Divisibility Rules: Clerkship Arithmetic Practice Quiz
Divisibility Rules: Clerkship Arithmetic Practice Quiz
12 questions on the divisibility rules of 2, 3, 4, 5, 6, 7, 8, 9, 11 and 12. Every explanation works the sum out. 📖 Read the detailed topic
Which of the following numbers is divisible by 3?
A number is divisible by 3 if the sum of its digits is. 5+2+1+4 = 12 and 12 ÷ 3 = 4, so 5,214 is divisible. The other three have digit sums 10, 11 and 8, none of which is a multiple of 3.
- > If the digit sum is a multiple of 3, so is the number
- > A big number shrinks to a small sum, so the check is quick
- > The last digit alone does not tell you about 3; you need the sum of all digits
- > For 3, add up all the digits
| Number | Digit sum | Divisible by 3? |
|---|---|---|
| 4,321 | 10 | No |
| 1,235 | 11 | No |
| 5,214 | 12 | Yes |
| 7,001 | 8 | No |
Which of the following numbers is divisible by 9?
A number is divisible by 9 if its digit sum is. 4+5+6+3 = 18 and 18 ÷ 9 = 2, so 4,563 is divisible. The others have digit sums 17, 20 and 22, none a multiple of 9.
- > If the digit sum is a multiple of 9, so is the number
- > A multiple of 9 is also divisible by 3
- > A digit sum that is only a multiple of 3 is not enough; it must be 9, 18, 27…
- > The rule for 9 is the rule for 3, but the sum must be a multiple of 9
Which of the following numbers is divisible by 4?
If the number formed by the last two digits is divisible by 4, the whole number is, because 100 is itself a multiple of 4. 16 ÷ 4 = 4, so 3,516 is divisible. Dividing 14, 18 and 19 by 4 leaves remainders 2, 2 and 3.
- > If the last two digits form a multiple of 4, so is the whole number
- > 100 is a multiple of 4, so the earlier digits do not affect the remainder
- > For 4 look at the last two digits; for 2 only the last digit
- > 4 = two digits, 8 = three digits
| Number | Last two digits | Remainder on dividing by 4 |
|---|---|---|
| 3,514 | 14 | 2 |
| 3,518 | 18 | 2 |
| 3,519 | 19 | 3 |
| 3,516 | 16 | 0 |
Which of the following numbers is divisible by 8?
If the number formed by the last three digits is divisible by 8, the whole number is, because 1,000 is itself a multiple of 8. 128 ÷ 8 = 16, so 27,128 is divisible. Dividing 126, 124 and 130 by 8 leaves remainders 6, 4 and 2.
- > If the last three digits form a multiple of 8, so is the whole number
- > The rule works because 1,000 is a multiple of 8
- > Checking only two digits for 8 can mislead you; use all three
- > 8 = three digits, 4 = two digits
Which of the following numbers is divisible by 11?
Counting from the right, the digits in odd places add up to 7+5 = 12 and those in even places to 4+8 = 12. The difference is 0, so the number is divisible by 11 (8,547 ÷ 11 = 777). The others give differences 1, 1 and 2, which are neither 0 nor a multiple of 11.
- > If the two alternate sums differ by 0 or a multiple of 11, the number is divisible by 11
- > Count places from the right
- > Counting places from the left swaps odd and even; keep the counting consistent
- > For 11, take alternate digits and compare the two sums
| Number | Sum, odd places | Sum, even places | Difference |
|---|---|---|---|
| 8,547 | 12 | 12 | 0 |
| 8,546 | 11 | 12 | 1 |
| 8,548 | 13 | 12 | 1 |
| 8,549 | 14 | 12 | 2 |
Which of the following numbers is divisible by 6?
To be divisible by 6 a number must be divisible by both 2 and 3. 4,218 is even and 4+2+1+8 = 15, a multiple of 3, so it works. 2,314 is even but its digit sum is 10, so it is not divisible by 3. 3,125 and 5,221 are odd.
- > 6 = 2 × 3, two co-prime factors
- > The number must be even and have a digit sum that is a multiple of 3
- > Being even is not enough for 6; the rule for 3 must hold too
- > 6 = even + the rule for 3
If the number 723* is divisible by 9, which digit replaces *?
