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Tests for divisibilty

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visibility 77update 7 months agobookmarkshare

🎯 In this topic you will

  • Use divisibility tests to find factors of large numbers
 

🧠 Key Words

  • divisible
  • tests for divisibility
Show Definitions
  • divisible: A number is divisible by another if it can be divided exactly with no remainder.
  • tests for divisibility: Simple rules used to check whether one number divides evenly into another.
 

📗 Understanding divisibility and remainders

2, 3 and 5 are all factors of 30.

You say that ‘30 is divisible by 2’ because 30 ÷ 2 does not have a remainder.

30 is divisible by 3 and 30 is divisible by 5.

30 is not divisible by 4 because 30 ÷ 4 = 7 with remainder 2 (which can be written as 7 r 2).

 

🔎 Reasoning Tip

Divisibility by 2: A whole number is divisible by 2 when 2 is a factor of that number.

 

🔍 Divisibility rules for numbers

87 654 is a large number.

Is 87 654 divisible by 2? By 3? By 4? By 5?

Here are some rules for divisibility:

  • A number is divisible by 2 when the last digit is 0, 2, 4, 6 or 8.
    87 654 is divisible by 2 because the last digit is 4.
  • A number is divisible by 3 when the sum of the digits is a multiple of 3.
    8 + 7 + 6 + 5 + 4 = 30 and 30 = 10 × 3, so 87 654 is divisible by 3.
  • A number is divisible by 4 when the number formed by the last two digits is divisible by 4.
    The last two digits of 87 654 are 54 and 54 ÷ 4 = 13 r 2. So 87 654 is not divisible by 4.
  • A number is divisible by 5 when the last digit is 0 or 5.
    The last digit of 87 654 is 4, so it is not divisible by 5.
  • A number is divisible by 6 when it is divisible by 2 and 3.
    87 654 is divisible by 6.
  • To test for divisibility by 7, remove the last digit, 4, to leave 8765
    • Subtract twice the last digit from 8765, that is:
      8765 – 2 × 4 = 8765 – 8 = 8757
    • If this number is divisible by 7, so is the original number.
      8757 ÷ 7 = 1252 with no remainder and so 87 654 is divisible by 7.
  • A number is divisible by 8 when the number formed by the last three digits is divisible by 8.
    654 ÷ 8 = 81 r 6, so 87 654 is not divisible by 8.
  • A number is divisible by 9 when the sum of the digits is divisible by 9.
    8 + 7 + 6 + 5 + 4 = 30 and 30 is not divisible by 9. So 87 654 is not divisible by 9.
  • A number is divisible by 10 when the last digit is 0.
    The last digit of 87 654 is 4, so 87 654 is not divisible by 10.
  • A number is divisible by 11 when the difference between the sum of the odd digits and the sum of the even digits is 0 or a multiple of 11.
    The sum of the odd digits of 87 654 is 4 + 6 + 8 = 18.
    The sum of the even digits of 87 654 is 5 + 7 = 12.
    18 – 12 = 6, so 87 654 is not a multiple of 11.
 
📘 Worked example

The number *7258 has one digit missing.

a) Find the missing digit when:

i) the number is divisible by 6
ii) the number is divisible by 11

b) A number is divisible by 66 when it is divisible by both 6 and 11. Could *7258 be divisible by 66? Give a reason for your answer.

Answer

a i) The number must be a multiple of 2 and 3.
The last digit is 8, so the number is divisible by 2.
The sum of the digits is $* + 7 + 2 + 5 + 8 = * + 22$
If this is a multiple of 3, then * must be 2, 5, or 8.
So there are three possible values for *.

a ii) The sum of the odd-position digits is $8 + 2 + * = 10 + *$
The sum of the even-position digits is $5 + 7 = 12$
When $* = 2$, the difference is $|10 + 2 - 12| = 0$, so 27258 is divisible by 11.

b) The answer to part a) shows that the number is divisible by 66 when $* = 2$.
This is the only possible value that satisfies both conditions.

A number divisible by 6 must be divisible by both 2 and 3. A number divisible by 11 must have alternating digit sums with zero difference.

The only digit that works for both is 2, so *7258 becomes 27258 — divisible by both 6 and 11, and therefore divisible by 66.

 

EXERCISES

1a. Show that the number 28 572 is divisible by 3 but not by 9.

👀 Show answer
Sum of digits = 2 + 8 + 5 + 7 + 2 = 24
24 is divisible by 3 → so 28 572 is divisible by 3.
24 is not divisible by 9 → so 28 572 is not divisible by 9.

1b. Change the final digit of 28 572 to make a number that is divisible by 9.

👀 Show answer
To be divisible by 9, the digit sum must be a multiple of 9.
We need the sum to be 27 (next multiple after 24).
So increase the digit sum by 3 → change final digit from 2 to 5.
New number: 28 575

2a. Show that 57 423 is divisible by 3 but not by 6.

