$$2,547 ÷ 8 = 318 \text{ R } 3$$ because $$318 × 8 = 2,544$$ and $$2,547 - 2,544 = 3$$. Choice B has an incorrect quotient (308 instead of 318). Choice C has an impossible remainder (11 is greater than the divisor 8). Choice D has an incorrect quotient that's too large.

$$1,845 ÷ 5 = 369$$ exactly because $$369 × 5 = 1,845$$. There is no remainder. Choice B has the wrong quotient and impossible remainder. Choice C has an impossible remainder (5 cannot be a remainder when dividing by 5). Choice D has the wrong quotient.

$$596 ÷ 8 = 74$$ remainder $$4$$ because $$74 × 8 = 592$$ and $$596 - 592 = 4$$. So Maria gets $$500 + 74 = 574$$ remainder $$4$$ for the complete problem. Choice B has wrong quotient. Choice C has wrong remainder. Choice D has both wrong quotient and remainder.

Since $$2,049 × 3 = 6,147$$ exactly matches the dividend, Roberto's division is completely correct with quotient 2,049 and remainder 0. Choice B incorrectly suggests a remainder when none exists. Choice C is redundant since remainder 0 is typically not written. Choice D provides an incorrect quotient.

This question tests 4th grade ability to find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, properties of operations, and/or the relationship between multiplication and division (CCSS.4.NBT.6). Division separates a total into equal groups—you can think of it as sharing (divide total among groups to find size of each) or grouping (divide total into groups of certain size to find how many groups). The standard algorithm divides place by place from left to right: divide, multiply, subtract, bring down next digit, repeat. When a number doesn't divide evenly, there is a remainder (the amount left over), which must always be less than the divisor. To divide 3,276 ÷ 9, students use place value to break into parts and divide each, finding quotient 364. Choice A is correct because using the standard algorithm gives quotient 364 with no remainder, and checking with multiplication confirms: 9 × 364 = 3,276. Choice C represents stopping too early or a place value error, which happens when students forget to divide all places. To help students: Use standard algorithm steps: (1) Divide—how many times does divisor go into current dividend? (2) Multiply—quotient digit times divisor, (3) Subtract—from current dividend, (4) Bring down next digit, (5) Repeat. Connect to multiplication: what times 9 equals 3,276? Check answers with multiplication. Use place value: 3,276 ÷ 9 = (2,700 ÷ 9) + (540 ÷ 9) + (36 ÷ 9) = 300 + 60 + 4 = 364. Watch for forgetting to bring down digits or weak facts.

This question tests 4th grade ability to find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, properties of operations, and/or the relationship between multiplication and division (CCSS.4.NBT.6). Division separates a total into equal groups—you can think of it as sharing (divide total among groups to find size of each) or grouping (divide total into groups of certain size to find how many groups). The standard algorithm divides place by place from left to right: divide, multiply, subtract, bring down next digit, repeat. When a number doesn't divide evenly, there is a remainder (the amount left over), which must always be less than the divisor. To divide $945 \div 9$ into groups of 9, students use place value to break into parts and divide each, finding quotient 105 with no remainder. Choice B is correct because using multiplication: $9 \times 105 = 945$, which demonstrates understanding of division process and place value. Choice A represents a basic division fact error, which happens when students make an error in one division step. To help students: Use standard algorithm steps: (1) Divide—how many times does divisor go into current dividend? (2) Multiply—quotient digit times divisor, (3) Subtract—from current dividend, (4) Bring down next digit, (5) Repeat. For word problems with remainders, consider context—do you need another group (round up) or ignore leftovers (round down)? Check answers with multiplication: quotient × divisor + remainder should equal dividend ($105 \times 9 + 0 = 945$ ✓).

This question tests 4th grade ability to find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, properties of operations, and/or the relationship between multiplication and division (CCSS.4.NBT.6). Division separates a total into equal groups—you can think of it as sharing (divide total among groups to find size of each) or grouping (divide total into groups of certain size to find how many groups). The standard algorithm divides place by place from left to right: divide, multiply, subtract, bring down next digit, repeat. When a number doesn't divide evenly, there is a remainder (the amount left over), which must always be less than the divisor. To divide 4,096 ÷ 8, students use the standard algorithm: divide each place starting from left, finding quotient 512 with no remainder. Choice A is correct because using multiplication: 8 × 512 = 4,096, which demonstrates understanding of division process and place value. Choice B represents a basic division fact error, such as miscalculating in one step, which happens when students make an error in one division step. To help students: Use standard algorithm steps: (1) Divide—how many times does divisor go into current dividend? (2) Multiply—quotient digit times divisor, (3) Subtract—from current dividend, (4) Bring down next digit, (5) Repeat. Emphasize checking with multiplication: quotient × divisor + remainder should equal dividend (512 × 8 + 0 = 4,096 ✓), and use place value: 4,096 ÷ 8 = (3,200 ÷ 8) + (800 ÷ 8) + (96 ÷ 8) = 400 + 100 + 12 = 512.

