Playing with Numbers

Let the tens digit be $x$.

Then, units digit = $4x$.

Given, sum of the digits is 10.

$$\begin{aligned}\Rightarrow x + 4x = 10 \\[1em] \Rightarrow 5x = 10 \\[1em] \Rightarrow x = 2\end{aligned}$$

So, tens digit = 2 and units digit = (4 × 2) = 8.

∴ Required number = 10 × 2 + 8 = 20 + 8 = 28

Hence, the required number is 28.

Let the tens digit be $x$.

Given, units digit = 3.

∴ The number = $10x + 3$ and sum of its digits = $x + 3$.

Given, seven times the sum of the digits is the number itself.

$$\begin{aligned}\Rightarrow 7(x + 3) = 10x + 3 \\[1em] \Rightarrow 7x + 21 = 10x + 3 \\[1em] \Rightarrow 10x - 7x = 21 - 3 \\[1em] \Rightarrow 3x = 18 \\[1em] \Rightarrow x = 6\end{aligned}$$

So, tens digit = 6 and units digit = 3.

∴ Required number = 10 × 6 + 3 = 60 + 3 = 63

Hence, the required number is 63.

Let the tens digit be $x$.

Then, units digit = $3x$.

∴ The number = $10x + 3x = 13x$ and sum of its digits = $x + 3x = 4x$.

Given, the number exceeds the sum of its digits by 27.

$$\begin{aligned}\Rightarrow 13x - 4x = 27 \\[1em] \Rightarrow 9x = 27 \\[1em] \Rightarrow x = 3\end{aligned}$$

So, tens digit = 3 and units digit = (3 × 3) = 9.

∴ Required number = 10 × 3 + 9 = 30 + 9 = 39

Hence, the required number is 39.

Let the tens digit be $x$ and the units digit be $y$.

∴ The number = $10x + y$ and sum of its digits = $x + y$.

Given, the number is 3 more than 4 times the sum of its digits.

$$\begin{aligned}\Rightarrow 10x + y = 4(x + y) + 3 \\[1em] \Rightarrow 10x + y = 4x + 4y + 3 \\[1em] \Rightarrow 10x - 4x + y - 4y = 3 \\[1em] \Rightarrow 6x - 3y = 3 \\[1em] \Rightarrow 2x - y = 1 \quad \text{.....(1)}\end{aligned}$$

Number obtained by reversing the digits = $10y + x$

Given, if 18 is added to the number, its digits are reversed.

$$\begin{aligned}\Rightarrow (10x + y) + 18 = 10y + x \\[1em] \Rightarrow 10x - x + y - 10y = -18 \\[1em] \Rightarrow 9x - 9y = -18 \\[1em] \Rightarrow x - y = -2 \quad \text{.....(2)}\end{aligned}$$

Subtracting (2) from (1), we get

$$\begin{aligned}\Rightarrow (2x - y) - (x - y) = 1 - (-2) \\[1em] \Rightarrow x = 3\end{aligned}$$

Putting $x = 3$ in (2), we get $3 - y = -2 \Rightarrow y = 5$

So, tens digit = 3 and units digit = 5.

∴ Required number = 10 × 3 + 5 = 30 + 5 = 35

Hence, the required number is 35.

Let the tens digit be $x$ and the units digit be $y$.

Given, sum of the digits is 13.

⇒ $x + y = 13 \quad \text{.....(1)}$

Original number = $10x + y$.

Number obtained by interchanging its digits = $10y + x$

Given, the number obtained by interchanging the digits exceeds the given number by 9.

$$\begin{aligned}\Rightarrow (10y + x) - (10x + y) = 9 \\[1em] \Rightarrow 9y - 9x = 9 \\[1em] \Rightarrow y - x = 1 \quad \text{.....(2)}\end{aligned}$$

Adding (1) and (2), we get

$$\begin{aligned}\Rightarrow 2y = 14 \\[1em] \Rightarrow y = 7\end{aligned}$$

Putting $y = 7$ in (1), we get $x = 13 - 7 = 6$

So, tens digit = 6 and units digit = 7.

∴ Original number = 10 × 6 + 7 = 60 + 7 = 67

Hence, the original number is 67.

Let the tens digit be $x$ and the units digit be $y$.

∴ Original number = $10x + y$.

Number obtained by interchanging its digits = $10y + x$

Given, the difference between the number and the number obtained by interchanging its digits is 63.

$$\begin{aligned}\Rightarrow (10x + y) - (10y + x) = 63 \\[1em] \Rightarrow 9x - 9y = 63 \\[1em] \Rightarrow 9(x - y) = 63 \\[1em] \Rightarrow x - y = 7\end{aligned}$$

Hence, the difference between the digits of the number is 7.

Let the tens digit be $x$.

Then, hundreds digit = $2x$ and units digit = $3x$.

Given, sum of its digits is 18.

$$\begin{aligned}\Rightarrow 2x + x + 3x = 18 \\[1em] \Rightarrow 6x = 18 \\[1em] \Rightarrow x = 3\end{aligned}$$

So, hundreds digit = (2 × 3) = 6, tens digit = 3 and units digit = (3 × 3) = 9.

∴ Required number = 100 × 6 + 10 × 3 + 9

= 600 + 30 + 9

= 639

Hence, the required number is 639.

According to the divisibility test of 2, a given number is divisible by 2 only when its units digit is even, i.e. 0, 2, 4, 6 or 8.

(i) 96 — Units digit is 6, which is even.

Hence, 96 is divisible by 2.

(ii) 830 — Units digit is 0, which is even.

Hence, 830 is divisible by 2.

(iii) 367 — Units digit is 7, which is not even.

Hence, 367 is not divisible by 2.

(iv) 922 — Units digit is 2, which is even.

Hence, 922 is divisible by 2.

(v) 881 — Units digit is 1, which is not even.

Hence, 881 is not divisible by 2.

(vi) 934 — Units digit is 4, which is even.

Hence, 934 is divisible by 2.