👀 Show answer
Sum of digits = 5 + 7 + 4 + 2 + 3 = 21 → divisible by 3.
Last digit is 3 → not even → not divisible by 2, so not divisible by 6.

2b. The number 5742* is divisible by 6. Find the possible values of the digit *.

👀 Show answer
Must be divisible by 2 → * must be even: 0, 2, 4, 6, or 8
Try each digit: Sum of 5 + 7 + 4 + 2 + * must be divisible by 3
Sum without * = 18 → valid if * = 0, 3, 6, or 9 → only * = 0 and 6 are even
So possible values: 0 or 6

3a. Show that 25 764 is divisible by 2 and by 4.

👀 Show answer
Ends in 4 → even → divisible by 2.
Last two digits = 64 → divisible by 4 → divisible by both 2 and 4.

3b. Is 25 764 divisible by 8? Give a reason for your answer.

👀 Show answer
Check last 3 digits: 764
764 ÷ 8 = 95.5 → not whole
So 25 764 is not divisible by 8.

4a. Show that 3 and 4 are factors of 25 320.

👀 Show answer
Sum of digits = 2 + 5 + 3 + 2 + 0 = 12 → divisible by 3
Last two digits = 20 → divisible by 4 →
So 3 and 4 are both factors of 25 320

4b. Find two more factors of 25 320 that are between 1 and 12.

👀 Show answer
Try 5: ends in 0 → divisible → ✔️
Try 8: 320 is divisible by 8 → whole result → ✔️
Two more factors: 5 and 8

5a. Choose any four digits.

👀 Show answer
Example digits: 1, 2, 3, 6

5b. If it is possible, arrange your digits to make a number that is divisible by:
i) 2  ii) 3  iii) 4  iv) 5  v) 6

👀 Show answer
Using digits 1, 2, 3, 6:
i) 1236 → even → divisible by 2 ✔️
ii) digit sum = 12 → divisible by 3 ✔️
iii) ends in 36 → divisible by 4 ✔️
iv) ends in 6 → not divisible by 5 ❌
v) divisible by both 2 and 3 → ✔️

5c. Can you arrange your digits to make a number that is divisible by all five numbers in part a? If not, can you make a number that is divisible by four of the numbers?

👀 Show answer
No, because the number does not end in 0 or 5, so it's not divisible by 5.
But we can make one divisible by 2, 3, 4, and 6.
Example: 1236 is divisible by 2, 3, 4, and 6.

5d. Give your answers to a partner to check.

👀 Show answer
This is a classroom discussion task. Compare and explain your logic to your partner and confirm divisibility using rules or calculator.

6a. Show that 924 is divisible by 11.

👀 Show answer
Alternating sum = 9 - 2 + 4 = 11 → divisible by 11.
So 924 is divisible by 11.

6b. Is 161 084 divisible by 11? Give a reason for your answer.

👀 Show answer
Alternating sum = 1 - 6 + 1 - 0 + 8 - 4 = 0
0 is divisible by 11 → so 161 084 is divisible by 11 ✔️

7a. Use a test for divisibility to show that 2583 is divisible by 7.

👀 Show answer
2583 ÷ 7 = 369 → exact whole number.
So 2583 is divisible by 7.

7b. Use a test for divisibility to show that 3852 is not divisible by 7.

👀 Show answer
3852 ÷ 7 = 550.2857… not a whole number.
So 3852 is not divisible by 7.

8a. Show that only two numbers between 1 and 10 are factors of 22 599.

👀 Show answer
Try divisibility rules for 1–10:
22 599 is divisible by 1 and 3 only.
Sum of digits = 2+2+5+9+9 = 27 → divisible by 3 ✔️
So only factors between 1 and 10 are: 1 and 3

8b. What numbers between 1 and 10 are factors of 99 522?

👀 Show answer
Sum of digits = 9+9+5+2+2 = 27 → divisible by 3 and 9 ✔️
Ends in 2 → divisible by 2 ✔️
Not divisible by 4 (last two digits = 22) ❌
Not divisible by 5 (doesn’t end in 0 or 5) ❌
99 522 ÷ 6 = 16 587 → whole ✔️
Divisible by: 1, 2, 3, 6, 9

9. Copy and complete this table. The first line has been done for you.

Number Factors between 1 and 10
12 2, 3, 4, 6
123  
1234  
12345  
123456  
👀 Show answer
Number Factors between 1 and 10
12 2, 3, 4, 6
123 1, 3
1234 1, 2
12345 1, 3, 5
123456 1, 2, 3, 4, 6

10. Use the digits 4, 5, 6 and 7 to make a number that is a multiple of 11.
How many different ways can you find to do this?

👀 Show answer
Use the alternating sum rule for 11: (digit1 - digit2 + digit3 - digit4) must be divisible by 11.
Try: 7456 → 7−4+5−6 = 2 → ✖️
Try: 5476 → 5−4+7−6 = 2 → ✖️
Try: 6547 → 6−5+4−7 = -2 → ✖️
Try: 5742 → 5−7+4−2 = 0 → ✔️
One possible number: 5742
You can test all 24 permutations. Answers may vary based on pattern checking.

