How to Solve Cryptarithmetic Problems Basics
Cryptarithmetic Basics Rules
There are the basics of cryptarithmetic problems that one must know. It will take much time to understand these as these problems are hard so be patient and give time to the sections and the rules of Cryptarithmetic Basics, Cryptarithmetic Questions.
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 .
Cryptarithmetic Questions How to solve –
Instructions for Cryptarithmetic Problems
- Alphabets can take up only one distinct value.
- Alphabets can only take up values between 0-9.
- Decoded numbers can’t begin with 0, for example, 0813.
- Problems are uni-solutional.
- 19 is the max value with a carryover for two one-digit number in the same column addition
- Carryover can only be 1 for addition problems
- Be patient there is no specific rule, you will only learn how to solve when you see examples.
Methods to solve Cryptarithmetic Problems
- Cryptarithmetic Introduction
- Method 1 to solve (Basic Method)
- Alternate method 2 – Unit Digit(Only Try this once you understand this basic method on this page)
Cryptarithmetic Problems to practise
- Cryptarithmetic Addition problems
- Practice questions after finishing Rules and Hacks here
- Cryparithmetic Division Problems (Only try these after you know how to solve Multiplication problem by both normal and unit digit methods)
Types of Cryparithmetic Problems in eLitmus
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Cryptarithmetic Questions Hacks
Rule 1
- If A + B = A then the possible value of
- B = 0 (no carry)
- B = 9 (As 1 carry from the previous step)
A X
+ B Y
--------
A Z
Example for B = 0
5 6
+ 0 2 <- Example for 9
-------
5 8
Example for B = 9
(carry) 1
5 6
+ 9 7 <- Example for 9
-------
1 5 3
Basics
- Numbers can’t begin with 0 like in the below example –
09841 makes no sense => 9841
- If there are the following numbers A, B, C, D, E
- All values are unique thus for e.g in below example A = 1 thus B, C, D, E ≠ 1
- Values are distinct thus for e.g. if G=1 then if at a later point in time you get G=2 you’re solving wrongly.
Rule 2
A X
X
----
B X
- If X * X = _X. Therefore – A={1, 5, 6}
Since only,
1*1 = _1
5*5 = _5
6*6 = _6 Rule 3
A X
Y
----
B X- If X x Y= _ X
- then possible values of X and Y, then X = 5 and Y = 3, 7, 9.
- When X = 2, 4, 8 and Y = 6
- X x Y = _X
5 x 3 = _5 i.e 15
5 x 7 = _5 i.e 35
5 x 9 = _5 i.e 45- X x Y = _X
2 x 6 = _ 2 i.e 12
4 x 6 = _ 4 i.e 24
8 x 6 = _ 8 i.e 48 Learn Cryptarithmetic Addition First
Before you start with Cryptarithmetic Multiplication or division, one must learn Cryptarithmetic Addition first. Below is one video that will help you learn Cryptarithmetic addition –
Alternate Method to solve Cryptarithmetic Problem
- Alternate method 2 – Unit Digit(Only Try this once you understand this basic method on this page)
Cryptarithmetic Problems to practise
- Cryptarithmetic Addition problems
- Practice questions after finishing Rules and Hacks here
- Cryparithmetic Division Problems (Only try these after you know how to solve Multiplication problem by both normal and unit digit methods)
Types of Cryparithmetic Problems in eLitmus
[table id=52 /]
Cryptarithmetic Basics Example
- Based on Addition Cryptarithmetic Questions
- Suggestion – Use Pen and Paper to learn from the example it might take more than 30 mins to understand this but when you get the logic, you’ll be able to solve any problem within very less time.
BASE +BALL ------------ GAMES ------------
=> Since, Maximum Carryover = 1
=> G = 1
Now Considering only the unit digits and tens digit
SE +LL ------- ES -------
- => S+L = E – (i) or S+L = 10 + E(Where 1 is carry) – (ii)
- => Similary E + L = S – (iii) or E + L = 10 + S(Where 1 is carry) – (iv)
- => E = S – L or E = S – L + 10(using this)
- => Putting value of E in S + L = E => S + L = S – L + 10
- => 2L = 10 => L = 5
- => from equation (iii) E + L = S => S – E = L = 5
- => This gives us possible values for (S,E) as (0,5), (1,6), (2,7), (3,8) and (4,9) or (E,S) if we take equation (i)
But out of these (0,5) and (1,6) cannot be accepted as G=1 and L=5.(Alphabets can only have distinct values)
From cryptarithmetic basics we are left with the possibilities of (2,7) , (3,8) and (4,9). We can also infer that of S and E , E is the smaller value and S is the larger.because if E were larger, we would have a carry and then S+L=E would not be valid. This means that S+L=E has a carry over of 1.
To solve Cryptarithmetic Questions use the trial and error method substituting values for the letters keeping all the above points in mind.
Let us assume E=2 and S=7 and B=6. So we have,
1 6A72 +6A55 ----------- 1AM27 ------------
Now A can be either 2 or 3 depending on whether we have a carry from A+A or not. But since E=2, that means A must be 3 and therefore there is a carry. But replacing the other A’s in the equation with 2’s gives us two contradictions.
Firstly M shall become equal to 7 (S is already equal to 7) and A+A does not produce a carry. Therefore our assumptions were wrong and we will have to try again for different values.
(I shall skip to the combination which yields the solution, but you shall have to try for all possible values in between)
Now let us try for E=3 and S=8 and B=7. We have,
1 7A83 +7A55 ---------- 1AM38 -----------
This gives us A=4 or 5 based on whether there is a carry or not, but since already L=5, A must be equal to 4, therefore M=9. We have obtained values for all unknowns without any contradictions and hence this is the solution)
So finally we have
1 7483 +7455 --------- 14938 ---------
Therefore,
- G=1
- E=3
- A=4
- L=5
- B=7
- S=8
- M=9
Cryptarithmetic Questions Solved Example –
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