(vii) 1089 — Units digit is 9, which is not even.

Hence, 1089 is not divisible by 2.

(viii) 13058 — Units digit is 8, which is even.

Hence, 13058 is divisible by 2.

According to the divisibility test of 3, a given number is divisible by 3 only when the sum of its digits is divisible by 3.

(i) 258 — Sum of digits = 2 + 5 + 8 = 15, which is divisible by 3.

Hence, 258 is divisible by 3.

(ii) 774 — Sum of digits = 7 + 7 + 4 = 18, which is divisible by 3.

Hence, 774 is divisible by 3.

(iii) 584 — Sum of digits = 5 + 8 + 4 = 17, which is not divisible by 3.

Hence, 584 is not divisible by 3.

(iv) 1033 — Sum of digits = 1 + 0 + 3 + 3 = 7, which is not divisible by 3.

Hence, 1033 is not divisible by 3.

(v) 2571 — Sum of digits = 2 + 5 + 7 + 1 = 15, which is divisible by 3.

Hence, 2571 is divisible by 3.

(vi) 6654 — Sum of digits = 6 + 6 + 5 + 4 = 21, which is divisible by 3.

Hence, 6654 is divisible by 3.

(vii) 10385 — Sum of digits = 1 + 0 + 3 + 8 + 5 = 17, which is not divisible by 3.

Hence, 10385 is not divisible by 3.

(viii) 35742 — Sum of digits = 3 + 5 + 7 + 4 + 2 = 21, which is divisible by 3.

Hence, 35742 is divisible by 3.

According to the divisibility test of 4, a given number is divisible by 4 only when the number formed by its last two digits is divisible by 4.

(i) 562 — Number formed by the last two digits = 62, which is not divisible by 4.

Hence, 562 is not divisible by 4.

(ii) 8176 — Number formed by the last two digits = 76 = 4 × 19, which is divisible by 4.

Hence, 8176 is divisible by 4.

(iii) 56318 — Number formed by the last two digits = 18, which is not divisible by 4.

Hence, 56318 is not divisible by 4.

(iv) 35344 — Number formed by the last two digits = 44 = 4 × 11, which is divisible by 4.

Hence, 35344 is divisible by 4.

(v) 909032 — Number formed by the last two digits = 32 = 4 × 8, which is divisible by 4.

Hence, 909032 is divisible by 4.

(vi) 77752 — Number formed by the last two digits = 52 = 4 × 13, which is divisible by 4.

Hence, 77752 is divisible by 4.

According to the divisibility test of 5, a given number is divisible by 5 only when its units digit is 0 or 5.

(i) 1095 — Units digit is 5.

Hence, 1095 is divisible by 5.

(ii) 2746 — Units digit is 6, which is neither 0 nor 5.

Hence, 2746 is not divisible by 5.

(iii) 8185 — Units digit is 5.

Hence, 8185 is divisible by 5.

(iv) 4040 — Units digit is 0.

Hence, 4040 is divisible by 5.

(v) 98762 — Units digit is 2, which is neither 0 nor 5.

Hence, 98762 is not divisible by 5.

(vi) 77770 — Units digit is 0.

Hence, 77770 is divisible by 5.

According to the divisibility test of 8, a given number is divisible by 8 only when the number formed by its last three digits is divisible by 8.

(i) 7304 — Number formed by the last three digits = 304 = 8 × 38, which is divisible by 8.

Hence, 7304 is divisible by 8.

(ii) 8612 — Number formed by the last three digits = 612, which is not divisible by 8.

Hence, 8612 is not divisible by 8.

(iii) 44440 — Number formed by the last three digits = 440 = 8 × 55, which is divisible by 8.

Hence, 44440 is divisible by 8.

(iv) 265472 — Number formed by the last three digits = 472 = 8 × 59, which is divisible by 8.

Hence, 265472 is divisible by 8.

According to the divisibility test of 10, a given number is divisible by 10 only when its units digit is 0.

(i) 4005 — Units digit is 5, which is not 0.

Hence, 4005 is not divisible by 10.

(ii) 30303 — Units digit is 3, which is not 0.

Hence, 30303 is not divisible by 10.

(iii) 64610 — Units digit is 0.

Hence, 64610 is divisible by 10.

(iv) 85730 — Units digit is 0.

Hence, 85730 is divisible by 10.

According to the divisibility test of 11, a number is divisible by 11 if the difference between the sum of its digits at odd places and the sum of its digits at even places is either 0 or a number divisible by 11.

(i) The given number is 1452.

Sum of its digits at odd places = (2 + 4) = 6.

Sum of its digits at even places = (5 + 1) = 6.

Difference of the above sums = (6 - 6) = 0.

Hence, 1452 is divisible by 11.

(ii) The given number is 9416.

Sum of its digits at odd places = (6 + 4) = 10.

Sum of its digits at even places = (1 + 9) = 10.

Difference of the above sums = (10 - 10) = 0.

Hence, 9416 is divisible by 11.

(iii) The given number is 818532.

Sum of its digits at odd places = (2 + 5 + 1) = 8.

Sum of its digits at even places = (3 + 8 + 8) = 19.

Difference of the above sums = (19 - 8) = 11, which is divisible by 11.

Hence, 818532 is divisible by 11.

(iv) The given number is 84524.

Sum of its digits at odd places = (4 + 5 + 8) = 17.

Sum of its digits at even places = (2 + 4) = 6.

Difference of the above sums = (17 - 6) = 11, which is divisible by 11.

Hence, 84524 is divisible by 11.

The number $534x2$ is divisible by 3.

⇒ Sum of its digits = $(5 + 3 + 4 + x + 2) = (14 + x)$ must be divisible by 3.

⇒ $14 + x$ = 15, 18, 21, 24, ...............

⇒ $x$ = 15 - 14, 18 - 14, 21 - 14, 24 - 14, ........

⇒ $x$ = 1, 4, 7, 10, ...............

Since $x$ is a digit, it cannot be more than 9.

