Sum of the digits of 27x
= 2 + 7 + x = 9 + x
Since, 27x is a multiple of 3,
∴ (9 + x) is divisible by 3.
⇒ 9 + x must be equal to 9 or 12 or 15
When 9 + x = 9 ⇒  x = 0
When 9 + x = 12 ⇒  x = 3
When 9 + x = 15 ⇒  x = 6
When 9 + x = 18 ⇒  x = 9
When 9 + x = 21 ⇒  x = 12, which is not possible. [∵ x is a digit]
∴ x can equal to 0, 3, 6 or 9.

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.

Let the original consumption be 100 kg and its original price be Rs 100.
New price of 100 kg of tea = Rs 120
∵ price is increased by 20%
∵ For Rs 120, we get tea = 100 kg
ஃ For Rs 100, we get tea
\(= \frac{100}{120} \times 100 kg =\frac{250}{3}kg\)
∴ Reduction in consumption
\(= (100- \frac{250}{3})%=\frac{50}{3}%\)
=\(16 \frac{2}{3}%\)%.

A = 5, B = 3

A = 8, b = 1, C = 7

39

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

It is given that, 1 AB + CCA = 697 and there is no carry-over in addition. We can write this expression as:
\(\begin{matrix} \quad \quad 1\ A\ B \\ +\quad C\ C\ A \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad 6\ 9\ 7 \\ \ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
In the first column (unit's place)
B + A = 7                                         ...........................(i)
In second column (ten's place)
A + C = 9                                        ............................(ii)
In third column (hundred's place)
1 + C = 6                                       ........................(iii)
C = 6 - 1 = 5
It is clear, C = 5.
Put C = 5 in Eq. (ii), we get
A + 5 = 9 \(\Rightarrow\) A = 9 - 5 = 4
Hence, A = 4.
Put A = 4 in Eq. (i), we get
B + A = 7 \(\Rightarrow\) B + 4 = 7 \(\Rightarrow\) B = 7 - 4 = 3
Hence, B= 3.
Now, A = 4, B = 3 and C = 5
\(\therefore\) Sum of A, Band C = A + B + C = 4 + 3 + 5 = 12

The given number is 123123A4.
Sum of digits at odd places = 4 + 3 + 1+ 2 = 10
Sum of digits at even places = A + 2 + 3 + 1= A + 6
If given number is divisible by 11, difference between sum of digits at odd places and sum of digits at even places will be either 0 or a multiple of 11
Case 1
(A + 6) - 10 = 0 \(\Rightarrow\) A + 6 - 10 = 0
\(\Rightarrow\) A - 4 = 0 \(\Rightarrow\) A = 0 + 4 = 4
Case 2
(A + 6) -10 = Multiple of 11
A + 6 - 10 = 11 \(\Rightarrow\)A - 4 = 11
A = 11 + 4 = 15
which is not possible because A is a 1-digit number. Hence, the value of A is 4.

The number 31 k2 is divisible by 6. It means, it is divisible by 2 and 3 both.
\(\because\) The unit's place of this number is 2, which is an even number. Hence, this number is divisible by 2. Now, to check the divisibility by 3. We have to add all the digits of this number.
3 + 1+ k + 2 will be divisible by 3. The possible answers are
6 + k = 6 \(\Rightarrow\) k = 6 - 6 = 0
6 + k = 9 \(\Rightarrow\) k = 9-6 = 3
6 + k = 12 \(\Rightarrow\) k = 12 - 6 = 6
6 + k = 15 \(\Rightarrow\) k = 15 - 6 = 9
\(\therefore\) Possible values of k are 0, 3, 6 and 9.

It is given that, 48101B095 is divisible by 33 = 3 x11.
Hence, it will be divisible by 3 and 11.
If 48101B095 is divisible by 3. Sum of digits
= 4 + 8 + 1 + 0 + 1 + B + 0 + 9 + 5 = 28 + B
(28 + B) is a multiple of 3.
If 481018095 is divisible by 11,
Sum of digits at odd places = 5 + 0 + 1 + 1+ 4 = 11
Sum of digits at even places = 9 + B + 0 + 8
= 17 + B
Difference = 17 + B -11 = 6 + B       (6 + 8) is a multiple of 11.
Least value of 6 + B = 11
B=11-6=5
Since, (28 + B) is a multiple of 3.
Least possible value of 28 +B = 30
8 = 30 - 28 = 2
Which is impossible because we have obtained
B = 5.
Now, we take next value, 28 + B = 33
B = 33 -28 =5
Hence, B = 5 is the solution.

\(\begin{matrix} \quad \quad 2\quad A\quad B \\ +\quad A\quad B\quad 1 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad B\quad 1\quad 8 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Here, we have two letters A and B whose values are to be found.
Studying the addition in the one's column, we have B + 1 which gives 8, therefore B must be 7.
Then, the puzzle becomes
\(\begin{matrix} \quad \quad 2\quad A\quad 7 \\ +\quad A\quad 7\quad 1 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad 7\quad 1\quad 8 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Studying the addition in ten's digit column. We have A + 7 which gives 1, i.e. a number whose unit's digit is 1.
So, A must be 4.
Then, the puzzle is solved as shown below:
\(\begin{matrix} \quad \quad 2\quad 4\quad 7 \\ +\quad 4\quad 7\quad 1 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad 7\quad 1\quad 8 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Hence, A = 4 and B = 7.

\(\begin{matrix} \quad \quad A\quad 1 \\ +\quad 1\quad B \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad B\quad 0 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Here, we have two letters A and B whose values are to be found.
Studying the addition in one's column, we have 1+ B and we get 0 from this i.e. a number whose one's digit is 0.
So, B must be 9. Then, for addition of ten's column, we have 1 + A + 1= 9
Then, A must be 7, then the puzzle is solved as shown below:
\(\begin{matrix} \quad \quad 7\quad 1 \\ +\quad 1\quad 9 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad 9\quad 0 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Hence, A = 7 and B = 9.

\(\begin{matrix} \quad \quad A\quad B \\ \quad \times \quad 6 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ B\quad B\quad B \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Here, we have two letters A and B whose values are to be found, since one's digit of B X 6 is 8, so B must be 2 or 4 or 6 or 8.
Then, possible values of BBB are 222, 444, 666 or 888.
If we divide these numbers by 6, then quotient should be A2 or A4 or A6 or A8.
Now, 222 \(\div\) 6 = 37, remainder = 0
But the quotient is not of the form A2, so B = 2 is not possible.
444 \(\div\) 6 = 74, remainder = 0
Also, quotient is of the form A4, which clearly works well.
Then, the puzzle is solved as shown below:
\(\begin{matrix} \quad \quad 7\quad 4 \\ \quad \times \quad 6 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ 4\quad 4\quad 4 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
Hence, A = 7 and B = 4

\(\begin{matrix} \quad \quad A\quad B \\ \quad \times \quad 5 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ C\quad A\quad B \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \end{matrix}\)
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