Divisibility Challenge Quiz: Master Rules from 2–10

A number is divisible by 2 if it ends in 0, 2, 4, 6, or 8. The number 34 ends in 4, which is an even digit, so it is divisible by 2. The number 13 ends in 3, 27 ends in 7, and 45 ends in 5, all of which are odd digits. A number ending in an odd digit cannot be divided evenly by 2. Ending in an even digit is the only check needed for the divisibility rule for 2.

The answer is False. A number is divisible by 10 only if it ends in 0. Ending in 5 only guarantees divisibility by 5. For example, 25 ends in 5 and divides evenly by 5, but 25 divided by 10 leaves a remainder of 5, so it is not divisible by 10. To pass the divisibility rule for 10, a number must end specifically in 0, not just in 5.

Each number in the sequence increases by 4, so the next number is 16 plus 4, which equals 20. All numbers in this sequence are multiples of 4, and 20 is the next multiple of 4 after 16. The number 18 is not a multiple of 4, 24 skips a step in the pattern, and 22 is not a multiple of 4. Only 20 continues the sequence correctly and passes the divisibility rule for 4.

A number is divisible by 5 if it ends in 0 or 5. A number is only divisible by 10 if it ends in 0. The number 35 ends in 5, so it passes the rule for 5, but it does not end in 0, so it fails the rule for 10. The numbers 30, 20, and 10 all end in 0, making them divisible by both 5 and 10. Only 35 fits the condition of being divisible by 5 but not by 10.

To check divisibility by 3, add up the digits of the number. If the digit sum is divisible by 3, the whole number is too. For 21, the digits are 2 and 1, and 2 plus 1 equals 3. Since 3 is divisible by 3, the number 21 passes the rule. For 16, the digit sum is 7. For 23, it is 5. For 14, it is 5. Since 7 and 5 are not divisible by 3, none of those numbers pass the rule.

The answer is True. The divisibility rule for 5 includes numbers ending in 0 or 5. The divisibility rule for 10 requires a number to end in 0. A number ending in 0 satisfies both rules at the same time. For example, 40 divided by 5 equals 8 with no remainder, and 40 divided by 10 equals 4 with no remainder. Any number ending in 0 will always pass both rules.

To check divisibility by 7, divide the number and see if there is no remainder. Dividing 56 by 7 gives exactly 8 with no remainder, so 56 is divisible by 7. Dividing 45 by 7 gives 6 with a remainder of 3. Dividing 38 by 7 gives 5 with a remainder of 3. Dividing 61 by 7 gives 8 with a remainder of 5. Only 56 divides evenly by 7, making it the correct answer.

To use the divisibility rule for 9, add up the digits of the number. If the sum is divisible by 9, the whole number is too. For 81, the digits are 8 and 1, and 8 plus 1 equals 9. Since 9 is divisible by 9, the number 81 passes the rule. For 34, the digit sum is 7. For 56, the digit sum is 11. For 47, the digit sum is 11. None of those sums are divisible by 9, so only 81 qualifies.

To check divisibility by 4, look at the last two digits and see if they are divisible by 4. For 24, dividing 24 by 4 gives 6 with no remainder, so 24 passes. For 36, dividing 36 by 4 gives 9 with no remainder, so 36 also passes. The number 30 divided by 4 gives 7 with a remainder of 2, so it fails. The number 22 divided by 4 gives 5 with a remainder of 2, so it also fails. Only 24 and 36 are divisible by 4.

The answer is False. To check if a number is divisible by 9, add its digits together. For 42, the digits are 4 and 2, and 4 plus 2 equals 6. Since 6 is not divisible by 9, the number 42 is not divisible by 9. A number is only divisible by 9 when its digit sum equals 9 or a multiple of 9, such as 18 or 27. The digit sum of 42 does not meet that condition.

To be divisible by both 2 and 3, a number must pass both rules. For the rule of 2, the number must end in an even digit. For the rule of 3, the digit sum must be divisible by 3. For 18, the last digit is 8, which is even, and the digit sum is 1 plus 8 equals 9, which is divisible by 3. Both rules pass for 18. The number 14 is even but its digit sum is 5. The number 21 has a digit sum of 3 but ends in an odd digit. The number 25 fails both rules.

The divisibility rule for 4 says that if the last two digits of a number form a number divisible by 4, the whole number is divisible by 4. The last two digits here are 16, and 16 divided by 4 equals 4 with no remainder. This confirms the whole number is divisible by 4. The last two digits do not determine divisibility by 3, 5, or 7, since those rules work differently.

To be divisible by both 3 and 5, a number must pass both rules. For the rule of 5, the number must end in 0 or 5. For the rule of 3, the digit sum must be divisible by 3. For 45, the last digit is 5, passing the rule for 5, and the digit sum is 4 plus 5 equals 9, which is divisible by 3, passing the rule for 3. The number 20 is divisible by 5 but its digit sum is 2. The number 18 is divisible by 3 but ends in 8. The number 32 fails both rules.

To check if a number is a multiple of 9, add its digits and see if the sum is divisible by 9. In option B, the digit sums of 9, 18, 27, and 36 are all 9, confirming all four are multiples of 9. Option A includes 28, whose digit sum is 10. Option C includes 19, whose digit sum is 10. Option D includes 35, whose digit sum is 8. None of those digit sums are divisible by 9, so only option B forms a valid sequence of multiples of 9.

The answer is True. The number 6 is made by multiplying 2 and 3 together. Any number that divides evenly by 6 must also divide evenly by both 2 and 3. For example, 24 divided by 6 equals 4, and 24 is also divisible by 2 (giving 12) and by 3 (giving 8). This relationship holds for every multiple of 6 because 6 itself is built from those two factors.

To check divisibility by 3, add the digits of each number. For 27, the digit sum is 2 plus 7 equals 9, which is divisible by 3, so 27 passes. For 48, the digit sum is 4 plus 8 equals 12, which is divisible by 3, so 48 passes. For 32, the digit sum is 3 plus 2 equals 5, which is not divisible by 3. For 55, the digit sum is 5 plus 5 equals 10, which is not divisible by 3. Only 27 and 48 are divisible by 3.

Each number in the sequence increases by 9, so the next number is 36 plus 9, which equals 45. All numbers in this sequence are multiples of 9. To verify that 45 belongs, add its digits: 4 plus 5 equals 9, which is divisible by 9. The numbers 44, 43, and 47 are not multiples of 9, and none of their digit sums are divisible by 9.

To check divisibility by 8, divide the number by 8 and see if there is no remainder. Dividing 56 by 8 gives exactly 7 with no remainder, confirming 56 is divisible by 8. Dividing 42 by 8 gives 5 with a remainder of 2. Dividing 52 by 8 gives 6 with a remainder of 4. Dividing 50 by 8 gives 6 with a remainder of 2. Only 56 passes the divisibility rule for 8.

The number 30 passes several divisibility rules. It ends in 0, so it is divisible by both 5 and 10. It is even, so it is divisible by 2. Its digit sum is 3 plus 0 equals 3, which is divisible by 3. Since 30 passes all four rules, option A is the only complete and correct answer. The other options each name just one rule, making them incomplete and incorrect.

The answer is False. Many numbers are divisible by more than one rule at the same time. For example, the number 30 is divisible by 2, 3, 5, and 10 all at once. The number 18 is divisible by both 2 and 3. The number 81 is divisible by both 3 and 9. Divisibility rules are independent checks, and passing one rule does not stop a number from passing others as well.