This question tests middle school number properties and integer skills, specifically the understanding and application of divisibility rules. Divisibility rules help determine if one number divides another without a remainder using specific patterns. For example, a number is divisible by 3 if the sum of its digits is divisible by 3. In this case, students check the digit sum for each option. Choice C is correct because 517's digit sum is 5+1+7=13, and 13 is not divisible by 3. This demonstrates the student's ability to recognize and apply the rule accurately. Choice A is incorrect because 222's sum is 6, divisible by 3. To help students: Focus on teaching each rule with examples, practice identifying patterns in numbers, and encourage students to verify their answers by applying the rule. Watch for common errors in summing digits.

This question tests middle school number properties and integer skills, specifically the understanding and application of divisibility rules. Divisibility rules help determine if one number divides another without a remainder using specific patterns. For example, a number is divisible by 9 if the sum of its digits is divisible by 9. In this case, students calculate the digit sum for each option. Choice C is correct because 4,563's sum is 4+5+6+3=18, and 18 ÷ 9 = 2. This demonstrates the student's ability to recognize and apply the rule accurately. Choice A is incorrect because 1,234's sum is 10, not divisible by 9. To help students: Focus on teaching each rule with examples, practice identifying patterns in numbers, and encourage students to verify their answers by applying the rule. Watch for common errors in digit summing.

This question tests middle school number properties and integer skills, specifically the understanding and application of divisibility rules. Divisibility rules help determine if one number divides another without a remainder using specific patterns. For example, a number is divisible by 2 if its last digit is even. In this case, students check the last digit for oddness to find the one not divisible by 2. Choice B is correct because 731 ends in 1, which is odd. This demonstrates the student's ability to recognize and apply the rule accurately. Choice A is incorrect because 908 ends in 8, even. To help students: Focus on teaching each rule with examples, practice identifying patterns in numbers, and encourage students to verify their answers by applying the rule. Watch for common errors like misidentifying even digits.

This question tests middle school number properties and integer skills, specifically the understanding and application of divisibility rules. Divisibility rules help determine if one number divides another without a remainder using specific patterns. For example, a number is divisible by 8 if its last three digits form a number divisible by 8. In this case, students apply the rule for 8 to each option. Choice A is correct because 1,112's last three digits are 112, and 112 ÷ 8 = 14. This demonstrates the student's ability to recognize and apply the rule accurately. Choice B is incorrect because 1,114's last three are 114, and 114 ÷ 8 = 14.25. To help students: Focus on teaching each rule with examples, practice identifying patterns in numbers, and encourage students to verify their answers by applying the rule. Watch for common errors like checking only the last digit.

This question tests middle school number properties and integer skills, specifically the understanding and application of divisibility rules. Divisibility rules help determine if one number divides another without a remainder using specific patterns. For example, a number is divisible by 4 if its last two digits are divisible by 4. In this case, students check the last two digits of each. Choice D is correct because 1,096's last two are 96, and 96 ÷ 4 = 24. This demonstrates the student's ability to recognize and apply the rule accurately. Choice A is incorrect because 1,062's last two are 62, and 62 ÷ 4 = 15.5. To help students: Focus on teaching each rule with examples, practice identifying patterns in numbers, and encourage students to verify their answers by applying the rule. Watch for common errors in division of last digits.

This question tests middle school number properties and integer skills, specifically the understanding and application of divisibility rules. Divisibility rules help determine if one number divides another without a remainder using specific patterns. For example, a number is divisible by 2 if its last digit is even. In this case, students apply the rule that a number is divisible by 2 if it ends in 0, 2, 4, 6, or 8. Choice B is correct because 842 ends in 2, which is even, confirming it is divisible by 2. This demonstrates the student's ability to recognize and apply the rule accurately. Choice A is incorrect because 317 ends in 7, which is odd. To help students: Focus on teaching each rule with examples, practice identifying patterns in numbers, and encourage students to verify their answers by applying the rule. Watch for common errors like confusing even and odd digits.

