Unit Digit Method for Cryptarithmetic Problems
Cryptarithmetic Unit Digit Method
When to use Unit Digit Method?
- When you don’t know how to solve the problem and just can’t get started.
- You think the problem is really tough to solve.
- In this case you have to start hit and trial with the possible values of unit digit of the multiplication problem.
Time to solve whole problem – 10 mins
Suggestion – You must know how to solve Basic Crypt arithmetic questions firsts
The Cryparithmetic methods discussed below are developed by PrepInsta only and available on 2 PrepInsta owned websites only. Anyone copying the method will be legally sued as these are not open source but are PrepInsta’s Proprietary methods.
Steps to Solve-
- Break the problem in to subcomponents for eg -If it is a 3 x 3 Cryptarithmetic Problem, then you have to convert it into 3 x 1 problem.
- After dividing the problem in three parts, analyse all the parts very closely and try to collect some clue. Now, you have to choose one among 3 which has maximum number of clues.
- You have to start hit and trial with the possible values of the variable which is present at the unit digit. {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
At each step, you have to check, whether the values satisfies Basic Cryptarithmetic Rule
Example
W B A
x B P W
----------
C X R F
F X A X
A A C C
-----------------
A P C A B F
Firstly divide the problem in three parts. As,
(1) W B A (2) W B A (3) W B A
x W x P x B
----------- --------- ---------
C X R F F X A X A A C C
Now, you have to select one from three which has maximum number of clues.i.e. maximum number of variables getting repeated.
In this case take,
(3) W B A
x B
----------
A A C C
Now, you have to start hit and trial with the possible values C i.e.
C={1, 2, 3, 4, 5, 6, 7, 8, 9}
Firstly take C=1 and check further
Possible ways of getting 1 as Unit Digit
(3) W B A
x B
----------
A A C C
then
Case 1 C=1 A=1 B=1 Rejected As A = B = C [violates the Basic Cryptarithmetic Rules.]
Case 2 C=1 A=7 B=3 Needs to be checked further
Case 3 C=1 A=3 B=7 Needs to be checked further
Case 4 C=1 A=9 B=9 Rejected As A = B [violates the basic Cryptarithmetic Rules.]
Now check for Case 2 and Case 3 only.
(3) W B A
x B
----------
A A C C
W 3 7
x 3
----------
7 7 1 1
Rejected as, even after taking the value of W=9, You will never get 77.
W 3 7
x 3
----------
7 7 1 1
Now, you have to check with case 3
C=1, A=3, B=7
W B A
x B
----------
A A C C
W 7 3
x 7
---------
3 3 1 1
Here you can easily predict the value of W=4
Hence W=4, C=1, A=3, B=7 put these values in main problem
4 7 3
x 7 P 4
----------
1 X R F
F X 3 X
3 3 1 1
----------------
3 P 1 3 7 F
Now, you can see
4 7 3
x 4
1 8 9 2 [ 1 X R F]
If you compare side by side you will get X=8, R=9 and F=2
Put these values in main problem, and rewrite again
4 7 3
x 7 P 4
----------
1 8 9 2
2 8 3 8
3 3 1 1
----------------
3 P 1 3 7 2
Now, you can easily predict the value of P=6
Hence,
4 7 3
x 7 6 4
----------
1 8 9 2
2 8 3 8
3 3 1 1
----------------
3 6 1 3 7 2
[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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