To prove that 51 is not prime, we need to find a factor other than 1 and 51 itself. Let's check if 51 is divisible by 3 using the sum of digits rule: 5 + 1 = 6, and since 6 is divisible by 3, then 51 is divisible by 3. Indeed, 51 ÷ 3 = 17, so 3 × 17 = 51. This means 3 is a factor of 51, proving it is composite. The smallest factor that proves 51 is not prime is 3.

Let's check each number: 77 = 7 × 11 (composite); 79 is not divisible by 2,3,5,7 and since √79 < 9, we only need to check up to 7, so 79 is prime; 81 = 3⁴ = 3 × 27 (composite); 83 is not divisible by 2,3,5,7 and since √83 < 10, we only need to check up to 9, so 83 is prime. Therefore, exactly 2 numbers (79 and 83) are prime.

A number with exactly 2 prime factors (counting repeats) means either p×q where p,q are distinct primes, or p² where p is prime. Let's check numbers 40-50: 41(prime), 42=2×3×7(3 factors), 43(prime), 44=2²×11(3 factors), 45=3²×5(3 factors), 46=2×23(2 factors), 47(prime), 48=2⁴×3(5 factors), 49=7²(2 factors). So 46=2×23 and 49=7×7 both have exactly 2 prime factors counting repeats.

We need the smallest composite number that is not divisible by 2, 3, 5, or 7. Let's check 121 = 11 × 11: Is 121 divisible by 2? No (it's odd). By 3? 1+2+1=4, not divisible by 3. By 5? No (doesn't end in 0 or 5). By 7? 121÷7=17.28..., so no. Since 121 = 11² is composite but not divisible by 2, 3, 5, or 7, it proves Tommy wrong. 121 is the smallest such number.

First, find all factors of 60: 1,2,3,4,5,6,10,12,15,20,30,60. Now identify which are multiples of 4 (numbers that can be divided by 4 evenly): 4÷4=1✓, 12÷4=3✓, 20÷4=5✓, 60÷4=15✓. The factors of 60 that are also multiples of 4 are: 4, 12, 20, and 60. Therefore, 4 factors of 60 are multiples of 4.

When you see a question about factors and multiples, you need to systematically list all factors first, then carefully identify which ones meet the removal condition.

To find all factors of $$72$$, look for pairs of numbers that multiply to give $$72$$: $$1 \times 72 = 72$$, $$2 \times 36 = 72$$, $$3 \times 24 = 72$$, $$4 \times 18 = 72$$, $$6 \times 12 = 72$$, and $$8 \times 9 = 72$$. This gives us the complete list: $$1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72$$ (which is indeed $$12$$ factors total).

Next, identify which factors are multiples of $$8$$. A multiple of $$8$$ means $$8$$ times some whole number: $$8 \times 1 = 8$$, $$8 \times 2 = 16$$, $$8 \times 3 = 24$$, etc. From our factor list, only $$8$$, $$24$$, and $$72$$ are multiples of $$8$$ (since $$72 = 8 \times 9$$).

Removing these three factors ($$8$$, $$24$$, and $$72$$) leaves us with: $$1, 2, 3, 4, 6, 9, 12, 18, 36$$. Counting these gives us $$9$$ remaining factors.

Choice A ($$8$$ factors) miscounts the remaining factors. Choice B ($$11$$ factors) likely only removed $$8$$ itself, forgetting that $$24$$ and $$72$$ are also multiples of $$8$$. Choice C ($$10$$ factors) probably missed removing one of the multiples of $$8$$.

The answer is D: $$9$$ factors remain.

Remember: when removing multiples of a number, check every factor systematically—multiples can be larger numbers that aren't obvious at first glance.

For a number to have exactly 4 factors, it must be either the cube of a prime (like 8 = 2³) or the product of two distinct primes (like 6 = 2×3). Multiples of 6 between 30 and 50 are: 36, 42, 48. Let's check their factors: 36 has factors 1,2,3,4,6,9,12,18,36 (9 factors); 42 has factors 1,2,3,6,7,14,21,42 (8 factors); 48 has factors 1,2,3,4,6,8,12,16,24,48 (10 factors). None of these has exactly 4 factors.

