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# How to solve Cryptarithmetic Division asked in eLitmus and other exams

## How to solve Cryptarithmetic Division Problems in eLitmus

**The tricks to solve eLitmus Cryparithmetic Division problem is to convert the problem into multiple Cryparithmetic Multiplication problems.**

**Please go to our Cryptarithmetic multiplication page first and then come back here.**

## How to solve Cryptarithmetic Division Problems in eLitmus

- Cryptarithmetic Problem – 2
- Cryptarithmetic Problem – 3
- Cryptarithmetic Problem – 4
- Cryptarithmetic Problem – 5
- Cryptarithmetic Problem – 6
- Cryptarithmetic Problem – 7

Let us try to understand this with an example –

Consider this original problem here –

Also remember what we had learnt in our school, there are the following four terms –

- Divisor – RPB
- Dividend – USERS
- Quotient – RB
- Remainder – BBR

**Now, Quotient x Divisor + Remainder = Dividend**

Now the above question can be written as -and

R P B

x R

------ 1

B R S R

and

R P B

x B

------- - 2

P T C B

Now we will take the first one as it has lesser Alphabets thus,

it must be easier to solve.

R P B

x R

-------

B R S R

Now, B x R = _R

Now according to Rule 3/Hack 3 (check this page for rule)

There are two possibilities in Hack 3 -

1) B = 3 or 7 or 9 when, R =5

2) B = 6 when, R = 2 or 4 or 8

First we go ahead with case 1.1 ie R = 5 and B = 3 -

5 P 3

x 5

-------

3 5 S 5

Now if you watch closely, this is not possible as,

we will never get 3 in 1000's place. Since,

First lets check max carry from 100's position, max value of P = 9

now, 9 x 5 or P x R in 100's position, can be max = 45

Thus producing 4 as carry in thousands position.

Now, for 1000's position 5 x 5 + carry i.e R x R + carry = 25 + 5 = 29

Thus there will be 2 at max in 1000's position.

Read this Bit again, Cryparithmetic problems are very very hard and we suggest giving atleast 1 week for cryparithmetic problems only. It will take you

a lot of time to understand this but when you do it will increase your percentile by as much as 50%.

we go ahead with case 1.2

Now, lets Try B = 7 and R = 5

5 P 7

x 5

-------

7 5 S 5

Now, again this is not possible as we will never,

get 7 in 1000's position as

Max carry when P = 9 will be 4( 9 x 5 =45)

and 5 x 5 + carry(4) = 29 which gives 2 not 7. so this is not possible again.

we go ahead with case 1.3

Now, repeated this for B = and R = 5

5 P 9

x 5

-------

9 5 S 5

As Max carry = 4 again ( P x 5 for 9 is 45)

and max multiplication at 1000's is 5 x 5 = 25

29 is the answer again which has 2 in 1000's place not 9

we go ahead with case 2.1

So, lets try other one,

B = 6 and R = 2

2 P 6

x 2

-------

6 2 S 2

Now, this is also not possible as -

In 100's place Max value of P = 9 thus max carry it can generate is -

9 x 2 = 18 thus 1 carry

and at 1000's position 2 x 2 + carry(1) gives us 5 not 6.

Lets try case 2.2

Where B = 6 and R = 4

4 P 6

x 4

-------

6 4 S 4

Now, this is also rejected. Please try yourself, if you dont understand ask in comments.

Now, case 2.3 Where B = 6 and R = 8

This is accepted as -

8 P 6

x 8

-------

6 8 S 8

In units place 6 x 8 = 48 thus carry 4.

In 1000's place 8 x 8 = 64 but in the answer's left

most two digits its 68. Thus carry for 4 is also coming from

P x 8 as - ( 8 x 8) + 4 = 68

Now, value of P to give 4 as carry in 1000's place can be -

only 5 or 6 as 5 x 8 = 40 and 6 x 8 = 48

Now, 6 is not possible as there are existing carry of 4 from unit's position

so, 48 + 5 = 52 thus 5 carry and also value of B is already assumed to be 6, so P can't be 6

Now, we have gotten the following values -

B = 6, R = 8, P = 5 and obv value of s will be 4 as 8 x 5 + carry(4) = 44 4 carry goes to next

and 4 goes to S value.

R P B

x R

------

B R S R

8 5 6

x 8

-------

6 8 4 8

Now, trying to solve the 2nd part -

R P B

x B

-------

P T C B

Replacing

8 5 6

x 6

-------

P T C 6

Now, at unit's 6 x 6 = 36, justifying B value as 6 and 3 carry,

5 x 6 = 30 and 3 carry = 33 thus C = 3 and next step 3 carry.

Now, 8 x 6 + Carry(3) = 51

Thus P = 5 and T = 1

Do Comment if you have any queries related to Cryptarithmetic Division.

____________

R P B | U S E R S

| B R S R

--------

P R D S

P T C B

-------

B B R

Values :

____________

8 5 6 | U 4 E 8 4

| 6 8 4 8

--------

5 8 D 4

5 1 3 6

--------

6 6 8

Let us now, create a subtraction problem-

U 4 E 8

- 6 8 4 8

---------

5 8 D

- Clearly, D = 0 ( as 8 – 8)
- As E – 4 = 8
- E must be taking 1 borrow from previous step
- So 12 – 4= 8 is happening
- So, E = 2

- Now, 4 becomes 3
- 3 – 8 = 5

- Thus, 13 – 8 = 5
- 1 borrows from 7

- Thus, (U -1) – 6 = 0
- U = 7

8 6

____________

8 5 6 | 7 4 2 8 4

| 6 8 4 8

--------

5 8 0 4

5 1 3 6

--------

6 6 8

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