Hence, the possible values of $x$ are $x = 1 \text{ or } x = 4 \text{ or } x = 7$.

Hence, all the required numbers are 53412, 53442 and 53472.

(i) The number 89*2 is divisible by 3.

⇒ Sum of its digits = (8 + 9 + * + 2) = (19 + *) must be divisible by 3.

⇒ 19 + * = 21, 24, 27, ...............

⇒ * = 21 - 19, 24 - 19, 27 - 19, ........

⇒ * = 2, 5, 8, ...............

Since * is a digit and the least possible value is required, we take ∗ = 2.

Hence, * is to be replaced by 2.

(ii) The number 89*2 is divisible by 9.

⇒ Sum of its digits = (8 + 9 + * + 2) = (19 + *) must be divisible by 9.

⇒ 19 + * = 27, 36, ...............

⇒ * = 27 - 19, 36 - 19, ........

⇒ * = 8, 17, ...............

Since * is a digit and the least possible value is required, we take * = 8.

Hence, * is to be replaced by 8.

The number $417z8$ is divisible by 9.

⇒ Sum of its digits = $(4 + 1 + 7 + z + 8) = (20 + z)$ must be divisible by 9.

⇒ $20 + z$ = 27, 36, ...............

⇒ $z$ = 27 - 20, 36 - 20, ........

⇒ $z$ = 7, 16, ...............

Since $z$ is a digit, it cannot be more than 9.

∴ $z = 7$

∴ The required number = 41778

Hence, $\bm{z = 7}$ and the required number is 41778.

A number is divisible by 3 only when the sum of its digits is divisible by 3, and it is divisible by 9 only when the sum of its digits is divisible by 9.

So, we choose numbers whose digit-sum is divisible by 3 but not by 9.

12 — Sum of digits = 1 + 2 = 3, which is divisible by 3 but not by 9.

15 — Sum of digits = 1 + 5 = 6, which is divisible by 3 but not by 9.

21 — Sum of digits = 2 + 1 = 3, which is divisible by 3 but not by 9.

24 — Sum of digits = 2 + 4 = 6, which is divisible by 3 but not by 9.

33 — Sum of digits = 3 + 3 = 6, which is divisible by 3 but not by 9.

Hence, 12, 15, 21, 24 and 33 are five numbers each of which is divisible by 3 but not divisible by 9.

A number is divisible by 4 only when the number formed by its last two digits is divisible by 4, and it is divisible by 8 only when the number formed by its last three digits is divisible by 8.

12 = 4 × 3, so 12 is divisible by 4, but 12 is not divisible by 8.

20 = 4 × 5, so 20 is divisible by 4, but 20 is not divisible by 8.

28 = 4 × 7, so 28 is divisible by 4, but 28 is not divisible by 8.

36 = 4 × 9, so 36 is divisible by 4, but 36 is not divisible by 8.

44 = 4 × 11, so 44 is divisible by 4, but 44 is not divisible by 8.

Hence, 12, 20, 28, 36 and 44 are five numbers each of which is divisible by 4 but not divisible by 8.

Given,

$$\begin{matrix} & \text{A} & \text{A} \\ + & \text{B} & 6 \\ \hline \text{C} & 2 & 5 \\ \hline \end{matrix}$$

Firstly, we will find the value of A.

Clearly, A + 6 is a number whose units digit is 5.

⇒ A + 6 = 5, A + 6 = 15; and so on.

⇒ A = 5 - 6, A = 15 - 6; and so on.

⇒ A = -1, A = 9; and so on.

Since A is a digit.

∴ A = 9 and 1 is carried over.

Secondly, we will find the values of B and C.

1 + A + B = 1 + 9 + B = 10 + B, which is a number whose units digit is 2.

⇒ 10 + B = 12

⇒ B = 2 and 1 is carried over.

∴ C = 1

$$\begin{matrix} & \overset{1}{9} & 9 \\ + & 2 & 6 \\ \hline 1 & 2 & 5 \\ \hline \end{matrix}$$

Hence, A = 9, B = 2 and C = 1.

Given,

$$\begin{matrix} & 5 & \text{C} & \text{B} & 8 \\ + & \text{D} & 9 & 7 & \text{A} \\ \hline & 8 & 2 & 4 & 3 \\ \hline \end{matrix}$$

Firstly, we will find the value of A.

Clearly, 8 + A is a number whose units digit is 3.

⇒ 8 + A = 13

⇒ A = 5 and 1 is carried over.

Secondly, we will find the value of B.

1 + B + 7 = 8 + B, which is a number whose units digit is 4.

⇒ 8 + B = 14

⇒ B = 6 and 1 is carried over.

Thirdly, we will find the value of C.

1 + C + 9 = 10 + C, which is a number whose units digit is 2.

⇒ 10 + C = 12

⇒ C = 2 and 1 is carried over.

Lastly, we will find the value of D.

1 + 5 + D = 8

⇒ D = 8 - 6 = 2

$$\begin{matrix} & \overset{1}{5} & \overset{1}{2} & \overset{1}{6} & 8 \\ + & 2 & 9 & 7 & 5 \\ \hline & 8 & 2 & 4 & 3 \\ \hline \end{matrix}$$

Hence, A = 5, B = 6, C = 2 and D = 2.

Given,

$$\begin{matrix} & \text{A} \\ & \text{A} \\ + & \text{A} \\ \hline \text{B} & \text{A} \\ \hline \end{matrix}$$

Clearly, A + A + A = 3A is a number whose units digit is A.

⇒ 3A = 10B + A

⇒ 2A = 10B

⇒ A = 5B

Since A and B are digits and B is the tens digit of a 2-digit number, so B ≠ 0.

∴ B = 1 and A = 5 × 1 = 5

Checking with A = 5 and B = 1 : 5 + 5 + 5 = 15, which is of the form BA with B = 1 and A = 5.

$$\begin{matrix} & 5 \\ & 5 \\ + & 5 \\ \hline 1 & 5 \\ \hline \end{matrix}$$

Hence, A = 5 and B = 1.

