A student needs to solve 4 × ? = 28. She tries subtracting: 28 − 4 = 24, and writes 24 as her answer. What is wrong with her approach?
ASubtraction is the right operation, but she should keep subtracting 4 until she reaches 0
BAn unknown factor problem asks what multiplies by 4 to give 28 — that's a division relationship, not subtraction. She should think: 28 ÷ 4 = ?
CShe should add 4 repeatedly until she reaches 28
DSubtraction works here, but she made an arithmetic error; she needs to subtract again
Unknown factor problems are division in disguise. '4 × ? = 28' asks: 'What number, multiplied by 4, gives 28?' This is exactly 28 ÷ 4 = ?. Subtracting once gives 24, which is not the answer — there is no meaningful connection between a single subtraction and finding a missing factor. The most direct path is to recall the multiplication fact: 4 × 7 = 28, so ? = 7.
Question 2 Multiple Choice
A student correctly solves ? × 6 = 42 and gets 7. Her classmate says she's wrong because 'the unknown must come second.' Is the classmate correct?
AYes — unknown factors must always appear in the second position for the equation to be solvable
BNo — multiplication is commutative, so ? × 6 and 6 × ? are equivalent; 7 is correct either way
CYes — when the unknown comes first, it is a different type of problem with a different answer
DNo, but she must rewrite it as 6 × ? = 42 and solve again to confirm
The commutative property guarantees that ? × 6 = 6 × ? for any value of ?. The position of the unknown — first or second — does not change the answer or the strategy. Verify: 7 × 6 = 42 ✓. If the first-position setup feels uncomfortable, mentally rewrite ? × 6 = 42 as 6 × ? = 42, then solve. The answer remains 7. Practicing both positions builds the flexibility that algebra will later require.
Question 3 True / False
The equation 3 × ? = 18 and the division problem 18 ÷ 3 = ? are two different questions with different answers.
TTrue
FFalse
Answer: False
They are the same question expressed two different ways. Both ask: 'What number, when multiplied by 3, gives 18?' The unknown factor approach uses multiplication knowledge (3 × 6 = 18). The division approach computes 18 ÷ 3 = 6. Both arrive at ? = 6 because an unknown factor problem IS division — the link between multiplication and division is the entire key insight of this topic.
Question 4 True / False
Knowing the multiplication fact 9 × 7 = 63 immediately gives you the answer to the unknown factor problem ? × 7 = 63.
TTrue
FFalse
Answer: True
? × 7 = 63 asks for the number that, multiplied by 7, gives 63. Since 9 × 7 = 63, the answer is ? = 9. The fact family connects everything: 9 × 7 = 63 and 7 × 9 = 63 both answer unknown factor problems for the same three numbers. This is why building multiplication fact fluency is the direct path to solving unknown factor problems — no additional procedures are needed.
Question 5 Short Answer
Explain why an unknown factor problem like 5 × ? = 40 is the same as a division problem. What is the connection between the two?
Think about your answer, then reveal below.
Model answer: An unknown factor problem asks: 'What number multiplied by 5 gives 40?' Division asks exactly the same thing: 40 ÷ 5 = ?. Both seek the same unknown. The connection is that multiplication and division are inverse operations. If 5 × 8 = 40, then 40 ÷ 5 = 8, and ? = 8 in 5 × ? = 40. They are three ways of expressing the same relationship among the numbers 5, 8, and 40.
This equivalence is the bridge from multiplication to division and eventually to algebra. Recognizing that 'find the missing factor' and 'divide' describe the same operation is a conceptual milestone. The same fact family (5 × 8 = 40, 8 × 5 = 40, 40 ÷ 5 = 8, 40 ÷ 8 = 5) captures all the relationships among those three numbers — a pattern that extends into every equation-solving situation students will encounter later.