Factors and Multiples Practice Worksheet
Unit 1 Practice Part 1: Prime, Composite & Vocabulary
Factors and Multiples Practice
Name:
Date:
Problem 1 Prime & Composite Numbers
4.OA.B.4
a. Is 33 a prime number or a composite number? Explain or show your reasoning.
b. Is 37 a prime number or a composite number? Explain or show your reasoning.
Problem 2 Factors & Multiples Vocabulary
4.OA.B.4
Select all true statements.
A
24 is a multiple of 6.
B
18 is a factor of 9.
C
90 is a multiple of 5.
D
The only factor pair of 25 is 1 and 25.
E
The factor pairs of 18 are 1 and 18, 2 and 9, and 3 and 6.
Turn to Page 2 for solutions and 2 ways to solve each problem → Page 1 of 4 (Practice)
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Solutions & 2 Ways to Solve: Problems 1 & 2
Step-by-step explanations and mathematical models
Answer Key
Problem 1 Prime & Composite Numbers Solutions
4.OA.B.4
1a. Is 33 prime or composite? Answer: Composite
Way 1: Factor Pairs List
\(1 \times 33 = 33\) and \(3 \times 11 = 33\). Since 33 has 4 factors (1, 3, 11, 33), it has more than one factor pair, making it composite.
Way 2: Rectangular Array / Division
\(33 \div 3 = 11\). 33 tiles can be arranged in a rectangle of 3 rows of 11 tiles with no remainder. A number that forms arrays other than \(1 \times 33\) is composite.
1b. Is 37 prime or composite? Answer: Prime
Way 1: Systematic Factor Testing
37 is not divisible by 2 (odd), 3 (\(3 \times 12 = 36\)), 4, 5 (doesn't end in 0/5), or 6 (\(6 \times 6 = 36\)). Its only factor pair is \(1 \times 37\).
Way 2: Array Model Proof
With 37 tiles, the only rectangle possible is a single straight line: \(1 \times 37\) or \(37 \times 1\). A number whose only array has a width of 1 is prime.
Problem 2 Factors and Multiples Vocabulary Solutions
True Statements: A, C, E
✓ A is TRUE: 24 is a multiple of 6 because \(6 \times 4 = 24\).
✗ B is FALSE: 18 is greater than 9; 18 is a multiple of 9, while 9 is a factor of 18.
✓ C is TRUE: 90 is a multiple of 5 because \(5 \times 18 = 90\) (it ends in 0).
✗ D is FALSE: 25 also has the factor pair 5 and 5, since \(5 \times 5 = 25\).
✓ E is TRUE: The factor pairs of 18 are exactly 1 and 18, 2 and 9, and 3 and 6.
Way 1: Equation & Definition Test
"\(x\) is a multiple of \(y\)" means \(y \times \text{whole number} = x\). Test with division: \(24 \div 6 = 4\) (whole number → True); \(90 \div 5 = 18\) (whole number → True).
Way 2: Listing & Skip Counting
Skip count by 6: 6, 12, 18, 24 (A is true). Factors of 9: {1, 3, 9} (18 is not on list, so B is false). Factors of 25: {1, 5, 25} (disproves D).
Turn to Page 3 for Problems 3 & 4 → Page 2 of 4 (Solutions)
Unit 1 Practice Part 2: Factor Pairs & Word Problem
Factors and Multiples Practice
Name:
Problem 3 Finding All Factor Pairs
4.OA.B.4
Find all factor pairs of 48.
Problem 4 Common Multiples Word Problem
4.OA.B.4
Elena is organizing craft beads for friendship bracelets. The total number of beads never changes, but the number of bracelets does.
a. The beads can be divided equally among 4 bracelets. Could there be 40 beads? Explain or show your reasoning.
b. The beads can be divided equally among 3 bracelets or 4 bracelets. Could there be 40 beads? Explain or show your reasoning.
c. The beads can be divided equally among 3 bracelets, 4 bracelets, or 5 bracelets. How many beads could there be? Explain or show your reasoning.
Turn to Page 4 for solutions and 2 ways to solve each problem → Page 3 of 4 (Practice)
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Solutions & 2 Ways to Solve: Problems 3 & 4
Step-by-step explanations and mathematical models
Answer Key
Problem 3 All Factor Pairs of 48
4.OA.B.4
Answer: 1 and 48, 2 and 24, 3 and 16, 4 and 12, and 6 and 8.
Way 1: Factor Rainbow / Partner Pairs
Start with 1 and work inward:
\(1 \times 48 = 48 \rightarrow\) 1 and 48
\(2 \times 24 = 48 \rightarrow\) 2 and 24
\(3 \times 16 = 48 \rightarrow\) 3 and 16
\(4 \times 12 = 48 \rightarrow\) 4 and 12
\(6 \times 8 = 48 \rightarrow\) 6 and 8
5 and 7 do not divide 48 evenly. Stop at 8!
Way 2: Systematic T-Chart Division
Test integers sequentially by division:
\(48 \div 1 = 48\) | \(48 \div 2 = 24\)
\(48 \div 3 = 16\) | \(48 \div 4 = 12\)
\(48 \div 5 = 9\text{ R }3\) (No) | \(48 \div 6 = 8\)
\(48 \div 7 = 6\text{ R }6\) (No) | Stop (8 already tested)
All 5 factor pairs systematically found.
Problem 4 Common Multiples in Context Solutions
4.OA.B.4
4a. Divided equally among 4 bracelets. Could there be 40 beads? Answer: Yes
Way 1 (Division): \(40 \div 4 = 10\) beads per bracelet with no beads left over.
Way 2 (Multiples): 40 is a multiple of 4 because \(4 \times 10 = 40\) (skip-counting by 4 lands on 40).
4b. Divided equally among 3 or 4 bracelets. Could there be 40 beads? Answer: No
Way 1 (Remainder): \(40 \div 3 = 13\text{ R }1\). There is a leftover bead, so beads cannot be divided equally.
Way 2 (Multiples Test): Multiples of 3 near 40 are 36, 39, 42. 40 is skipped; sum of digits (\(4+0=4\)) is not divisible by 3.
4c. Divided equally among 3, 4, or 5 bracelets. How many beads? Answer: 60 beads (or 120, etc.)
Way 1 (Common Multiples List): Multiples of 4 and 5 are 20, 40, 60. Testing 3: \(20 \div 3 = 6\text{ R }2\); \(40 \div 3 = 13\text{ R }1\); \(60 \div 3 = 20\). 60 is divisible by all three!
Way 2 (Factor Product / LCM): Multiply \(3 \times 4 \times 5 = 60\). Check divisibility: \(60 \div 3 = 20\), \(60 \div 4 = 15\), \(60 \div 5 = 12\). All share equally with zero remainder.
Illustrative Mathematics Grade 4 • Unit 1 Aligned Page 4 of 4 (Solutions)