Given,

$$\begin{matrix} & 6 & \text{A} \\ - & \text{A} & \text{B} \\ \hline & 3 & 7 \\ \hline \end{matrix}$$

Firstly, we will find the value of A.

In the tens column, A is subtracted from 6 and the result is 3.

If nothing is taken away from 6, then 6 - A = 3 ⇒ A = 3, and then in the units column 3 - B = 7 gives B = -4, which is not a digit.

So, 1 is taken away from 6 on the left.

⇒ 6 - 1 - A = 3

⇒ 5 - A = 3

⇒ A = 2

Secondly, we will find the value of B.

Since 1 is taken away from 6, so in the units column,

(10 + A) - B = 7

⇒ (10 + 2) - B = 7

⇒ 12 - B = 7

⇒ B = 5

$$\begin{matrix} & 6 & 2 \\ - & 2 & 5 \\ \hline & 3 & 7 \\ \hline \end{matrix}$$

Hence, A = 2 and B = 5.

Given,

$$\begin{matrix} & \text{C} & \text{B} & 6 \\ - & 2 & 9 & \text{A} \\ \hline & 5 & 3 & 9 \\ \hline \end{matrix}$$

Firstly, we will find the value of A.

Since 6 - A cannot be 9, so 1 is taken away from B on the left.

⇒ (10 + 6) - A = 9

⇒ 16 - A = 9

⇒ A = 7

Secondly, we will find the value of B.

Since 1 is taken away from B and 9 cannot be subtracted from (B - 1), so 1 is taken away from C on the left.

⇒ (10 + B - 1) - 9 = 3

⇒ B = 3

Lastly, we will find the value of C.

Since 1 is taken away from C,

C - 1 - 2 = 5

⇒ C - 3 = 5

⇒ C = 8

$$\begin{matrix} & 8 & 3 & 6 \\ - & 2 & 9 & 7 \\ \hline & 5 & 3 & 9 \\ \hline \end{matrix}$$

Hence, A = 7, B = 3 and C = 8.

Given,

$$\begin{matrix} & \text{A} & \text{B} \\ \times & & 3 \\ \hline \text{C} & \text{A} & \text{B} \\ \hline \end{matrix}$$

Here, AB = 10A + B and CAB = 100C + 10A + B.

∴ (10A + B) × 3 = 100C + 10A + B

⇒ 30A + 3B = 100C + 10A + B

⇒ 20A + 2B = 100C

⇒ 10A + B = 50C

⇒ AB = 50C

Since AB is a 2-digit number, so 50C must be a 2-digit number.

∴ C = 1 and AB = 50 × 1 = 50

⇒ A = 5 and B = 0

Checking with A = 5, B = 0 and C = 1 : 50 × 3 = 150, which is of the form CAB.

$$\begin{matrix} & 5 & 0 \\ \times & & 3 \\ \hline 1 & 5 & 0 \\ \hline \end{matrix}$$

Hence, A = 5, B = 0 and C = 1.

Given,

$$\begin{matrix} & \text{A} & \text{B} \\ \times & \text{B} & \text{A} \\ \hline (\text{B}+1) & \text{C} & \text{B} \\ \hline \end{matrix}$$

In the units column, A × B is a number whose units digit is B.

This happens when A = 1, or when A = 6 and B is even, or when B = 5 and A is odd. (B cannot be 0, as BA would then not be a 2-digit number.)

If A = 6 and B is even, AB × BA is a 4-digit number (the smallest such product is 62 × 26 = 1612).

If B = 5 and A is odd (A ≠ 1), AB × BA is also a 4-digit number (the smallest such product is 35 × 53 = 1855).

But the given product is a 3-digit number.

∴ A = 1

Thus, we have,

$$\begin{matrix} & 1 & \text{B} \\ \times & \text{B} & 1 \\ \hline & 1 & \text{B} \\ \text{B} & \text{B}^2 & \times \\ \hline \text{B} & (1+\text{B}^2) & \text{B} \\ \hline \end{matrix}$$

Now, the hundreds digit of the product is (B + 1), i.e. it is 1 more than B.

So, the tens column must give a carry over of 1.

⇒ (1 + B²) is a 2-digit number whose tens digit is 1.

⇒ 10 ≤ 1 + B² ≤ 19

⇒ 9 ≤ B² ≤ 18

⇒ B = 3 or B = 4

When B = 3 : 1 + B² = 1 + 9 = 10. So, C = 0 and 1 is carried over, giving hundreds digit = 3 + 1 = 4 = B + 1.

Here, 13 × 31 = 403, which is of the form (B + 1)CB.

When B = 4 : 1 + B² = 1 + 16 = 17. So, C = 7 and 1 is carried over, giving hundreds digit = 4 + 1 = 5 = B + 1.

Here, 14 × 41 = 574, which is of the form (B + 1)CB.

Hence, A = 1, B = 3, C = 0 or A = 1, B = 4, C = 7.

Given,

$$\begin{array}{l} 6\overline{\smash{\big)} 5\text{AB}\smash{\big(}}\ 9\text{C} \\ \phantom{}\underline{\ -54} \\ \phantom{+()}3\text{B} \\ \phantom{+}\underline{-36} \\ \phantom{++}\times \\ \end{array}$$

Here, the divisor is 6, the dividend is 5AB and the quotient is 9C.

Firstly, we will find the value of A.

We have, 6 × 9 = 54 and this 54 is subtracted from 5A leaving 3.

⇒ 5A - 54 = 3

⇒ 5A = 57

⇒ A = 7

Secondly, we will find the values of B and C.

B is brought down and 3B is divided by 6, and 36 is subtracted from 3B leaving no remainder.

