# Playing with Numbers - Class 8 - Mathematics

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## Exercise 5.1 - Playing with Numbers | R.D. Sharma | Mathematics | Class 8

Without performing actual addition and division write the quotient when the sum of 69 and 96 is divided by

(i) 11

(ii) 15

We know that:

The sum of any two digit number $\overline{a b}$ and the number $\overline{b a}$ by reversing its digits is completely divisible by

(i) the sum $a+b$ of its digits and the quotient is 11 .

(ii) 11 and the quotient is $a+b$ i.e. the sum of its digits.

Let's apply the provided rule to the numbers 69 and 96:

**Identify the digits $a$ and $b$:**For 69, $a = 6$ and $b = 9$.

When reversed to get 96, the digits are essentially interchanged but remain $a = 6$ and $b = 9$.

**Rule Application:**According to the rule, the sum of a two-digit number $\overline{ab}$ (69 here) and its reverse $\overline{ba}$ (96 here) is divisible:

(i) by the sum of its digits (a + b) (which is $6 + 9 = 15$) with a quotient of 11.

(ii) by 11 with a quotient of the sum of its digits (a + b) (which is $6 + 9 = 15$).

Applying these:

### Division by 11:

The quotient when the sum of 69 and 96 is divided by 11 is (a + b = 15).

### Division by 15:

The quotient when the sum of 69 and 96 is divided by 15 is 11.

Thus:

The quotient when the sum of 69 and 96 is divided by 11 is 15.

The quotient when the sum of 69 and 96 is divided by 15 is 11.

Without performing actual computations, find the quotient when $94-49$ is divided by

(i) 9

(ii) 5

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Sign up nowIf sum of the number 985 and two other numbers obtained by arranging the digits of 985 in cyclic order is divided by 111, 22 and 37 respectively. Find the quotient in each case.

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Sign up nowFind the quotient when the difference of 985 and 958 is divided by 9.

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Sign up now## Exercise 5.2 - Playing with Numbers | R.D. Sharma | Mathematics | Class 8

Given that the number $\overline{35 a 64}$ is divisible by 3 , where $a$ is a digit, what are the possible values of $a$ ?

To find the values for $a$ such that the number $\overline{35a64}$ is divisible by 3, we need to use the rule of divisibility for 3. According to this rule, a number is divisible by 3 if the sum of its digits is divisible by 3.

Let's calculate the sum of the known digits of the number $\overline{35a64}$ first: $$ 3 + 5 + 6 + 4 = 18 $$

Now we add the unknown digit $a$ to this sum: $$ 18 + a $$

This sum needs to be divisible by 3. Let's set up the divisibility condition: $$ (18 + a) \mod 3 = 0 $$

We can simplify the expression $18 + a$ modulo 3: $$ 18 \mod 3 = 0 \quad \text{(since 18 is divisible by 3)} $$

Therefore, the condition reduces to: $$ a \mod 3 = 0 $$

This means that $a$ itself must be a digit divisible by 3. The digits from 0 to 9 that are divisible by 3 are 0, 3, 6, and 9.

Hence, the possible values of $a$ are **0, 3, 6, and 9**.

If $x$ is a digit such that the number $\overline{18 x 71}$ is divisible by 3 , find possible values of $x$.

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If $x$ is a digit of the number $\overline{66784 x}$ such that it is divisible by 9 , find possible values of $x$.

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If $\overline{3 x}$ is a multiple of 11 , where $x$ is a digit, what is the value of $x$ ?

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If $\overline{98215 x 2}$ is a number with $x$ as its tens digit such that it is divisible by 4 . Find all possible values of $x$.

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If $x$ denotes the digit at hundreds place of the number $\overline{67 x 19}$ such that the number is divisible by 11. Find all possible values of $x$.

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Find the remainder when 981547 is divided by 5 . Do this without doing actual division.

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Find the remainder when 51439786 is divided by 3 . Do this without performing actual division.

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Without performing actual division, find the remainder when 928174653 is divided by 11 .

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Given an example of a number which is divisible by

(i) 2 but not by 4 .

(iii) 4 but not by 8 .

(ii) 3 but not by 6 .

(iv) both 4 and 8 but not by 32 .

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Which of the following statements are true?

(i) If a number is divisible by 3 , it must be divisible by 9 .

(ii) If a number is divisible by 9 , it must be divisible by 3 .

(iii) If a number is divisible by 4 , it must be divisible by 8 .

(iv) If a number is divisible by 8 , it must be divisible by 4 .

(v) A number is divisible by 18 , if it is divisible by both 3 and 6 .

(vi) If a number is divisible by both 9 and 10 , it must be divisible by 90 .

(vii) If a number exactly divides the sum of two numbers, it must exactly divide the numbers separately.

(viii) If a number divides three numbers exactly, it must divide their sum exactly.

(ix) If two numbers are co-prime, at least one of them must be a prime number.

(x) The sum of two consecutive odd numbers is always divisible by 4 .

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## Extra Questions - Playing with Numbers | R.D. Sharma | Mathematics | Class 8

Samir bought a box containing ball pens and gel pens. The number of ball pens was six, less than gel pens, and there were a total of 24 pens. How many gel pens did Samir buy?

Let's denote the number of ball pens as $x$ and the number of gel pens as $y$. According to the information provided, Samir bought a total of 24 pens. This can be expressed in equation form as:

$$ x + y = 24 $$

Additionally, it is stated that the number of ball pens was **6 less than** the number of gel pens, which can be translated into the equation:

$$ x = y - 6 $$

To find the values of $x$ and $y$, we can substitute the expression from the second equation into the first equation:

Replace $x$ with $(y - 6)$:

$$ (y - 6) + y = 24 $$

Simplify the equation:

$$ 2y - 6 = 24 $$

Add 6 to both sides to solve for $y$:

$$ 2y = 24 + 6 = 30 $$

Divide both sides by 2:

$$ y = \frac{30}{2} = 15 $$

So, the **number of gel pens (y)** Samir bought is **15**.

Now, to find the number of ball pens:

$$ x = y - 6 = 15 - 6 = 9 $$

Thus, the **number of ball pens (x)** Samir bought is **9**.

The statement "Product of 38 and 72 is a multiple of 4." can be proved with the help of which of the following statements?

A. Product of two odd natural numbers is always odd.

B. Product of two even natural numbers is always divisible by 4.

C. Product of an even and an odd natural number is always divisible by 2.

D. Product of two even natural numbers is always divisible by 2.