Tests of Divisibility (Algebraic Proof)
You already know divisibility rules — a number is divisible by 3 if the sum of its digits is divisible by 3, and so on. But why do these rules work?
Using the general form of numbers, we can prove these divisibility tests algebraically. By writing numbers as 10a + b or 100a + 10b + c, and rearranging terms, the reasons behind each rule become clear.
This chapter focuses on proving divisibility rules for 2, 3, 5, 9, and 10 using algebra.
What is Tests of Divisibility (Algebraic Proof)?
Definition: An algebraic proof of divisibility uses the general form of a number to show mathematically why a divisibility test works.
Key idea: Express the number in general form, then rewrite it as:
Number = (multiple of divisor) + (remainder expression)
If the remainder expression is divisible by the divisor, the whole number is divisible.
Methods
Proof: Divisibility by 9
Let the 3-digit number be 100a + 10b + c.
- Rewrite: 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.
Proof: Divisibility by 3
Same approach as above:
- 100a + 10b + c = 99a + 9b + (a + b + c)
- 99a + 9b = 3(33a + 3b), which is divisible by 3.
- So the number is divisible by 3 if and only if (a + b + c) is divisible by 3.
Proof: Divisibility by 2
- Number = 10a + b (2-digit) or 100a + 10b + c (3-digit).
- 10a and 100a + 10b are all divisible by 2 (since 10 = 2 × 5).
- So the number is divisible by 2 if and only if the last digit is divisible by 2.
Proof: Divisibility by 5
- 10a, 100a, 10b are all divisible by 5.
- So the number is divisible by 5 if and only if the last digit is 0 or 5.
Proof: Divisibility by 10
- 10a, 100a, 10b are all divisible by 10.
- So the number is divisible by 10 if and only if the last digit is 0.
Solved Examples
Example 1: Example 1: Proving for a specific number (divisibility by 9)
Problem: Prove that 576 is divisible by 9 using the general form.
Solution:
- 576 = 100(5) + 10(7) + 6
- = 99(5) + 9(7) + (5 + 7 + 6)
- = 99(5) + 9(7) + 18
- 99(5) + 9(7) = 9(55 + 7) = 9(62), divisible by 9.
- 18 = 9 × 2, divisible by 9.
Answer: Since both parts are divisible by 9, 576 is divisible by 9.
Example 2: Example 2: Proving for divisibility by 3
Problem: Check if 843 is divisible by 3 using algebraic proof.
Solution:
- 843 = 99(8) + 9(4) + (8 + 4 + 3)
- = 9(88 + 4) + 15
- 9(92) is divisible by 3.
- Sum of digits = 15, and 15 ÷ 3 = 5. Divisible by 3.
Answer: Yes, 843 is divisible by 3.
Example 3: Example 3: Not divisible by 9
Problem: Is 253 divisible by 9?
Solution:
- Sum of digits = 2 + 5 + 3 = 10
- 10 ÷ 9 gives remainder 1. Not divisible by 9.
- Algebraically: 253 = 9(27 + 5) + (2+5+3) = 9(32) + 10
- 9(32) = 288 is divisible by 9, but 10 is not.
Answer: No, 253 is not divisible by 9.
Example 4: Example 4: Divisibility by 11 (algebraic)
Problem: Prove the divisibility test for 11 using the general form of a 3-digit number.
Solution:
- Number = 100a + 10b + c
- 100 = 99 + 1 = 11(9) + 1, so 100a = 11(9a) + a
- 10 = 11 − 1, so 10b = 11b − b
- Number = 11(9a) + a + 11b − b + c = 11(9a + b) + (a − b + c)
- 11(9a + b) is divisible by 11.
- So the number is divisible by 11 if (a − b + c) is divisible by 11.
Answer: A 3-digit number abc is divisible by 11 if (a − b + c) is divisible by 11.
Example 5: Example 5: Checking divisibility by 11
Problem: Is 792 divisible by 11?
Solution:
- a = 7, b = 9, c = 2
- a − b + c = 7 − 9 + 2 = 0
- 0 is divisible by 11.
Answer: Yes, 792 is divisible by 11. (792 ÷ 11 = 72.)
Example 6: Example 6: Divisibility by 2 (algebraic)
Problem: Prove that 374 is divisible by 2.
Solution:
- 374 = 100(3) + 10(7) + 4
- 100(3) = 2 × 150, divisible by 2.
- 10(7) = 2 × 35, divisible by 2.
- Last digit = 4, which is even (divisible by 2).
Answer: Since all parts are divisible by 2, 374 is divisible by 2.
Example 7: Example 7: Two-digit proof
Problem: Prove that 72 is divisible by 9 using general form.
Solution:
- 72 = 10(7) + 2 = 9(7) + (7 + 2) = 9(7) + 9
- = 9(7) + 9(1) = 9(8)
- = 9 × 8, clearly divisible by 9.