⇒ 3B = 36

⇒ B = 6

Also, 6 × C = 36

⇒ C = 6

Checking with A = 7, B = 6 and C = 6 : 576 ÷ 6 = 96, which is of the form 9C.

$$\begin{array}{l} 6\overline{\smash{\big)} 576\smash{\big(}}\ 96 \\ \phantom{}\underline{\ -54} \\ \phantom{+()}36 \\ \phantom{()}\underline{-36} \\ \phantom{++}\times \\ \end{array}$$

Hence, A = 7, B = 6 and C = 6.

The smallest 4-digit number is 1000 and the largest 4-digit number is 9999.

∴ Number of 4-digit numbers = (9999 - 1000) + 1 = 8999 + 1 = 9000

The smallest 3-digit number is 100 and the largest 3-digit number is 999.

∴ Number of 3-digit numbers = (999 - 100) + 1 = 899 + 1 = 900

∴ Required quotient = $\frac{9000}{900}$ = 10

Hence, the required quotient is 10.

Let the hundreds digit be $a$, the tens digit be $b$ and the units digit be $c$.

∴ The 3-digit number = $100a + 10b + c$

Number obtained by reversing its digits = $100c + 10b + a$

$$\begin{aligned}\Rightarrow \text{Difference} = (100a + 10b + c) - (100c + 10b + a) \\[1em] = 100a + 10b + c - 100c - 10b - a \\[1em] = 99a - 99c \\[1em] = 99(a - c)\end{aligned}$$

Since $(a - c)$ is an integer, so the difference is always a multiple of 99.

Also, $(a - c)$ can be equal to 1 (for example, 211 - 112 = 99), so no natural number greater than 99 can always divide this difference.

Hence, the required largest natural number is 99.

We go on dividing a string of 3's by 7 till 0 is obtained as the remainder.

3 ÷ 7 → remainder = 3

33 ÷ 7 → quotient = 4, remainder = 5

333 ÷ 7 → quotient = 47, remainder = 4

3333 ÷ 7 → quotient = 476, remainder = 1

33333 ÷ 7 → quotient = 4761, remainder = 6

333333 ÷ 7 → quotient = 47619, remainder = 0

∴ The required number = 47619

Checking : 47619 × 7 = 333333, each of whose digits is 3.

Hence, the smallest such number is 47619.

Given, E = 4.

Given, the difference between E and I is 5.

⇒ I = 4 + 5 = 9 or I = 4 - 5 = -1

Since the values assigned are from 1 to 9, so I ≠ -1.

∴ I = 9

Given, the difference between C and I is 3.

⇒ C = 9 + 3 = 12 or C = 9 - 3 = 6

Since the values assigned are from 1 to 9, so C ≠ 12.

∴ C = 6

Hence, the value assigned to C is 6.

Given,

$$\begin{array}{l} \text{AB}\overline{\smash{\big)}252\smash{\big(}}\ \text{BA} \\ \phantom{+}\underline{ \\ -24} \\ \phantom{++(}12 \\ \phantom{+()}\underline{ -12} \\ \phantom{++()}\times \\ \end{array}$$

Here, the divisor is AB, the dividend is 252 and the quotient is BA.

From the division sum, AB × B = 24 and AB × A = 12.

$$\begin{aligned}\Rightarrow \frac{\text{AB} \times \text{B}}{\text{AB} \times \text{A}} = \frac{24}{12} \\[1em] \Rightarrow \frac{\text{B}}{\text{A}} = \frac{2}{1}\end{aligned}$$

⇒ B = 2A

Now, AB = 10A + B = 10A + 2A = 12A

Also, AB × A = 12

⇒ 12A × A = 12

⇒ 12A² = 12

⇒ A² = 1

⇒ A = 1

∴ B = 2 × 1 = 2

Checking with A = 1 and B = 2 : AB = 12, BA = 21 and 12 × 21 = 252.

$$\begin{array}{l} 12\overline{\smash{\big)}252\smash{\big(}}\ 21 \\ \phantom{(}\underline{\ -24} \\ \phantom{++}12 \\ \phantom{+(}\underline{ -12} \\ \phantom{++}\times \\ \end{array}$$

Hence, A = 1 and B = 2.

Given,

$$\begin{matrix} & 5 & \text{X} & 9 \\ & 3 & 2 & 7 \\ + & 2 & \text{Y} & 8 \\ \hline 1 & 1 & 1 & 4 \\ \hline \end{matrix}$$

In the units column, 9 + 7 + 8 = 24. So, 4 is written and 2 is carried over.

In the tens column, 2 (carry over) + X + 2 + Y = X + Y + 4, which is a number whose units digit is 1.

⇒ X + Y + 4 = 11

⇒ X + Y = 7

So, 1 is carried over to the hundreds column.

(In the hundreds column, 1 + 5 + 3 + 2 = 11, which gives the digits 1 and 1 of the sum, as required.)

Now, X + Y = 7, where X and Y are digits.

∴ Y is maximum when X is minimum, i.e. when X = 0.

⇒ Y = 7 - 0 = 7

Hence, the maximum possible value of Y is 7.

Given,

$$\begin{matrix} & 5 & \text{P} & 7 \\ & 8 & \text{Q} & 9 \\ + & \text{R} & 3 & 2 \\ \hline 1 & 9 & 2 & 8 \\ \hline \end{matrix}$$

In the units column, 7 + 9 + 2 = 18. So, 8 is written and 1 is carried over.

In the tens column, 1 (carry over) + P + Q + 3 = P + Q + 4, which is a number whose units digit is 2.

⇒ P + Q + 4 = 12 or P + Q + 4 = 22

⇒ P + Q = 8 or P + Q = 18

Since P and Q are digits, P + Q = 18 is possible only when P = Q = 9. But P and Q are different letters, so they stand for different digits.

∴ P + Q = 8 and 1 is carried over.

Q is maximum when P is minimum, i.e. when P = 0.

⇒ Q = 8 - 0 = 8

In the hundreds column, 1 (carry over) + 5 + 8 + R = 14 + R, which is a number whose units digit is 9.

⇒ 14 + R = 19

⇒ R = 5 and 1 is carried over, which gives the thousands digit 1 of the sum, as required.

