Yes

0 or 5

it is given that
\(\begin{matrix} \quad \quad A \\ \quad \quad A \\ +\quad A \\ \_ \_ \_ \_ \_ \_ \_ \_ \\ B\quad A \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
We observe first column, i.e.
A + A + A = A
It is possible for A = 5.
5 + 5 + 5 = 15
Hence, B =1.
Now, this number puzzle can be solved as:
\(\begin{matrix} \quad \quad 5 \\ \quad \quad 5 \\ +\quad 5 \\ \_ \_ \_ \_ \_ \_ \_ \_ \\ 1\quad 5 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Hence, A = 5 and B =1.

The given number is 14567. Unit's digit of this number is an odd number. Hence, this number is not divisible by 2.

Yes, No

The given number is 108.
Sum of digits of 108 = 1+0 +8 = 9
Now, on dividing this sum by 3, we get
9 \(\div\) 3 = 3 and remainder = 0
which is divisible by 3.
Hence, 108 is divisible by 3.

If the number abc is divisible by 11, then (a - b + c) is either 11 or a multiple of 11. Yes, it is necessary that (a + c - b ) should be divisible by 11.

98
This number is divisible by 2 because the digit at unit's place is 8, which is an even number.

666

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 ]

(i) Let the awards given in 1958 be x% of the awards given in 1993. Then,
x% of 81=28
⇒ \(\frac{x}{100} \times 81=28\)
⇒ \(x=\frac{28 \times 100}{81}\)
= 34.5 (approx.)
Hence, the awards given is 1958 is 34.5 per cent of the awards given in 1993.
(h) Let the awards given in 1993 be x% of the awards given in 1958. Then,
x% of 28 = 81
⇒ \(\frac{x}{100} \times 28=81\)
⇒\(x=\frac{81 \times 100}{28}=\frac{81 \times 25}{7}\)
⇒ \(\frac{2025}{7}\)=289 (approx.)
Hence, the award given is 1993 is 289 per cent of the awards given in 1958.

(i) 10 x 34 + 5
(ii) 10 x 27 + 5

We have, a 3-digit number 2a3, when added to the number 326, gives a 3-digit number 5b9.
We can write 2a3 and 5b9 in generalised form as.
2a³ = 2 x 100 + a x10 + 3 = 200 + 10a + 3
5b9 = 5 x 100 + b x 10 + 9 = 500 + 10b + 9
Now, it is given that
2a3 + 326 = 5b9
200 + 10a + 3 + 326 = 500 + 10b+ 9
10a + (200 + 3 + 326) = 10b + (500 + 9)
10a + (200 + 329) = 10b+ 509
10a + 529 = 10b + 509
529 - 509 = 10b -10a
20 = 10 x (b-a)
b - a = \({20 \over 10}\)
b - a = 2                        .........(i)
Now, it is given that 5b9 is divisible by 9.
So, sum of digits of 5b9 will be divisible by 9, i.e
5 + b + 9 = Multiple of 9
14 + b =18 (nearest multiple of 9)
b = 18 - 14 = 4
Put b = 4 in Eq. (i), we get
b - a = 2 \(\Rightarrow\)4-a = 2
4 - 2 = a \(\Rightarrow\) 2 = a
\(\Rightarrow\) a = 2
Hence, a = 2, b = 4 and b - a = 2

\(\begin{matrix} \quad \quad A\ B \\ \quad \times \ \ \ \ \ 3 \\ \_ \_ \_ \_ \_ \_ \_\_ \\ \quad C\ A\ B \\ \_\_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
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

We have the sum of the digits of 51x3
= 5+1+x+3=9+x
Since, 51x3 is divisible by 9.
∴ (9 + x) must be divisible by 9.
∴ (9 + x) must be equal to 0 or 9 or 18 or 27
or ... But x is a digit, then
9+x=9 ⇒ x=0
9 + x = 18 ⇒ x = 9
x = 27 ⇒ x = 18, which is not possible.
∴ The required value of x = 0 or 9.

A = 5, B = 3

A = 5

Given, 21y 5 is a multiple of 9, i.e. 21y5 is divisible by 9.
We know that a number is divisible by 9 if the sum of its digits is divisible by 9.
So, sum of digits of 21y 5 = 2 + 1+ y + 5 = 8 + y
Therefore, (8 +y) should be 0, 9, 18,27, ... , etc.
Since x is a digit.
\(\therefore\) 8 + y = 9  \(\Rightarrow\) y = 9 - 8 \(\Rightarrow\) y=1
Hence, the value of y is 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

2 + 3 + 7 = 12 is divisible by 3.

In 19, 27, 99, one's digit is not divisible by 2.

 1 3
\(\times\) 3
^____
3 9
^____