Numbers in General Form
Every number can be written in an expanded form using place values. In algebra, we express numbers using variables to represent their digits. This is called the general form of a number.
For example, a 2-digit number with tens digit a and units digit b is written as 10a + b. A 3-digit number with digits a, b, c is written as 100a + 10b + c.
Writing numbers in general form helps us prove divisibility rules, solve number puzzles, and understand the structure of our number system algebraically.
What is Numbers in General Form?
Definition: The general form of a number is its representation as a sum of digit-values using place value notation.
2-digit: 10a + b
3-digit: 100a + 10b + c
Where:
- a, b, c are the digits (0 to 9)
- For a 2-digit number, a ≠ 0 (tens digit cannot be zero)
- For a 3-digit number, a ≠ 0 (hundreds digit cannot be zero)
Examples:
- 57 = 10 × 5 + 7 (a = 5, b = 7)
- 83 = 10 × 8 + 3 (a = 8, b = 3)
- 342 = 100 × 3 + 10 × 4 + 2 (a = 3, b = 4, c = 2)
- 709 = 100 × 7 + 10 × 0 + 9 (a = 7, b = 0, c = 9)
Methods
Writing a number in general form:
- Identify each digit and its place value.
- Multiply each digit by its place value.
- Write as a sum.
Example: Write 465 in general form.
- 4 is in hundreds place: 4 × 100 = 400
- 6 is in tens place: 6 × 10 = 60
- 5 is in units place: 5 × 1 = 5
- 465 = 100 × 4 + 10 × 6 + 5
Reversing a 2-digit number:
- Original number = 10a + b
- Reversed number = 10b + a
- Sum = (10a + b) + (10b + a) = 11a + 11b = 11(a + b)
- Difference = (10a + b) − (10b + a) = 9a − 9b = 9(a − b)
Important results:
- The sum of a 2-digit number and its reverse is always divisible by 11.
- The difference of a 2-digit number and its reverse is always divisible by 9.
Solved Examples
Example 1: Example 1: Writing in general form
Problem: Write 73 in general form.
Solution:
- 73 = 10 × 7 + 3
- Here a = 7, b = 3
Answer: 73 = 10(7) + 3.
Example 2: Example 2: Three-digit number
Problem: Write 528 in general form.
Solution:
- 528 = 100 × 5 + 10 × 2 + 8
- Here a = 5, b = 2, c = 8
Answer: 528 = 100(5) + 10(2) + 8.
Example 3: Example 3: Sum of number and its reverse
Problem: Find the sum of 47 and its reverse 74. Show it is divisible by 11.
Solution:
- 47 = 10(4) + 7
- 74 = 10(7) + 4
- Sum = 47 + 74 = 121
- Using formula: 11(a + b) = 11(4 + 7) = 11 × 11 = 121 ✓
- 121 ÷ 11 = 11. Divisible by 11.
Answer: Sum = 121, which is divisible by 11.
Example 4: Example 4: Difference of number and reverse
Problem: Find the difference between 85 and 58. Show it is divisible by 9.
Solution:
- 85 − 58 = 27
- Using formula: 9(a − b) = 9(8 − 5) = 9 × 3 = 27 ✓
- 27 ÷ 9 = 3. Divisible by 9.
Answer: Difference = 27, divisible by 9.
Example 5: Example 5: Three-digit reversal
Problem: Find the difference between 843 and 348. Show it is divisible by 99.
Solution:
- 843 − 348 = 495
- Using formula: 99(a − c) = 99(8 − 3) = 99 × 5 = 495 ✓
- 495 ÷ 99 = 5. Divisible by 99.
Answer: Difference = 495, divisible by 99.
Example 6: Example 6: Finding digits
Problem: A 2-digit number is 7 times the sum of its digits. If the digits are reversed, the new number is 27 less. Find the number.
Solution:
- Let number = 10a + b
- Condition 1: 10a + b = 7(a + b) → 10a + b = 7a + 7b → 3a = 6b → a = 2b
- Condition 2: (10a + b) − (10b + a) = 27 → 9(a − b) = 27 → a − b = 3
- From a = 2b: 2b − b = 3 → b = 3, a = 6
Answer: The number is 63.
Example 7: Example 7: Proving divisibility by 3
Problem: Prove that a 2-digit number is divisible by 3 if the sum of its digits is divisible by 3.
