask you a question again

What will be the value of E in the following question ?
A G E
*O A T
---------------
S O A E
H O E
G O T E
----------------
G E C O I R

No wonder again. Yes, E will be 0.

Do you again think eLitmus will ask you so simple question ? No, I don't think so.

So like in the previous step, just proceed assuming that the first(from the first bullet point) and the last digit in multiplicand will not be 0

Now we have the following result for non zeroes: None of these will be 0: A, O, S, H, G, T, E
Step 2: Figure out the characters which can not take value 1
• No digit in multiplier will be 1

OK, answer my this question and you will get to know why so ?
A G E
*O A T
---------------
A G E
H O G
G O T O
----------------
G E C O I R

OR you can even see this one:
A G E
*O A T
---------------
S O A R
A G E
G O T O
----------------
G E C O I R

Again eLitmus won't ask you so simple question.

So stick to the rule that no digit in multiplier will be 1. Following this rule in our case none of these will be 1: O, A, T
• The trailing digit in multiplicand will now be 1. An example to prove the above point:
A G E
*O A T
---------------
S O A T
H O A
G O T O
----------------
G E C O I T

So following this rule we find that E will not be 1

By far digits which can't be 1 are: O, A, T, E
Following the above two steps we find:

Non 0 digits: A, O, S, H, G, T, E and non 1 digits: O, A, T, E

So digits which may be 0 are: C, I and digits which may me 1 are: S, H, G,
C, I
Let's proceed further.

Step 3: Identify 6 or 5

Let's see some mathematical property:
• 6*0 = 0
• 6*2 = 12
• 6*4 = 24
• 6*6 = 36
• 6*8 = 48
• 5*1 = 5
• 5*3 = 15
• 5*5 = 25
• 5*7 = 35
• 5*9 = 45
Notice when 6 is multiplied with an even digit, the unit place digit of the result has the same even digit.
And similarly when 5 is multiplied with an odd digit, the unit place digit of the result is 5

So what do we get from these rules ?

Let's apply these rules to solve our problem.

A G E
*O A T
---------------
S O A R
H O G
G O T O
----------------
G E C O I R

Focus on the digits highlighted in blue. Let's write it in this format for better understanding. C is the carry, so just ignore it.

E
* O
----
C O

From the above rule we discussed we can say these:
O is 5 and E is odd
OR

O is even and E is 6

So we have two ways starting from here. Follow one path(any one) and see if you are able to solve the problem. If not, take another path and proceed further.

Case 1: O is 5 and E is odd

A G E
*5 A T
---------------
S 5 A R
H 5 G
G 5 T 5
----------------
G E C 5 I R

Now ?? What next ?? Let me admit one thing here. I can't teach you from here how to solve this. No one can tell you what will be the next step from here. It is just you who can solve this by hit and trial and applying your common sense.
Anyhow I will proceed with my way but remember one thing you have to solve it by yourself from here keeping in mind the results of step 1 and 2 and applying own logic.
Follow my steps and you can learn many tricks from my experience.

Remember we found just two character which may be 0: C and I. This question is a very good example where we can use this.
Now if I was 0, there would have been a carry to the next summation but notice carefully the third column of summation from right (X is just a carry, so ignore it)

5
5
+ 5
---
X
There is no carry from the last step for this summation, that means the summation from 2nd column from right did not generate any carry. That implies that I cannot be 0.
So we can conclude C is 0.

A G E
*5 A T
---------------
S 5 A R
H 5 G
G 5 T 5
----------------
G E 0 5 I R

Since E is odd

A G 7
*5 A T
---------------
S 5 A R
H 5 G
G 5 T 5
----------------
G 7 0 5 I R

Let's analyze something interesting. Notice multiplication:

A G 7
* 5
---------
G 5 T 5

When 5 is multiplied with 7, it gives 3 as a carry.
When 5 is multiplied with G, it gives either 0 or 5 depending on whether G is even or odd. So T will be either 3 or 8 and there will be some carry too.
What about that carry ? That carry can be either 0(when G = 1), 1(when G = 2 or 3), 2(when G = 4 or 5), 3(when G = 6 or 7), or maximum 4(when G = 8 or 9).
But notice that no carry is passed to the next multiplication because 5 * A  has a result having 5 at unit place. If there was any carry from 1 to 4, it would not have been 5.
So we can conclude G =  1 and  T = 8. Which also implies A = 3
So putting all these values we get:
3 1 7
*5 3 8
---------------
S 5 3 R
H 5 1
1 5 8 5
----------------
1 7 0 5 I R

Now just multiply and you have the result in your hand:
3 1 7
*5 3 8
---------------
2 5 3 6
9 5 1
1 5 8 5
----------------
1 7 0 5 4 6

We had two cases possible. We followed the first and landed on the solution with the help of some data analysis. If we were stuck somewhere with wrong result then we would have selected the other case to proceed with. As a homework please proceed with the other case and see how you get stuck and how will you back track. BLOGGER: 1
Name

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MRA Job Seekers | Bangalore Job Seekers Hub: cryptarithmetic problems - elitmus cryptarithmetic problems PART 2
cryptarithmetic problems - elitmus cryptarithmetic problems PART 2