11a. Show that 2521 is not divisible by any integer between 1 and 12.

👀 Show answer
Check divisibility by 2–12. None divide exactly into 2521.
So 2521 has no factors between 2 and 12. ✔️

11b. Rearrange the digits of 2251 to make a number divisible by 5.

👀 Show answer
Ends in 0 or 5 → try: 2155 ✖️, 2251 ✖️, 1525 ✔️
Possible answer: 1525

11c. Rearrange the digits of 2251 to make a number divisible by 4.

👀 Show answer
Last 2 digits must form a number divisible by 4.
Try: 1252 → last two = 52 ✖️
Try: 1224 → last two = 24 ✔️
Valid example: 1224

11d. Rearrange the digits of 2251 to make a number divisible by 8.

👀 Show answer
Last 3 digits must form a number divisible by 8.
Try: 512 → ✔️
Possible arrangement: 2512

11e. Find the smallest integer larger than 2521 that is divisible by 6.

👀 Show answer
2526 ÷ 6 = 421 → ✔️
Answer: 2526

11f. Find the smallest integer larger than 2521 that is divisible by 11.

👀 Show answer
2530 ÷ 11 = 230 → ✔️
Answer: 2530

12a. Explain why any positive integer where every digit is 4 must be divisible by 2 and by 4.

👀 Show answer
Ends in 4 → divisible by 2
Last 2 digits = 44 → divisible by 4
So all such numbers are divisible by 2 and 4 ✔️

12b. Every digit is 4. It is divisible by 5. Explain why this is impossible.

👀 Show answer
To be divisible by 5, number must end in 0 or 5.
But all digits are 4 → cannot end in 0 or 5 → not divisible by 5.

12c.i. Every digit is 4. It is divisible by 3. Find a number with both these properties.

👀 Show answer
444 → sum = 12 → divisible by 3 → ✔️

12c.ii. Is there more than one possible number? Give a reason.

👀 Show answer
Yes. Any number made of only 4s whose digit sum is divisible by 3 will work.
Example: 444, 444444 (sum = 24) ✔️

12d.i. Every digit is 4. It is divisible by 11. Find a number with both these properties.

👀 Show answer
44 → alternating sum = 4 − 4 = 0 → divisible by 11 ✔️

12d.ii. Is there more than one possible number? Give a reason for your answer.

👀 Show answer
Yes. 4444 → alternating sum = 4−4+4−4 = 0 ✔️
Any number with even number of 4s will give alternating sum of 0 ✔️
 

🧠 Think like a Mathematician

Question: When can you tell if a number is divisible by a product like 8, 10 or 15? What rules help you decide?

Equipment: Pencil, paper

  1. a)$2 \times 4 = 8$
    Look at this statement:
    A number is divisible by 8 when it is divisible by 2 and by 4.
    Do you think the statement is correct? Give evidence to justify your answer.
  2. b)$2 \times 5 = 10$
    Look at this statement:
    A number is divisible by 10 when it is divisible by 2 and by 5.
    Do you think the statement is correct? Give evidence to justify your answer.
  3. c)$3 \times 5 = 15$
    Look at this statement:
    A number is divisible by 15 when it is divisible by 3 and by 5.
    Do you think the statement is correct? Give evidence to justify your answer.

Follow-up Questions:

1. Is being divisible by two factors always enough to be divisible by their product?
2. Does it depend on whether the two factors share a common factor?
3. Can you explain the general rule?
👀 show answer
  • a: Not always correct. A number divisible by both 2 and 4 is not necessarily divisible by 8. For example, 12 is divisible by 2 and 4, but not by 8.
  • b: Yes, because 2 and 5 have no common factors. If a number is divisible by both, it will be divisible by their product 10.
  • c: Yes, the statement is correct. If a number is divisible by both 3 and 5 (which are coprime), then it is divisible by 15.
  • General Rule: A number is divisible by the product of two integers $a \times b$ if it is divisible by both $a$ and $b$, and$a$ and $b$ have no common factors (they are coprime).
 

📘 What we've learned

  • We explored what it means for a number to be divisible — having no remainder when divided.
  • We used divisibility rules to test large numbers for divisibility by 2, 3, 4, 5, 6, 7, 8, 9, 10, and 11.
  • We learned specific tests: e.g., divisible by 2 if the last digit is even; by 3 if the digit sum is a multiple of 3; by 5 if it ends in 0 or 5.
  • We practiced applying these rules to multi-digit numbers and evaluated missing digits using logic.
  • We used divisibility to identify all the factors of large numbers between 1 and 10.
  • We explored when a number is divisible by a product (e.g., 6, 10, 15) based on its divisibility by each factor.
  • We worked backwards to build numbers satisfying given divisibility conditions, including multiple rules at once.
  • We used reasoning to compare digit positions, sums, and patterns to test divisibility, including for 11.
  • We applied divisibility to real-world number sense and problem solving, including building custom numbers.

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