Exercise 5.3
Solve each of the following Cryptarithms:
Solution:
Two possible values of A are:
(i) If 7 + B < 9, 3 + A = 9
∴ A = 6
But if A = 6, 7 + B must be larger than 9.
Hence, it is impossible.
(ii) If 7 + B > 9
∴ 1 + 3 + A = 9
=> A = 5
1f A = 5 and 7 + B = 5,
B must be 8
∴ A = 5, B = 8
Solution:
Two possibilities of A are:
(i) If B + 7 < 9,
A = 6
But clearly, if A = 6,
B + 7 > 9;
it is impossible
(ii) If B + 7 > 9,
A = 5 and B + 7 = 5
Clearly, B = 8
∴ A = 5, B = 8
Solution:
If 1 + B = 0 Surely, B = 9
If 1 + A + 1 = 9 Surely, A = 7
Solution:
B + 1 = 8, B = 7A + B = 1, A + 7 = 1, A = 4
So, A = 4, B = 7
Solution:
A + B = 9 as the sum of two digits can never be 192 + A = 0, A must be 8A + B = 9, 8 + B = 9, B = 1
So, A = 8, B = 1
Solution:
If A + B = 8, A + B > 9 is possible only if A = B = 9 But from 7 + B = A, A = B = 9 is impossible. Surely, A + B = 8, A + B < 9
So, A + 7 = 9, Surely A = 27 + B = A, 7 + B = 2, B = 5
So, A = 2, B = 5
Solution:
0 is the only unit digit number, which gives the same 0 at the unit digit when multiplied by 4. So, the possible value of B is 0. Similarly, for A also, 0 is the only possible digit. But then A, B and C will all be 0, and if A, B and C become 0, these numbers cannot be of two — digit or three — digit. Therefore, both will become a one — digit number. Thus, there is no solution