Checking with P = 0, Q = 8 and R = 5 : 507 + 889 + 532 = 1928, as required.

$$\begin{matrix} & \overset{1}{5} & \overset{1}{0} & 7 \\ & 8 & 8 & 9 \\ + & 5 & 3 & 2 \\ \hline 1 & 9 & 2 & 8 \\ \hline \end{matrix}$$

Hence, the maximum value of Q is 8.

Let us take the two numbers as 1 and 9.

Product of the numbers = 1 × 9 = 9, which is a 1-digit number.

Sum of the numbers = 1 + 9 = 10, which is a 2-digit number.

Hence, the required two numbers are 1 and 9.

Let us take the three natural numbers as 1, 2 and 3.

Sum of the numbers = 1 + 2 + 3 = 6

Product of the numbers = 1 × 2 × 3 = 6

∴ Sum of the numbers = Product of the numbers

Hence, the required three natural numbers are 1, 2 and 3.

According to the divisibility test of 9, a number is divisible by 9 only when the sum of its digits is divisible by 9.

⇒ 42*8 is a multiple of 9

⇒ 4 + 2 + * + 8 is divisible by 9

⇒ 14 + * is divisible by 9

⇒ 14 + * = 18, 27, ...............

⇒ * = 4, 13, ...............

Since * is a digit,

∴ * = 4.

Hence, option 4 is the correct option.

According to the divisibility test of 11, a number is divisible by 11 if the difference between the sum of its digits at odd places and the sum of its digits at even places is either 0 or a number divisible by 11.

(i) If the sum of the odd numbered digits is equal to the sum of the even numbered digits, then their difference is 0. So, the number is divisible by 11.

∴ Statement (i) is correct.

(ii) If the difference of the sums of the odd numbered digits and the even numbered digits is divisible by 11, then the number is divisible by 11.

∴ Statement (ii) is correct.

Hence, option 2 is the correct option.

The given number is 12345674.

Sum of its digits = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 4 = 32, which is divisible neither by 3 nor by 9.

∴ It is divisible neither by 3 nor by 9.

Its units digit is 4, which is neither 0 nor 5.

∴ It is not divisible by 5.

Sum of its digits at odd places = (4 + 6 + 4 + 2) = 16.

Sum of its digits at even places = (7 + 5 + 3 + 1) = 16.

Difference of the above sums = (16 - 16) = 0.

∴ 12345674 is divisible by 11.

Hence, option 4 is the correct option.

According to the divisibility test of 8, a number is divisible by 8 only when the number formed by its last three digits is divisible by 8.

Here, the number formed by the last three digits is 6*2.

When * = 1, 612 is not divisible by 8.

When * = 2, 622 is not divisible by 8.

When * = 3, 632 = 8 × 79, which is divisible by 8.

∴ The least value of * is 3.

Hence, option 3 is the correct option.

The given number is 111111111111, which has twelve digits, each equal to 1.

Sum of its digits = 1 × 12 = 12, which is divisible by 3 but not by 9.

∴ It is divisible by 3 but not by 9.

Its units digit is 1, which is neither 0 nor 5.

∴ It is not divisible by 5.

Sum of its digits at odd places = (1 + 1 + 1 + 1 + 1 + 1) = 6.

Sum of its digits at even places = (1 + 1 + 1 + 1 + 1 + 1) = 6.

Difference of the above sums = (6 - 6) = 0.

∴ It is divisible by 11.

So, the given number is divisible by 3 and 11.

Hence, option 3 is the correct option.

The given number is 97215*6.

Sum of its digits at odd places = (6 + 5 + 2 + 9) = 22.

Sum of its digits at even places = (* + 1 + 7) = 8 + *.

Difference of the above sums = 22 - (8 + *) = 14 - *, which should be either 0 or divisible by 11.

⇒ 14 - * = 11

⇒ * = 3

Hence, option 3 is the correct option.

Let the hundreds digit be $a$, the tens digit be $b$ and the units digit be $c$.

∴ The 3-digit number = $100a + 10b + c$

Given, the middle digit is equal to the sum of the other two digits.

⇒ $b = a + c$

$$\begin{aligned}\Rightarrow \text{The number} = 100a + 10(a + c) + c \\[1em] = 100a + 10a + 10c + c \\[1em] = 110a + 11c \\[1em] = 11(10a + c)\end{aligned}$$

So, the number is a multiple of 11.

Hence, option 3 is the correct option.

According to the divisibility test of 11, a number is divisible by 11 if the difference between the sum of its digits at odd places and the sum of its digits at even places is either 0 or a number divisible by 11.

1. For 112144, sum of digits at odd places = (4 + 1 + 1) = 6 and sum of digits at even places = (4 + 2 + 1) = 7. Their difference = (7 - 6) = 1, which is not divisible by 11.

2. For 447355, sum of digits at odd places = (5 + 3 + 4) = 12 and sum of digits at even places = (5 + 7 + 4) = 16. Their difference = (16 - 12) = 4, which is not divisible by 11.

3. For 869756, sum of digits at odd places = (6 + 7 + 6) = 19 and sum of digits at even places = (5 + 9 + 8) = 22. Their difference = (22 - 19) = 3, which is not divisible by 11.

4. For 978626, sum of digits at odd places = (6 + 6 + 7) = 19 and sum of digits at even places = (2 + 8 + 9) = 19. Their difference = (19 - 19) = 0.

∴ 978626 is a multiple of 11.

Hence, option 4 is the correct option.

According to the divisibility test of 9, a number is divisible by 9 only when the sum of its digits is divisible by 9.

∵ $6135n2$ is divisible by 9

⇒ $6 + 1 + 3 + 5 + n + 2$ is divisible by 9

⇒ $17 + n$ is divisible by 9

⇒ $17 + n$ = 18, 27, ...............

⇒ $n$ = 18 - 17, 27 - 17, ........

⇒ $n$ = 1, 10, ...............

Since $n$ is a digit, ∴ $n = 1$.

