Cryptarithmetic Division Problem 5

PT

Converting into Multiplication problem
  T T N 
  x   R
-------   - 1
Q T P A

  T T N
  x   R
-------   - 2
Q T P S

  T T N
  x   P
-------  - 3
B T Q P

Taking the third eqn -

  T T N
  x   P
-------  - 3
B T Q P

N x P = _P is same has P x N = _P 
thus, the two possible cases according rule
mentioned in hack 3 on this page will be -

Case 1 - N = {3, 7, 9} P = 5
Case 2 - P = 6, and N = {2, 4, 8}

Case 1.1 - N = 3 and P = 5.

  T T 3
  x   5
-------
B T Q 5

Now at unit's 3 x 5 = 15 thus 1 is carry to next operation.
Now T x 5 + 1(carry) possible values are -> {1, 6}.
as multiplication with 5 will give 0 or 5 and 
add 1 to both so 1 and 6

Q = {1,6}
Taking Q = 1

  T T 3
  x   5
-------
B T 1 5

For getting 1 at 10's place, possible value for T can be
T = {2, 4, 6, 8}

For T -> 2

  2 2 3
  x   5
-------
B T 1 5

Now in 10's position 5 x 2 + 1(carry) = 11.
1 carry to 100's
Now again at 100's 5 x 2 + 1 carry = 11. So value for T in row 3
is 1. T = 1. But, T value was already assumed to be 2. So this is also rejected.

Now, lets take T = 4

  T T 3
  x   5
-------
B T 6 5

Now, here again if you solve like previous. T value in row 3,
will come out to be 2. But, T value was assumed to be 2, so rejected.

similarly for T = 6 it will be rejected as T will come out as 3
And for T = 8 it will come out to be 4 so rejected

Now, lets try with Q = 6.(Q had two values 1 and 6)

To get 6 at tens place T value can only be 7 or 9
ie T = {7, 9}

Let's try for 7
  7 7 3
  x   5
-------
B T 6 5 

units -> 3 x 5 = 15 1 carry
10's 7 x 5 +1(carry) = 36 3 carry
100's 7 x 5 + 3(carry) = 38

So T value is 8 but was assumed to be 7 so rejected

Now lets use T = 9

  9 9 3
  x   5
-------
B T 6 5 

In this case T will come out to be 9 only
and B =4. We can find all values from here. Solved !

PT