This question tests 4th grade ability to find all factor pairs for whole numbers 1-100, recognize that a number is a multiple of each of its factors, determine if a number is a multiple of a given one-digit number, and determine whether a number is prime or composite (CCSS.4.OA.4). A factor is a whole number that divides another number evenly with no remainder—if a × b = n, then both a and b are factors of n, and (a, b) is a factor pair. To find all factor pairs, systematically check: does 1 divide n? does 2? does 3? and so on up to √n, stopping when factors start repeating. A prime number has exactly 2 factors (1 and itself), while a composite number has more than 2 factors. The number 1 is special—it's neither prime nor composite (only 1 factor). For the number 40, to find which of 5,6,8,10 are factors: check divisibility—40÷5=8 (yes), 40÷6≈6.67 (no), 40÷8=5 (yes), 40÷10=4 (yes). Choice B is correct because it lists all that divide evenly: 5,8,10. This demonstrates systematic factor finding and understanding of prime/composite definitions. Choice D represents including non-factors, which happens when students include numbers that don't divide evenly. To help students: For finding factor pairs, use systematic approach—start with 1 (always works), check 2, 3, 4, 5, etc., stop when you start seeing the same pairs reversed. Example for 24: 1×24 ✓, 2×12 ✓, 3×8 ✓, 4×6 ✓, 5 doesn't work, 6×4 (already have 4×6, stop). For prime/composite, count factors: exactly 2 = prime, more than 2 = composite. Remember: 1 is neither (only 1 factor), 2 is only even prime. For multiples, divide: if no remainder, it IS a multiple (48 ÷ 6 = 8 R0 ✓). Use arrays to visualize: 12 objects can arrange as 1×12, 2×6, 3×4—each arrangement shows a factor pair. Connect: if f is a factor of n, then n is a multiple of f (inverse relationship). Watch for: missing factor pairs, including non-factors, calling 1 prime, thinking all odd numbers are prime (9, 15, 21 are composite), confusing factors with multiples, and not checking systematically.

This question tests 4th grade ability to find all factor pairs for whole numbers 1-100, recognize that a number is a multiple of each of its factors, determine if a number is a multiple of a given one-digit number, and determine whether a number is prime or composite (CCSS.4.OA.4). A factor is a whole number that divides another number evenly with no remainder—if a × b = n, then both a and b are factors of n, and (a, b) is a factor pair. To find all factor pairs, systematically check: does 1 divide n? does 2? does 3? and so on up to √n, stopping when factors start repeating. A prime number has exactly 2 factors (1 and itself), while a composite number has more than 2 factors. The number 1 is special—it's neither prime nor composite (only 1 factor). For the number 18, to find all factors: systematically check divisibility starting from 1 up to 18, listing all that divide evenly: 1,2,3,6,9,18. Choice A is correct because all factors are listed (checked all numbers 1 through 18) and none are missing or extra. This demonstrates systematic factor finding and understanding of prime/composite definitions. Choice B represents including non-factors, which happens when students include numbers that don't divide evenly. To help students: For finding factor pairs, use systematic approach—start with 1 (always works), check 2, 3, 4, 5, etc., stop when you start seeing the same pairs reversed. Example for 24: 1×24 ✓, 2×12 ✓, 3×8 ✓, 4×6 ✓, 5 doesn't work, 6×4 (already have 4×6, stop). For prime/composite, count factors: exactly 2 = prime, more than 2 = composite. Remember: 1 is neither (only 1 factor), 2 is only even prime. For multiples, divide: if no remainder, it IS a multiple (48 ÷ 6 = 8 R0 ✓). Use arrays to visualize: 12 objects can arrange as 1×12, 2×6, 3×4—each arrangement shows a factor pair. Connect: if f is a factor of n, then n is a multiple of f (inverse relationship). Watch for: missing factor pairs, including non-factors, calling 1 prime, thinking all odd numbers are prime (9, 15, 21 are composite), confusing factors with multiples, and not checking systematically.

This question tests 4th grade ability to find all factor pairs for whole numbers 1-100, recognize that a number is a multiple of each of its factors, determine if a number is a multiple of a given one-digit number, and determine whether a number is prime or composite (CCSS.4.OA.4). A factor is a whole number that divides another number evenly with no remainder—if a × b = n, then both a and b are factors of n, and (a, b) is a factor pair. To find all factor pairs, systematically check: does 1 divide n? does 2? does 3? and so on up to √n, stopping when factors start repeating. A prime number has exactly 2 factors (1 and itself), while a composite number has more than 2 factors. The number 1 is special—it's neither prime nor composite (only 1 factor). For the number 45, to find all factors: systematically check divisibility starting from 1 up to √45≈6.7, listing 1,3,5,9,15,45 since they divide evenly. Choice A is correct because all factors are listed (checked all numbers 1 through √45) and none are missing. Choice B represents including non-factors, which happens when students include numbers that don't divide evenly. To help students: For finding factor pairs, use systematic approach—start with 1 (always works), check 2, 3, 4, 5, etc., stop when you start seeing the same pairs reversed. Example for 24: 1×24 ✓, 2×12 ✓, 3×8 ✓, 4×6 ✓, 5 doesn't work, 6×4 (already have 4×6, stop). For prime/composite, count factors: exactly 2 = prime, more than 2 = composite. Remember: 1 is neither (only 1 factor), 2 is only even prime. For multiples, divide: if no remainder, it IS a multiple (48 ÷ 6 = 8 R0 ✓). Use arrays to visualize: 12 objects can arrange as 1×12, 2×6, 3×4—each arrangement shows a factor pair. Connect: if f is a factor of n, then n is a multiple of f (inverse relationship). Watch for: missing factor pairs, including non-factors, calling 1 prime, thinking all odd numbers are prime (9, 15, 21 are composite), confusing factors with multiples, and not checking systematically.