Answer: 72 = 9 × 8, so it is divisible by 9.
Example 8: Example 8: Finding missing digit
Problem: A number 2_6 (3-digit) is divisible by 9. Find the missing digit.
Solution:
- Let the missing digit be b.
- Sum of digits = 2 + b + 6 = 8 + b
- For divisibility by 9: 8 + b must be divisible by 9.
- 8 + b = 9 → b = 1 (since b is a single digit)
- 8 + b = 18 → b = 10 (not a single digit, rejected)
Answer: The missing digit is 1. The number is 216.
Example 9: Example 9: General proof for any number of digits
Problem: Prove the divisibility test for 9 works for a 4-digit number.
Solution:
- Number = 1000a + 100b + 10c + d
- = 999a + 99b + 9c + (a + b + c + d)
- = 9(111a + 11b + c) + (a + b + c + d)
- 9(111a + 11b + c) is divisible by 9.
- So the number is divisible by 9 if (a + b + c + d) is divisible by 9.
Answer: The rule works for 4-digit numbers because 1000 = 999 + 1, 100 = 99 + 1, 10 = 9 + 1.
Example 10: Example 10: Multiple divisibility checks
Problem: Check if 630 is divisible by 2, 3, 5, 9, and 10.
Solution:
- By 2: Last digit = 0 (even). ✓ Divisible.
- By 3: Sum = 6 + 3 + 0 = 9, 9 ÷ 3 = 3. ✓ Divisible.
- By 5: Last digit = 0. ✓ Divisible.
- By 9: Sum = 9, 9 ÷ 9 = 1. ✓ Divisible.
- By 10: Last digit = 0. ✓ Divisible.
Answer: 630 is divisible by 2, 3, 5, 9, and 10.
Real-World Applications
Real-world applications:
- Checking calculations: Quick divisibility checks help verify arithmetic without a calculator.
- ISBN and barcode validation: Book ISBNs use divisibility by 11 as a check digit system.
- Bank account numbers: Modular arithmetic and divisibility checks are used to validate account numbers.
- Computer science: Hash functions use modular arithmetic principles based on divisibility.
- Competitive mathematics: Proving divisibility results algebraically is a common topic in olympiads.
Key Points to Remember
- The general form of numbers is the key to proving divisibility rules.
- 10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1 — this pattern drives the digit-sum tests.
- Divisibility by 3 and 9: check sum of digits.
- Divisibility by 2, 5, 10: check the last digit.
- Divisibility by 11: check the alternating sum of digits.
- To prove a rule, rewrite the number as (multiple of divisor) + (test expression).
- These proofs work for numbers with any number of digits.
- Finding missing digits uses the same approach — set digit sum equal to a multiple of the divisor.
Practice Problems
- Prove that 468 is divisible by 9 using the general form.
- Prove the divisibility test for 3 for a 2-digit number.
- Is 1234 divisible by 9? Use the algebraic method.
- A number 5_4 is divisible by 3. Find all possible values of the missing digit.
- Prove that a 3-digit number abc is divisible by 2 if c is even.
- Check if 2574 is divisible by 11 using the algebraic test.
Frequently Asked Questions
Q1. Why does the digit-sum test work for 9?
Because 10 = 9+1, 100 = 99+1, etc. So any number = 9(something) + sum of digits. The 9(something) part is always divisible by 9, so only the digit sum matters.
Q2. Does the digit-sum test work for all numbers, not just 3 and 9?
The digit-sum test only works for 3 and 9 because 10 ≡ 1 (mod 3) and 10 ≡ 1 (mod 9). For other divisors, different tests are needed.
Q3. How does the divisibility test for 11 work?
Since 10 ≡ −1 (mod 11), the alternating sum of digits (a − b + c − d + ...) determines divisibility by 11.
Q4. Can these proofs be extended to larger numbers?
Yes. The pattern works for any number of digits because 10ⁿ always equals (a multiple of 9) + 1.
Q5. What is modular arithmetic?
Modular arithmetic deals with remainders. 10 ≡ 1 (mod 9) means 10 leaves remainder 1 when divided by 9. This is the foundation of divisibility proofs.
Q6. How do you find a missing digit using divisibility?
Set up the divisibility condition (e.g., digit sum = multiple of 9) and solve for the unknown digit, keeping in mind it must be 0-9.
Q7. Why doesn't the last-digit test work for 3?
Because 10 is not divisible by 3 — it leaves remainder 1. So the tens place contributes to the remainder. You need the sum of ALL digits.
Q8. Is the algebraic proof the same as the short-cut rule?
The short-cut rule IS the result of the algebraic proof. The proof explains WHY the rule works; the rule is simply the conclusion stated without the proof.
