Two other numbers with the same digits are 741 adn 174
Sum of the three number
\(\begin{array}{r} 417 \\ +741 \\ +174 \\ \hline 1332 \\ \hline \end{array}\)
We have 1332 + 37 = 36, remainder = 0

This also has two letters A and B whose values are to be found.
Since the ones digit of 3 x A is A, it must be that A = 0 or A = 5.
Now look at B. If B = 1, then BA x B3 would at most be equal to 19 × 19; that is, it would at most be equal to 361. But the product here is 57A, which is more than 500. So we cannot have B = 1.
If B = 3, then BA × B3 would be more than 30 x 30; that is, more than 900. But 57A is less than 600. So, B can not be equal to 3.
Putting these two facts together, we see that B = 2 only. So the multiplication is either 20 x 23, or 25 x 23.
The first possibility fails, since 20 x 23 = 460. But, thesecond one works out correctly, since 25 x 23 = 575.
So the answer is A = 5, B = 2.

The usual form of 10000a + 100b+ c
= 10000a+ 1000 x 0 + 100b + 10 x 0 + c = a 0 b 0 c

If the number abed is divisible by 11, then [(b+d)-(a+c)] must be divisible by 11, i.e [(b + d) - (a + c)] must be 11 or a multiple of 11.

Yes, if a number is divisible by any number m, then it will also be divisible by each of the factors of m.
e.g. 63 is divisible by 21.
Factors of 21 are 1, 3 and 7.
Therefore, 63 is also divisible by 1, 3 and 7.

Here, the remainder = 0. So, N is an even number, i.e. its one's digit is even. Therefore, the one's digit must be 0, 2, 4, 6 or 8.

The one's digit, when divided by 5 leaves a remainder 3. So, the one's digit must be either 3 or 8.

A number is said to be in a generalised form, if it is expressed as the sum of the products of its digits with their respective place values as ab = a x 10 + b and abc = a x 100 + b x 10 + c.
302 = 100 x 3 + 10 x 0 + 1 x 2   [ \(\because\)abc =  100a + 10b + c ]

(a) Total amount received as pocket money is given by 2x + 3x + 5x + 7x
\(\Rightarrow\)  (2+ 3+ 5+ 7)x = 680
\(\Rightarrow\) 17x = 680 \(\Rightarrow\) x = 40
Hence, each of them receives amount Rs 80, Rs120, Rs 200, and Rs 280 as pocket money, respectively.
(b) The boys are very sensible and know the importance of saving money. They have interest in reading inspirational books like autobiographies of great personalities.

P = 9, Y = 1 and Z = 8

\(\begin{matrix} \quad \quad A\ B \\ \quad \times \ \ \ \ \ 3 \\ \_ \_ \_ \_ \_ \_ \_\_ \\ \quad C\ A\ B \\ \_\_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Here, we have three letters A, Band C whose values all to be found.
Since unit's digit of 3 x B is B. So, it must be B = 0 or B = 5
When B = 0, then
\(\begin{matrix} \quad \quad A\ 0 \\ \quad \times \quad 3 \\ \_ \_ \_ \_ \_ \_ \_ \\ C\ A\ 0 \\ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
and when B = 5, then
\(\begin{matrix} \quad \ A\ 5 \\ \quad \times \ 3 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ C\quad A\quad 5 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Now, unit's digit of 3 x A is A. So, A must be 0 or 5 but A cannot be 0 because if A = 0, then AB becomes the one-digit number. So, A must be 5 and multiplication is either 50 x 3 or 55 x 3.
Here, the second possibility fails, since 55 x 3 = 165 but the first possibility is correct.
\(\therefore\)  50 x 3 = 150
Therefore, the puzzle is solved as shown below:
\(\begin{matrix} \quad \quad 5\quad 0 \\ \quad \times \quad 3 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ 1\quad 5\quad 0 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Hence, A = 5, B = 0 and C = 1

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

Given, 24x is a multiple of 3.
We know that a number is divisible by 3 if the sum of its digits is divisible by 3. So, the sum of the digits of 
24x = 2 + 4 + x = 6 + x
Which should be a multiple of 3.
Therefore, 6 + x should be 0, 3, 6, 9, 12, 15, ... , etc
Since, x is a digit of 6 + x, so it can only be 6 or 9 or 12 or 15
\(\therefore\) 6 + x = 6 \(\Rightarrow\) x = 6 - 6 \(\Rightarrow\) x = 0
or 6 + x = 9 \(\Rightarrow\) x = 9 - 6 \(\Rightarrow\) x = 3
or 6 + x = 12 \(\Rightarrow\) x = 12 - 6 \(\Rightarrow\) x = 6
or 6 + x = 15 \(\Rightarrow\) x = 15 - 6 \(\Rightarrow\) x = 9
Hence, the value of x can be 0 or 3 or 6 or 9.

\(\begin{matrix} \quad \quad 1\quad 2\quad A \\ +\quad 6\quad A\quad B \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad A\quad 0\quad 9 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Here, we have two letters A and B whose values are to be found.
Studying the addition in ten's column. We have 2 + A and we get 0 from this. Therefore, A must be 8, since A will be one digit number, then the puzzle becomes
\(\begin{matrix} \quad \quad 1\quad 2\quad 8 \\ +\quad 6\quad 8\quad B \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad 8\quad 0\quad 9 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Now, studying the addition in one's column. We have 8 + B and we get 9 from this. So, B must be 1.
Thus, the puzzle is solved as shown below:
\(\begin{matrix} \quad \quad 1\quad 2\quad 8 \\ +\quad 6\quad 8\quad 1 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad 8\quad 0\quad 9 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Hence, A = 8 and B = 1.

\(\begin{matrix}\quad 4\quad A \\ +\quad 9\quad 8 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad C\quad B\quad 3 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Here, we have three letters A, B, and C whose values are to be found.
Studying the addition in the one's column, we have A + 8 and we get 3 from this, i.e. a number whose one's digit is 3, for this A has to be 5.
\(\because\) A + 8 = 5 + 8 = 13
Now, for sum in ten's column, we have
1 + 4 + 9 = CB \(\Rightarrow\) 14 = CB
Here, B = 4 and C = 1
Therefore, the puzzle is solved as shown below:
\(\begin{matrix}\quad \quad 4\quad 5 \\ +\quad 9\quad 8 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad 14\quad 3 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Hence, A = 5, B = 4 and C = 1

9 + 2 + 7 = 18 is divisible by 9

   5 1
  \(\times\) 2
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  102
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