Solution:
- Number = 10a + b = 9a + (a + b)
- 9a is always divisible by 3 (since 9 = 3 × 3).
- So the number is divisible by 3 if and only if (a + b) is divisible by 3.
Answer: Since 10a + b = 9a + (a + b) and 9a is divisible by 3, the number is divisible by 3 when a + b is divisible by 3.
Example 8: Example 8: Proving divisibility by 9
Problem: Prove that a 3-digit number is divisible by 9 if the sum of its digits is divisible by 9.
Solution:
- Number = 100a + 10b + c = 99a + 9b + (a + b + c)
- 99a + 9b = 9(11a + b), which is divisible by 9.
- So the number is divisible by 9 if and only if (a + b + c) is divisible by 9.
Answer: Since 100a + 10b + c = 9(11a + b) + (a + b + c), the number is divisible by 9 when a + b + c is divisible by 9.
Example 9: Example 9: Number puzzle
Problem: The sum of digits of a 2-digit number is 12. If the digits are reversed, the new number is 36 more. Find the number.
Solution:
- a + b = 12
- Reversed is bigger: (10b + a) − (10a + b) = 36 → 9(b − a) = 36 → b − a = 4
- From a + b = 12 and b − a = 4: 2b = 16, b = 8, a = 4
Answer: The number is 48.
Example 10: Example 10: General form of a 4-digit number
Problem: Write 3075 in general form.
Solution:
- 3075 = 1000 × 3 + 100 × 0 + 10 × 7 + 5
- = 1000(3) + 100(0) + 10(7) + 5
Answer: 3075 = 1000(3) + 10(7) + 5.
Real-World Applications
Real-world applications:
- Proving divisibility rules: General form is used to algebraically prove why divisibility tests work.
- Number puzzles: Many puzzles ("find the 2-digit number such that...") use general form to set up equations.
- Cryptarithmetic: Letters-for-digits puzzles rely on general form.
- Computer science: Understanding number representation in different bases uses the same principle.
- Algebra: Expressing numbers algebraically builds the foundation for polynomial expressions.
Key Points to Remember
- A 2-digit number with digits a, b = 10a + b.
- A 3-digit number with digits a, b, c = 100a + 10b + c.
- The notation ab (overline) means the number, NOT the product a × b.
- Sum of a 2-digit number and its reverse = 11(a + b).
- Difference of a 2-digit number and its reverse = 9(a − b).
- Difference of a 3-digit number and its reverse = 99(a − c).
- General form is used to prove divisibility rules for 3, 9, and 11.
- The leading digit of a number is always non-zero.
Practice Problems
- Write 96 and 609 in general form.
- Show that the sum of 38 and 83 is divisible by 11.
- Show that the difference of 71 and 17 is divisible by 9.
- A 2-digit number is 4 times the sum of its digits. The digit in the units place is 3 more than the digit in the tens place. Find the number.
- Prove that a 3-digit number abc is divisible by 3 if a + b + c is divisible by 3.
- The sum of digits of a 2-digit number is 9. If the digits are reversed, the number increases by 27. Find the number.
Frequently Asked Questions
Q1. What is the general form of a number?
The general form expresses a number using its digits and place values. A 2-digit number ab = 10a + b. A 3-digit number abc = 100a + 10b + c.
Q2. Why is the general form useful?
It helps prove divisibility rules, solve number puzzles, and understand the structure of numbers algebraically.
Q3. What does ab (with overline) mean?
It means the 2-digit number with tens digit a and units digit b. It equals 10a + b, NOT a × b.
Q4. Why is the sum of a 2-digit number and its reverse always divisible by 11?
Because (10a + b) + (10b + a) = 11(a + b), which is clearly a multiple of 11.
Q5. Why is the difference always divisible by 9?
Because (10a + b) − (10b + a) = 9(a − b), which is clearly a multiple of 9.
Q6. Can the leading digit be 0?
No. In a 2-digit number 10a + b, a must be from 1 to 9 (not 0), otherwise it becomes a 1-digit number.
Q7. How is general form related to expanded form?
They are the same concept. Expanded form = general form. Both express a number as a sum of digit × place value.
Q8. Can general form be used for 4-digit numbers?
Yes. A 4-digit number abcd = 1000a + 100b + 10c + d.