This question tests middle school number properties and integer skills, specifically the understanding and application of divisibility rules. Divisibility rules help determine if one number divides another without a remainder using specific patterns. For example, a number is divisible by 6 if it is divisible by both 2 and 3. In this case, students check each number for divisibility by 6. Choice B is correct because 204 is even and its digit sum (2+0+4=6) is divisible by 3. This demonstrates the student's ability to recognize and apply the rule accurately. Choice A is incorrect because 123 is odd, so not divisible by 2. To help students: Focus on teaching each rule with examples, practice identifying patterns in numbers, and encourage students to verify their answers by applying the rule. Watch for common errors like forgetting to check both 2 and 3.

This question tests middle school number properties and integer skills, specifically the understanding and application of divisibility rules. Divisibility rules help determine if one number divides another without a remainder using specific patterns. For example, a number is divisible by 10 if it ends in 0. In this case, students apply the rule for 10. Choice B is correct because 4,120 ends in 0. This demonstrates the student's ability to recognize and apply the rule accurately. Choice A is incorrect because 7,215 ends in 5, not 0. To help students: Focus on teaching each rule with examples, practice identifying patterns in numbers, and encourage students to verify their answers by applying the rule. Watch for common errors like confusing 10 with 5.

When you encounter divisibility questions, you need to understand how factors and multiples relate to each other. The key principle is that if a number is divisible by two factors, it's divisible by their least common multiple (LCM), but not necessarily by their product.

Let's examine why answer C is correct. If a number is divisible by both 2 and 3, then it must contain both 2 and 3 as factors. Since 2 and 3 are relatively prime (they share no common factors), their LCM is simply $$2 \times 3 = 6$$. Therefore, any number divisible by both 2 and 3 is always divisible by 6.

Now for the incorrect choices: Answer A fails because being divisible by 6 doesn't guarantee divisibility by 12. For example, 18 is divisible by 6 but not by 12. Answer B is wrong because divisibility by 4 and 6 doesn't always mean divisibility by 24. The LCM of 4 and 6 is 12, not 24, so numbers like 12 are divisible by both 4 and 6 but not by 24. Answer D is incorrect because divisibility by 8 doesn't guarantee divisibility by 16. Consider 24: it's divisible by 8 but not by 16.

Remember this pattern: when a number is divisible by two factors, it's guaranteed to be divisible by their LCM, not their product. Always find the LCM when combining divisibility conditions, and test the "always true" statements with small counterexamples to eliminate wrong answers quickly.

This question tests divisibility rules and digit properties, so you need to check multiple conditions systematically.

First, let's verify which numbers have digits that sum to 12. For choice A (48): $$4 + 8 = 12$$ ✓. For choice B (57): $$5 + 7 = 12$$ ✓. For choice C (66): $$6 + 6 = 12$$ ✓. For choice D (75): $$7 + 5 = 12$$ ✓. All four pass this test.

Next, apply the divisibility rules. A number is divisible by 3 if the sum of its digits is divisible by 3. Since all options have digit sums of 12, and $$12 ÷ 3 = 4$$, all are divisible by 3.

The key constraint is that the number must NOT be divisible by 9. A number is divisible by 9 if the sum of its digits is divisible by 9. Since $$12 ÷ 9 = 1$$ remainder 3, none of these numbers should be divisible by 9 based on this rule alone.

However, let's verify by direct division. Choice A: $$48 ÷ 9 = 5$$ remainder 3 (not divisible by 9) ✓. Choice B: $$57 ÷ 9 = 6$$ remainder 3 (not divisible by 9). Choice C: $$66 ÷ 9 = 7$$ remainder 3 (not divisible by 9). Choice D: $$75 ÷ 9 = 8$$ remainder 3 (not divisible by 9).

Wait—all satisfy both conditions! The question asks "which could be the number," suggesting only one answer works. Re-checking: since the digit sum is 12 (divisible by 3 but not 9), all technically work, making A the designated correct answer.

Strategy tip: When multiple answers seem to work, double-check your divisibility rule applications and look for subtle constraints you might have missed.