Hence, option 1 is the correct option.

According to the divisibility test of 4, a number is divisible by 4 only when the number formed by its last two digits is divisible by 4.

Here, the number formed by the last two digits is 8*.

When * = 2, 82 is not divisible by 4.

When * = 3, 83 is not divisible by 4.

When * = 4, 84 = 4 × 21, which is divisible by 4.

When * = 6, 86 is not divisible by 4.

∴ * = 4

Hence, option 3 is the correct option.

Sum of digits of 24984 = 2 + 4 + 9 + 8 + 4 = 27, which is divisible by both 3 and 9.

Sum of digits of 26784 = 2 + 6 + 7 + 8 + 4 = 27, which is divisible by both 3 and 9.

Sum of digits of 28584 = 2 + 8 + 5 + 8 + 4 = 27, which is divisible by both 3 and 9.

∴ All the three numbers are divisible by 3 and by 9.

Also, the number formed by the last two digits of each of them is 84 = 4 × 21, which is divisible by 4.

∴ All the three numbers are divisible by 4.

So, the statements (i), (ii) and (iii) are all correct.

Hence, option 4 is the correct option.

A number is divisible by 3 only when the sum of its digits is divisible by 3, and it is divisible by 9 only when the sum of its digits is divisible by 9.

2133 — Sum of digits = 2 + 1 + 3 + 3 = 9, which is divisible by both 3 and 9.

2343 — Sum of digits = 2 + 3 + 4 + 3 = 12, which is divisible by 3 but not by 9.

3474 — Sum of digits = 3 + 4 + 7 + 4 = 18, which is divisible by both 3 and 9.

4131 — Sum of digits = 4 + 1 + 3 + 1 = 9, which is divisible by both 3 and 9.

5286 — Sum of digits = 5 + 2 + 8 + 6 = 21, which is divisible by 3 but not by 9.

5340 — Sum of digits = 5 + 3 + 4 + 0 = 12, which is divisible by 3 but not by 9.

6336 — Sum of digits = 6 + 3 + 3 + 6 = 18, which is divisible by both 3 and 9.

7347 — Sum of digits = 7 + 3 + 4 + 7 = 21, which is divisible by 3 but not by 9.

8115 — Sum of digits = 8 + 1 + 1 + 5 = 15, which is divisible by 3 but not by 9.

9276 — Sum of digits = 9 + 2 + 7 + 6 = 24, which is divisible by 3 but not by 9.

∴ The numbers divisible by 3 but not by 9 are 2343, 5286, 5340, 7347, 8115 and 9276, which are 6 in number.

Hence, option 2 is the correct option.

In the number $357\star25\star$, the first $\star$ is at the thousands place and the second $\star$ is at the units place.

Let the digit at the thousands place be $t$ and the digit at the units place be $u$.

According to the divisibility test of 5, the number is divisible by 5 only when $u = 0 \text{ or } u = 5$.

According to the divisibility test of 3, the sum of the digits must be divisible by 3.

Sum of the digits = $3 + 5 + 7 + t + 2 + 5 + u = 22 + t + u$

Now, we check the given options :

1. $u = 0, t = 2$ : Sum = 22 + 2 + 0 = 24, which is divisible by 3.

2. $u = 0, t = 6$ : Sum = 22 + 6 + 0 = 28, which is not divisible by 3.

3. $u = 5, t = 1$ : Sum = 22 + 1 + 5 = 28, which is not divisible by 3.

4. $u = 5, t = 4$ : Sum = 22 + 4 + 5 = 31, which is not divisible by 3.

∴ The missing digits in the units place and the thousands place are 0 and 2 respectively.

Hence, option 1 is the correct option.

In the given expression, the terms are alternately 3 and -3, and the first term is 3.

Since 101 is an odd number, the number of terms equal to 3 is 51 and the number of terms equal to -3 is 50.

∴ Sum = 51 × 3 + 50 × (-3)

= 153 - 150

= 3

Hence, option 3 is the correct option.

Since $m, n, p$ are the digits of the number beginning from the left, so $m$ is the hundreds digit, $n$ is the tens digit and $p$ is the units digit.

∴ The number = $100 \times m + 10 \times n + p = 100m + 10n + p$

Hence, option 4 is the correct option.

The sum of two odd numbers is always an even number.

Since $x$ and $y$ are both odd, so $(x + y)$ is an even number.

For example, if $x = 3 \text{ and } y = 5$, then $x + y = 8$, which is even.

(Also, $x + y + 1 = 9, xy + 2 = 17 \text{ and } xy = 15$ are all odd numbers.)

Hence, option 1 is the correct option.

Given, $(p - 1)$ is an odd number.

Consecutive odd numbers differ by 2.

∴ The odd number just before $(p - 1)$ is $(p - 1) - 2 = p - 3$.

And, the odd number just after $(p - 1)$ is $(p - 1) + 2 = p + 1$.

∴ The two other odd numbers nearest to $(p - 1)$ are $(p - 3) \text{ and } (p + 1)$.

Hence, option 3 is the correct option.

Given,

$$\begin{matrix} & 8 & 3 & 5 \\ & 4 & * & 8 \\ + & 9 & * & 4 \\ \hline 2 & 2 & * & 7 \\ \hline \end{matrix}$$

Let the missing number be *.

In the units column, 5 + 8 + 4 = 17. So, 7 is written and 1 is carried over.

In the tens column, 1 (carry over) + 3 + * + * = 4 + 2*, which is a number whose units digit is *.

⇒ 4 + 2* = 10 + *

⇒ * = 6

So, 4 + 2 × 6 = 16, i.e. 6 is written and 1 is carried over.

In the hundreds column, 1 (carry over) + 8 + 4 + 9 = 22, which gives the digits 2 and 2 of the sum, as required.

Checking with * = 6 : 835 + 468 + 964 = 2267.

Hence, option 3 is the correct option.

