# Cryptarithmetic Division Problem 4 You must know Cryptarithmetic Multiplication first to be able to solve these, so we suggest studying Multiplication problems from our website first.

The first thing to do in this problem to break down into Multiplication and Subtraction problem as follows –

We have not been able to solve this question, please provide answer in comments or you can try other problems listed on the main page of Cryparithmetic Division Problem all others are solved completely.

`Breaking down the problem -  T T N  x   R------- - 1Q T P A  T T N  x   Q------- - 2P T P S  R P P M- Q T P A--------- - 3    Q D ANow lets take equation number 3From 10's digit we see that P - P = D,Thus it is clear that value of D = 0and possible values of P can be = {1,2,3,4,5,6,7,8,9}Now, Learn this Cryparithmetic Subtraction Rule -If, X - A = A,Then X = {2, 4, 6, 8}A = {1, 2, 3, 4}Thus, in our case M - A = ASo, they will have values,M = {2, 4, 6, 8}A = {1, 2, 3, 4}Now, lets do hit and trial method,D = 0, M = 2, A =1So, possible values for P = {1,2,3,5,6,7,9}R is not 1 as R > Q and if R is 1 then there is no value of Q.Let us assume P = 5  R 5 5 2- Q T 5 1---------    Q 0 1Since, R - Q is 0. Thus, R and Q must be consecutive.i.e. R = Q + 1Possible values for R and Q can be = {(7,6)}As, (1,2) not possible as (M,A) are there.As, (2,3) not possible as M is 2Next consecutive value are (5,6)P is already 5So (R,Q) cant be (6,5)If (P,Q) = (7,6)  7 5 5 8- 6 T 5 4---------    6 0 4Now, for 5 - T = 6T = 9, thus possible.Putting these in Eq - 1  9 9 Nx     7-------6 9 5 1Since, N x 7 = _1. Thus, N = 3(as 21)Putting all these values       _________9 9 3 |7 5 5 2 0 1       6 9 5 1       -------         6 0 1 0         5 9 5 S        --------             5 2 1Clearly S = 8Thus, we have found all values.`

### 2 comments on “Cryptarithmetic Division Problem 4”

• Subrat Hemant

T T N = 993
R Q = 76
Q T P A = 6 9 5 1
P T P S = 5 9 5 8
R P P M D A= 7 5 5 2 0 1 0
• Pinu Gargote 0