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How to solve Cryptarithmetic Division asked in eLitmus and other exams

Cryptarithmetic Division Elitmus

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.

Cryparithmetic Division Questions with Answers

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