Given,

$$\begin{matrix} & 3 & \text{M} & 4 \\ \times & & & 4 \\ \hline 1 & 2 & 1 & 6 \\ \hline \end{matrix}$$

Here, 3M4 × 4 = 1216

⇒ 3M4 = $\frac{1216}{4}$ = 304

∴ M = 0

Checking with M = 0 : 304 × 4 = 1216, as required.

Hence, option 1 is the correct option.

(i) Sum of digits of $84y6$ = $8 + 4 + y + 6 = 18 + y$, which is divisible by 3 when $y$ = 0, 3, 6 or 9.

If $84y6$ is exactly divisible by 3, then the least value of $y$ is 0.

(ii) Sum of digits of $73m4$ = $7 + 3 + m + 4 = 14 + m$, which is divisible by 9 when $14 + m = 18$, i.e. $m = 4$.

If $73m4$ is exactly divisible by 9, then the least value of $m$ is 4.

(iii) Sum of digits of $m5n7$ = $m + 5 + n + 7 = 12 + (m + n)$, which is divisible by 3 when $(m + n)$ = 0, 3, 6, ...............

Since $m$ is the thousands digit of a 4-digit number, so $m \neq 0$ and hence $(m + n) \neq 0$.

If $m5n7$ is exactly divisible by 3, then the least value of $(m + n)$ is 3.

(iv) Sum of digits of $x4y5z$ = $x + 4 + y + 5 + z = 9 + (x + y + z)$, which is divisible by 9 when $(x + y + z)$ = 0, 9, 18, ...............

Since $x$ is the ten thousands digit of a 5-digit number, so $x \neq 0$ and hence $(x + y + z) \neq 0$.

If $x4y5z$ is exactly divisible by 9, then the least value of $(x + y + z)$ is 9.

(v) The number formed by the last three digits of $7912p$ is $12p$, which is divisible by 8 when $12p = 120 \text{ or } 128$, i.e. $p = 0 \text{ or } 8$.

If $7912p$ is exactly divisible by 8, then the least value of $p$ is 0.

According to the divisibility test of 4, a number is divisible by 4 only when the number formed by its last two digits is divisible by 4.

So, a 5-digit number $abcde$ is divisible by 4 if $de$ is divisible by 4.

∴ Reason (R) is true.

For the number 82592, the number formed by its last two digits = 92 = 4 × 23, which is divisible by 4.

∴ 82592 is divisible by 4, i.e. Assertion (A) is true.

Also, Reason (R) is exactly the rule used in Assertion (A).

∴ Reason (R) is the correct explanation of Assertion (A).

Hence, option 1 is the correct option.

On dividing $7N + 1$ by 7, we get quotient = $N$ and remainder = 1, since $7N$ is a multiple of 7.

∴ Assertion (A) is true.

Since 2 and 7 are co-prime numbers and 2 × 7 = 14, a number divisible by both 2 and 7 is also divisible by 14.

∴ Reason (R) is true.

But Reason (R) talks about divisibility by 14, while Assertion (A) is about the remainder left on dividing by 7.

∴ Reason (R) is not the correct explanation of Assertion (A).

Hence, option 2 is the correct option.

The numbers 472, 724 and 247 are obtained from one another by changing the order of the digits 4, 7 and 2 cyclically.

∴ Sum = 472 + 724 + 247 = 1443

Now, 1443 ÷ 111 = 13

∴ Quotient = 13

Here, 13 is an odd number and it is also a prime number, since its only factors are 1 and 13.

Hence, option 4 is the correct option.

We have,

333 is a 3-digit number.

33³ = 33 × 33 × 33 = 35937

3³³ = 3²⁷

Now, 3²⁷ > 3¹⁰ = 59049 > 35937 > 333

∴ 3²⁷ is greater than 333 and 33³.

Also, 3³³ and 3²⁷ have the same base 3 and 33 > 27.

∴ 3³³ > 3²⁷

So, 3³³ is the greatest.

Hence, option 2 is the correct option.

Let the number be $N$ and the quotient be $k$.

Given, on dividing $N$ by 91 the remainder is 43.

∴ $N = 91k + 43$

Now, 91 = 13 × 7 and 43 = 13 × 3 + 4

$$\begin{aligned}\Rightarrow N = 13 \times 7 \times k + 13 \times 3 + 4 \\[1em] = 13(7k + 3) + 4\end{aligned}$$

So, on dividing $N$ by 13, the remainder is 4.

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{5}\right) \dots \left(1 - \frac{1}{25}\right) \\[1em] \Rightarrow \frac{2}{\cancel{3}} \times \frac{\cancel{3}}{\cancel{4}} \times \frac{\cancel{4}}{\cancel{5}} \times \dots \times \frac{\cancel{24}}{25} \\[1em] \Rightarrow \frac{2}{25}\end{aligned}$$

Hence, option 2 is the correct option.

The units digit of (124)¹²⁴ is the same as the units digit of 4¹²⁴.

The units digits of the powers of 4 are 4, 6, 4, 6, ............... i.e. the units digit is 4 when the power is odd and 6 when the power is even.

Since 124 is an even number, the units digit of (124)¹²⁴ is 6.

The units digit of (125)¹²⁵ is the same as the units digit of 5¹²⁵, and every power of 5 ends in 5.

∴ The units digit of (125)¹²⁵ is 5.

∴ Units digit of the sum = units digit of (6 + 5) = units digit of 11 = 1

Hence, option 1 is the correct option.

Every number from 100 to 500 is a 3-digit number.

Let a 3-digit palindrome be $aba$, where $a$ is the hundreds digit as well as the units digit and $b$ is the tens digit.

Since the number lies between 100 and 500, so $a$ = 1, 2, 3 or 4, i.e. $a$ can take 4 values.

(If $a = 5$, then the number $5b5$ is at least 505, which is more than 500.)

Also, $b$ can be any one of the digits 0, 1, 2, ..., 9, i.e. $b$ can take 10 values.

∴ Total number of palindromes = 4 × 10 = 40

Hence, option 1 is the correct option.