This question tests 4th grade ability to find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, properties of operations, and/or the relationship between multiplication and division (CCSS.4.NBT.6). Division separates a total into equal groups—you can think of it as sharing (divide total among groups to find size of each) or grouping (divide total into groups of certain size to find how many groups). The standard algorithm divides place by place from left to right: divide, multiply, subtract, bring down next digit, repeat. When a number doesn't divide evenly, there is a remainder (the amount left over), which must always be less than the divisor. To solve 7 × ? = 1,176, students use the relationship to multiplication: 7 times what equals 1,176?, or divide 1,176 ÷ 7, finding quotient 168 with no remainder. Choice C is correct because using multiplication: 7 × 168 = 1,176, which demonstrates understanding of division process and place value. Choice A represents a basic division fact error, which happens when students make an error in one division step. To help students: Connect to multiplication: 1,176 ÷ 7 asks 'what times 7 equals 1,176?' Use standard algorithm steps: (1) Divide—how many times does divisor go into current dividend? (2) Multiply—quotient digit times divisor, (3) Subtract—from current dividend, (4) Bring down next digit, (5) Repeat. Check answers with multiplication: quotient × divisor + remainder should equal dividend (168 × 7 + 0 = 1,176 ✓), and watch for weak division facts preventing fluency.

This question tests 4th grade ability to find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, properties of operations, and/or the relationship between multiplication and division (CCSS.4.NBT.6). Division separates a total into equal groups—you can think of it as sharing (divide total among groups to find size of each) or grouping (divide total into groups of certain size to find how many groups). The standard algorithm divides place by place from left to right: divide, multiply, subtract, bring down next digit, repeat. When a number doesn't divide evenly, there is a remainder (the amount left over), which must always be less than the divisor. To divide 1,509 ÷ 5, students use the standard algorithm: divide each place starting from left, finding quotient 301 with remainder 4. Choice A is correct because using multiplication: 5 × 301 + 4 = 1,505 + 4 = 1,509, which demonstrates understanding of division process and place value. Choice C represents a remainder ≥ divisor (invalid), which happens when students don't divide enough times (remainder should be less than divisor). To help students: Use standard algorithm steps: (1) Divide—how many times does divisor go into current dividend? (2) Multiply—quotient digit times divisor, (3) Subtract—from current dividend, (4) Bring down next digit, (5) Repeat. Emphasize remainder must be LESS than divisor—if remainder equals or exceeds divisor, divide one more time, and check with multiplication: quotient × divisor + remainder should equal dividend (301 × 5 + 4 = 1,509 ✓).

This question tests 4th grade ability to find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, properties of operations, and/or the relationship between multiplication and division (CCSS.4.NBT.6). Division separates a total into equal groups—you can think of it as sharing (divide total among groups to find size of each) or grouping (divide total into groups of certain size to find how many groups). The standard algorithm divides place by place from left to right: divide, multiply, subtract, bring down next digit, repeat. When a number doesn't divide evenly, there is a remainder (the amount left over), which must always be less than the divisor. To divide 672 ÷ 6, students use the standard algorithm: divide each place starting from left, finding the quotient. Choice B is correct because using the standard algorithm, 6 goes into 67 eleven times (6×11=66, subtract to get 1), bring down 2 to make 12, 6 goes into 12 two times (6×2=12, subtract to get 0), but adjusting places gives 112; checking with multiplication confirms 6×112=672. Choice A represents a basic division fact error, like miscounting in one step, which happens when students make an error in one division step. To help students: Use standard algorithm steps: (1) Divide—how many times does divisor go into current dividend? (2) Multiply—quotient digit times divisor, (3) Subtract—from current dividend, (4) Bring down next digit, (5) Repeat. Use place value: break into parts and divide each, and check answers with multiplication.