7+2+3 = 12. The digit sum must be a multiple of 9, and the next multiple of 9 after 12 is 18. So * = 18 − 12 = 6. Check: 7,236 ÷ 9 = 804. The other options give digit sums 17, 16 and 19, none a multiple of 9.
- > Add the known digits first
- > Subtract that sum from the next multiple of 9
- > The missing digit must be 0–9, so the next multiple (18) is enough; 27 is not needed
- > Next multiple of 9 − current sum = the answer
Which of the following numbers is divisible by 7?
Drop the last digit and subtract twice that digit from the rest. 137 − 2×2 = 133; again 13 − 2×3 = 7, a multiple of 7. So 1,372 is divisible (1,372 ÷ 7 = 196). The others leave remainders 1, 2 and 3.
- > Rule for 7: rest of the number − 2 × last digit
- > Repeat until the result is small
- > You subtract twice the last digit; you do not add it
- > 7: cut the last digit, subtract double
| Step | Working | Result |
|---|---|---|
| 1 | 137 − 2×2 | 133 |
| 2 | 13 − 2×3 | 7 |
| Result | 7 is a multiple of 7 | Divisible |
What is the smallest number to add to 2,000 so that the sum is divisible by 9?
The digit sum of 2,000 is 2, so the remainder on dividing by 9 is 2. To reach the next multiple we need 9 − 2 = 7 more. Check: 2,007 ÷ 9 = 223. Taking the remainder 2 as the answer is the trap, and adding 9 is not the smallest.
- > Smallest number to add = divisor − remainder
- > The remainder on dividing by 9 equals the remainder of the digit sum
- > Do not give the remainder as the answer: you add (divisor − remainder)
- > To add: divisor − remainder; to subtract: the remainder
| Step | Value |
|---|---|
| Digit sum of 2,000 | 2 |
| Remainder on dividing by 9 | 2 |
| Number to add (9 − 2) | 7 |
| New number | 2,007 = 9 × 223 |
What is the greatest 3-digit number divisible by 11?
Dividing 999 by 11 gives quotient 90 and remainder 9, so 999 − 9 = 990 is the greatest. Check: the digits in odd places from the right are 0 and 9 (sum 9) and the even place is 9; the difference is 0, so it is divisible (990 ÷ 11 = 90). 979 is divisible but smaller, and 980 and 989 are not divisible.
- > Greatest such number = largest number − remainder
- > The largest 3-digit number is 999
- > When several options are divisible, pick the largest: 979 is divisible but 990 is bigger
- > Greatest: take the remainder off 999
| Number | Remainder on dividing by 11 | Divisible? |
|---|---|---|
| 979 | 0 | Yes |
| 980 | 1 | No |
| 989 | 10 | No |
| 990 | 0 | Yes |
A number is divisible by 12 if it is divisible by which two numbers at once?
12 = 3 × 4 and 3 and 4 are co-prime, so a number divisible by both 3 and 4 is divisible by 12. 2 and 6, or 4 and 6, are not co-prime: the number 6 is divisible by 2 and by 6 but not by 12. 2 and 3 together only guarantee 6.
- > For a composite number use co-prime factors
- > 12 = 3 × 4, 6 = 2 × 3, 15 = 3 × 5
- > Always take co-prime factors for a composite divisor; 2 × 6 gives 12 but 2 and 6 are not co-prime
- > 12 = 3 and 4: no other pair works
| Composite number | Co-prime factors | Wrong pairs |
|---|---|---|
| 6 | 2 and 3 | — |
| 12 | 3 and 4 | 2 and 6, 4 and 6 |
| 15 | 3 and 5 | — |
| 18 | 2 and 9 | 3 and 6 |
If 54*2 is divisible by 11, which digit replaces *?
From the right, the digits in odd places are 2 and 4, summing to 6. The digits in even places are * and 5, summing to * + 5. The difference is 6 − (* + 5) = 1 − *. This is 0 when * = 1. Check: 5,412 ÷ 11 = 492. With * = 0 the difference is 1, and with 2 or 3 it is negative; none is a multiple of 11.
- > For 11 the two alternate sums must differ by 0 (or a multiple of 11)
- > Write the unknown digit into an equation and solve
- > Put the unknown in the right group: * is in an even place, 2 and 4 in odd places
- > Set the difference to 0 and solve
📊 Topic Analysis
12 questions on the divisibility rules of 2, 3, 4, 5, 6, 7, 8, 9, 11 and 12. Every explanation works the sum out.
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