666

100 x 1+ 10 x 3 + 1 x 9

100 x 7 + 10 x 1 + 8 = 700 + 10 + 8 = 718

Let the 2-digit number be ab and the number obtained by reversing the digits is ba.
Let the sum be a 3-digit number dad.
\(\therefore\)  ab + ba = dad
\(\Rightarrow\) 10a + b + 10b + a = dad
\(\Rightarrow\) 11a + 11b = dad
\(\Rightarrow\) 11(a+b) = dad
The sum a + b cannot exceed 18 because the greatest 2-digit number is 99 and 9 + 9 = 18
Since, 11(a + b) = dad, so dad is the multiple of 11
Also, 99 + 99 = 198
So, dad is less than 198.
All the 3-digit numbers which are multiples of 11 upto 198 are 110, 121, 132, 143, 154, 165, 176, 187 and 198.
Clearly, dad = 121
Hence, a = 2 and d = 1

A = 5, B = 3

Let a 2-digit number be 10x + y.
It is given that a two digit number exceeds the sum of the digits of that number by 18.
(10x + y)-(x + y) = 18                ................(i)
It is also given that digits at the unit's place is double the digit at the ten's place.
y = 2x                                ..................(ii)
Now, from Eq. (i), we get
(10x + y) - (x + y)=18
10x + y- x - y = 18 \(\Rightarrow\)9x = 18
x = \({18\over 9}\) = 2
Put x = 2 in Eq. (ii), we get y = 2 x 2 = 4
Now, the required number =10x + y
= 10x² +  4 = 20 + 4 = 24

\(\begin{matrix} \quad \quad A\quad B \\ \quad \times \quad 5 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ C\quad A\quad B \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Here, we have three letters A, Band C whose values are to be found.
Since unit's digit of B x 5 is B, so B must be 0 or 5.
When B = 0, then we have
\(\begin{matrix} \quad \quad A\quad 0 \\ \quad \times \quad 5 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ C\quad A\quad 0 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
When B = 5, then we have
\(\begin{matrix} \quad \quad A\quad 5 \\ \quad \times \quad 5 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ C\quad A\quad 5 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Now, unit's digit of 5 X A is A, so A must be 0 or 5.
But A \(\neq \) 0 as in the answer, there are three letters, so A must be 5. So, multiplication is either 50 xX 5 or 55 x 5.
Here, the second possibility fails, since 55 x 5 = 275 but the first possibility is correct, since 50 x 5 = 250.
Therefore, the puzzle is solved as shown below:
\(\begin{matrix} \quad \quad 5\quad 0 \\ \quad \times \quad 5 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ 2\quad 5\quad 0 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Hence, A = 5, B = 0 and C = 2