Playing With Numbers Class 8 Notes

Core Glossary & Terms

Generalized Form

Writing a number as the sum of the products of its digits with their respective place values. For example, $ab = 10a + b$.

Cryptarithms

Number puzzles where letters are used instead of digits. The goal is to find which letter corresponds to which digit.

Divisibility

The property of an integer being completely divisible by another integer without leaving a remainder.

Units Digit

The rightmost digit of a whole number, representing the ones place.

Tens Digit

The second digit from the right in a whole number, representing the tens place.

Generalized Form of Numbers Explained

In arithmetic, we write numbers using place value notation. For example, 52 is written as $50 + 2 = 10 \times 5 + 2$. Generally, a two-digit number having tens digit $a$ and units digit $b$ is represented in its generalized form as $10a + b$. Similarly, a three-digit number $abc$ is written as $100a + 10b + c$. Understanding this structure is essential for solving number games and reversing digit problems. For instance, if you reverse the digits of $ab$ to get $ba$, its generalized form becomes $10b + a$. Adding these two gives $11(a+b)$, which is always a multiple of 11. This simple algebraic structure explains many fascinating mathematical puzzles and divisibility rules.

Tests of Divisibility: Step-by-Step Rules

1. Divisibility by 10 — Check the units digit of the number. If the units digit is 0, then the number is divisible by 10. Otherwise, it is not.
2. Divisibility by 5 — Observe the units digit of the number. If the units digit is either 0 or 5, the entire number is perfectly divisible by 5.
3. Divisibility by 2 — Check the units digit of the given number. If the units digit is even (0, 2, 4, 6, or 8), the number is divisible by 2.
4. Divisibility by 9 and 3 — Find the sum of all the digits of the number. If this sum is divisible by 9, the number is divisible by 9. If the sum is divisible by 3, the number is divisible by 3.

Solved Mini-Examples for Revision

• {"title":"Example 1: Cryptarithm Puzzle","content":"Find the values of $A$ and $B$ in the addition problem: $3A + 25 = B2$.\n\nSolution:\nLooking at the units column, we have $A + 5 = 2$ or a number ending in 2 (which must be 12). \nSo, $A + 5 = 12 \\Rightarrow A = 7$.\nWith a carry-over of 1 to the tens column, we get:\n$1 + 3 + 2 = B \\Rightarrow B = 6$.\nTherefore, $A = 7$ and $B = 6$."}
• {"title":"Example 2: Divisibility by 9","content":"Find the value of $y$ for which the 4-digit number $24y3$ is divisible by 9.\n\nSolution:\nFor a number to be divisible by 9, the sum of its digits must be divisible by 9.\nSum of digits $= 2 + 4 + y + 3 = 9 + y$.\nSince $y$ is a single digit (from 0 to 9), the sum $9 + y$ can be 9 or 18.\nIf $9 + y = 9$, then $y = 0$.\nIf $9 + y = 18$, then $y = 9$.\nHence, $y$ can be either 0 or 9."}

Must Remember Key Points

• A two-digit number $ab$ is algebraically represented as $10a + b$.
• A three-digit number $abc$ is algebraically represented as $100a + 10b + c$.
• The digits $a, b, c$ can only take single-digit integer values from 0 to 9.
• In any cryptarithm letter puzzle, the leftmost digit of a number cannot be 0 (e.g., $ab$ means $a \neq 0$).
• Each letter in a cryptarithm must stand for a unique digit.
• The difference between a 2-digit number $ab$ and its reversed form $ba$ is always a multiple of 9.
• The sum of a 2-digit number $ab$ and its reversed form $ba$ is always a multiple of 11.
• The sum of a 3-digit number $abc$ and its cyclic permutations ($bca$ and $cab$) is always divisible by 111, 37, and 3.

Pro Exam Tips for Scoring High

Avoid making the common mistake of writing a two-digit number $ab$ as $a \times b$ in algebraic equations. Always expand it to $10a + b$. When solving cryptarithms, start by finding clues in the ones column or looking at the carry-over values (which can only be 1 or 2 in single-digit addition). This narrows down your choices quickly and prevents tedious trial-and-error!

Practice Questions with Solutions

• If $31z5$ is a multiple of 3, where $z$ is a digit, what are the possible values of $z$? For divisibility by 3, the sum of digits must be a multiple of 3. Sum $= 3 + 1 + z + 5 = 9 + z$. Since $z$ is a single digit, $9 + z$ can be 9, 12, 15, or 18. Thus, the possible values for $z$ are 0, 3, 6, or 9.
• Solve the cryptarithm addition: $A + A + A = BA$. Since $A + A + A = 3A$, the units digit of $3A$ must be $A$. Checking digits from 0 to 9: $3 \times 5 = 15$ (ends in 5). So, $A = 5$. The tens digit $B$ becomes 1. Therefore, $A = 5$ and $B = 1$.
• Write the number 405 in generalized form. The generalized form of 405 is: $100 \times 4 + 10 \times 0 + 5$.
• Check if 1085 is divisible by 5 and 10. The units digit of 1085 is 5. Since it ends in 5, it is divisible by 5. However, since the units digit is not 0, it is not divisible by 10.

Frequently Asked Questions