Playing with Number Ex 16.2 Class 8 NCERT

Understanding the Logic Behind Divisibility Rules

In playing with number ex 16 2 class 8 ncert, we learn why the divisibility rules for 3 and 9 actually work. Let's express a three-digit number $abc$ in its generalized form: $100a + 10b + c$. We can rewrite this algebraically as $(99a + a) + (9b + b) + c$, which simplifies to $99a + 9b + (a + b + c)$. Since $99a$ and $9b$ are always multiples of both 3 and 9, the divisibility of the entire number depends entirely on the remaining part: $(a + b + c)$. This is the sum of the digits! Thus, a number is divisible by 3 or 9 if and only if the sum of its digits is divisible by 3 or 9 respectively.

How to Find the Missing Digit

1. Sum the Digits — Add all the given digits together with the variable. For example, in the number $21y5$, the sum is $2 + 1 + y + 5 = 8 + y$.
2. Identify Multiples — List the multiples of the divisor (3 or 9) that are greater than or equal to your basic sum. For $8 + y$ checking divisibility by 9, the potential multiples are 9, 18, 27...
3. Set Up Equations — Create simple linear equations: $8 + y = 9$ or $8 + y = 18$.
4. Filter Valid Single Digits — Solve for the variable. Since the variable represents a single digit in a number, its value must be a whole number between 0 and 9 inclusive. Reject any value that is negative or greater than 9.

Avoid These Common Mistakes in Ex 16.2

1. The Single Digit Rule: Remember, the unknown letter (like $x, y, z$) is a digit in a base-10 number. It can only be a single integer from $0$ to $9$. If your equation gives $y = 12$, that is incorrect for a single-digit placeholder.

2. Multiple Solutions for Divisibility by 3: Since 3 has more frequent multiples, equations involving divisibility by 3 often yield multiple valid answers (e.g., $x = 0, 3, 6, 9$). Write down all valid single-digit solutions in your exam to score full marks!

Practice Questions with Solutions

• Q: If $21y5$ is a multiple of 9, where $y$ is a digit, what is the value of $y$? A: Step 1: Find the sum of the digits: $2 + 1 + y + 5 = 8 + y$. Step 2: Since it is a multiple of 9, the sum $(8 + y)$ must be divisible by 9. Step 3: The nearest multiple of 9 equal to or greater than 8 is 9. Step 4: Set up the equation: $8 + y = 9 \implies y = 9 - 8 = 1$. Step 5: If we try the next multiple of 9, which is 18: $8 + y = 18 \implies y = 10$. Since $y$ must be a single digit, 10 is not possible. Final answer: $y = 1$.
• Q: If $31z5$ is a multiple of 3, where $z$ is a digit, what are the possible values of $z$? A: Step 1: Find the sum of the digits: $3 + 1 + z + 5 = 9 + z$. Step 2: Since it is a multiple of 3, $(9 + z)$ must be divisible by 3. Step 3: Set $(9 + z)$ equal to multiples of 3 that are $\ge 9$: - $9 + z = 9 \implies z = 0$ - $9 + z = 12 \implies z = 3$ - $9 + z = 15 \implies z = 6$ - $9 + z = 18 \implies z = 9$ - $9 + z = 21 \implies z = 12$ (not a single digit, so reject). Final answer: $z$ can be 0, 3, 6, or 9.
• Q: If $24x$ is a multiple of 3, where $x$ is a digit, find all possible single-digit values of $x$. A: Step 1: Add the digits together: $2 + 4 + x = 6 + x$. Step 2: For $24x$ to be a multiple of 3, $(6 + x)$ must be a multiple of 3. Step 3: Solve for possible single-digit values of $x$ (from 0 to 9): - $6 + x = 6 \implies x = 0$ - $6 + x = 9 \implies x = 3$ - $6 + x = 12 \implies x = 6$ - $6 + x = 15 \implies x = 9$ Final answer: The possible values of $x$ are 0, 3, 6, and 9.
• Q: If $31y2$ is a multiple of 9, where $y$ is a digit, find the value of $y$. A: Step 1: Find the sum of the digits: $3 + 1 + y + 2 = 6 + y$. Step 2: Since $31y2$ is a multiple of 9, $(6 + y)$ must be a multiple of 9. Step 3: Set up the equation with the nearest multiple of 9: - $6 + y = 9 \implies y = 9 - 6 = 3$. - If we check the next multiple 18: $6 + y = 18 \implies y = 12$ (invalid since it must be a single digit). Final answer: $y = 3$.

Frequently Asked Questions