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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

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------       1B R S Rand  R P B  x   B-------     - 2P T C BNow 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 RNow, B x R = _RNow 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 =52) B = 6 when, R = 2 or 4 or 8First we go ahead with case 1.1 ie R = 5 and B = 3 -   5 P 3  x   5-------3 5 S 5Now 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 = 9now, 9 x 5 or P x R in 100's position, can be max = 45Thus producing 4 as carry in thousands position.Now, for 1000's position 5 x 5 + carry i.e R x R + carry = 25 + 5 = 29Thus 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 5Now, again this is not possible as we will never,get 7 in 1000's position asMax 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 1.3Now, repeated this for B = and R = 5  5 P 9  x   5-------9 5 S 5As Max carry = 4 again ( P x 5 for 9 is 45)and max multiplication at 1000's is 5 x 5 = 2529 is the answer again which has 2 in 1000's place not 9we go ahead with case 2.1 So, lets try other one,B = 6 and R = 2  2 P 6  x   2-------6 2 S 2Now, 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 carryand at 1000's position 2 x 2 + carry(1) gives us 5 not 6.Lets try case 2.2Where B = 6 and R = 4  4 P 6  x   4-------6 4 S 4Now, this is also rejected. Please try yourself, if you dont understand ask in comments.Now, case 2.3 Where B = 6 and R = 8This is accepted as -   8 P 6  x   8-------6 8 S 8In 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 = 68Now, 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 = 48Now, 6 is not possible as there are existing carry of 4 from unit's positionso, 48 + 5 = 52 thus 5 carry and also value of B is already assumed to be 6, so P can't be 6Now, 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 nextand 4 goes to S value.  R P Bx     R------B R S R  8 5 6x     8-------6 8 4 8Now, trying to solve the 2nd part -   R P B  x   B-------P T C BReplacing  8 5 6  x   6-------P T C 6Now, 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) = 51Thus P = 5 and T = 1Do 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 RValues :       ____________8 5 6 | U 4 E 8 4      | 6 8 4 8       --------          5 8 D 4          5 1 3 6          --------            6 6 8Let 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`