Multiples and factors

1. Test 51: digit sum 5 + 1 = 6 is divisible by 3, so 51 = 3 \times 17. Not prime.
2. Test 57: digit sum 5 + 7 = 12 is divisible by 3, so 57 = 3 \times 19. Not prime.
3. Test 91: 91 \div 7 = 13, so 91 = 7 \times 13. Not prime.
4. Test 59: not even, digit sum 14 (not divisible by 3), does not end in 0 or 5, and 59 \div 7 \approx 8.4. Since 8^2 = 64 > 59, we only need to test primes up to 7. Nothing divides it.

The prime number is 59.

Check: the only factor pair of 59 is 1 \times 59. ✓

Tip: To test whether n is prime, only try primes whose square is at most n. Odd numbers like 51, 57 and 91 look prime but are not — always try 3 and 7.

1. Multiples of 8: 8, 16, 24, 32, 40, 48, \ldots
2. Check each one for divisibility by 6:
   • 8 \div 6 — no
   • 16 \div 6 — no
   • 24 \div 6 = 4 — yes!
3. The first common multiple is 24.

Prime-factor method: 6 = 2 \times 3, 8 = 2^3. Take every prime with its largest power: 2^3 \times 3 = 24.

Answer: 24.

Check: 24 = 6 \times 4 = 8 \times 3. ✓

Tip: Multiplying the two numbers (6 \times 8 = 48) always gives a common multiple, but not always the least one. It works only when the numbers share no factor other than 1.

1. 29: not divisible by 2, 3 (digit sum 11) or 5. Since 6^2 = 36 > 29, no more tests are needed. Prime.
2. 37: not divisible by 2, 3 (digit sum 10) or 5; 7^2 = 49 > 37. Prime.
3. 87: digit sum 8 + 7 = 15, which is divisible by 3. So 87 = 3 \times 29. Composite.
4. 41: not divisible by 2, 3 (digit sum 5) or 5; 7^2 = 49 > 41. Prime.

Answer: 87.

Check: 3 \times 29 = 87, so 87 has the factors 1, 3, 29, 87 — more than two. ✓

Tip: The digit-sum test for 3 catches many numbers that look prime: 51, 57, 87, 111. Always try it before deciding a number is prime.

1. Multiples of 15: 15, 30, 45, 60, \ldots
2. Test each for divisibility by 9:
   • 15 \div 9 — no
   • 30 \div 9 — no
   • 45 \div 9 = 5 — yes!
3. The first common multiple is 45.

Prime-factor method: 9 = 3^2 and 15 = 3 \times 5. Take each prime with its largest power: 3^2 \times 5 = 45.

Answer: 45.

Check: 45 = 9 \times 5 = 15 \times 3. ✓

Tip: 9 \times 15 = 135 is a common multiple, but because 9 and 15 share the factor 3, a smaller one exists. Divide the product by the GCF: 135 \div 3 = 45.

1. Hot dogs come in groups of 10, so possible totals are multiples of 10: 10, 20, 30, 40, 50, \ldots
2. Buns come in groups of 8: 8, 16, 24, 32, 40, 48, \ldots
3. The smallest number on both lists is 40, the LCM of 10 and 8.
4. Prime check: 10 = 2 \times 5, 8 = 2^3, so LCM = 2^3 \times 5 = 40.

She should buy 40 hot dogs (4 packages of hot dogs and 5 packages of buns).

Check: 4 \times 10 = 40 and 5 \times 8 = 40. ✓

Tip: Phrases like "the same number", "at the same time again" or "smallest amount to buy" in a word problem point to the LCM. Splitting things into equal groups points to the GCF.

1. Each bouquet is identical, so the number of bouquets must divide 48 and 60 — it is a common factor.
2. We want the greatest number of bouquets, so find the GCF.
3. 48 = 2^4 \times 3 and 60 = 2^2 \times 3 \times 5.
4. Shared primes with smaller powers: 2^2 \times 3 = 12.

She can make 12 bouquets, each with 48 \div 12 = 4 red and 60 \div 12 = 5 white roses.

Check: 12 \times 4 = 48 and 12 \times 5 = 60, so no roses are left. ✓

Tip: Read what the question asks for. The GCF (12) is the number of bouquets; 4 and 5 are the roses per bouquet. Many wrong answers come from answering a different quantity.

1. Split 180 = 18 \times 10.
2. Split 18 = 2 \times 9 = 2 \times 3 \times 3.
3. Split 10 = 2 \times 5.
4. Collect the primes: 2 \times 2 \times 3 \times 3 \times 5.
5. Write with exponents: 2^2 \times 3^2 \times 5.

Answer: 2^2 \times 3^2 \times 5.

Check: 4 \times 9 \times 5 = 36 \times 5 = 180. ✓

Tip: After you finish, multiply your answer back out. It must equal the original number and every factor must be prime. Both checks catch most mistakes.

A number is divisible by 6 when it is divisible by both 2 and 3.

1. 2,145 bottles: digit sum 2+1+4+5 = 12 ✓ for 3, but it is odd ✗ for 2. Not divisible by 6.
2. 1,334 bottles: even ✓, but digit sum 1+3+3+4 = 11 ✗ for 3. Not divisible by 6.
3. 1,236 bottles: even ✓ and digit sum 1+2+3+6 = 12 ✓. Divisible by 6.
4. 1,516 bottles: even ✓, digit sum 1+5+1+6 = 13 ✗. Not divisible by 6.

Answer: 1,236 bottles (206 full cases).

Check: 206 \times 6 = 1{,}236. ✓

Tip: For 6, test both 2 and 3. Ending in 6 does not matter — 1{,}516 ends in 6 but fails the test for 3.

1. The number of kits must divide 36 and 54, and be as large as possible, so find the GCF.
2. 36 = 2^2 \times 3^2 and 54 = 2 \times 3^3. GCF = 2 \times 3^2 = 18. She can make 18 kits.
3. Pencils per kit: 36 \div 18 = 2.
4. (Erasers per kit: 54 \div 18 = 3.)

Answer: 2 pencils in each kit.

Check: 18 kits \times 2 pencils = 36 and 18 \times 3 erasers = 54; nothing is left over. ✓

Tip: GCF problems hide three different numbers: the GCF (number of groups) and the amount of each item per group. Underline what the question asks for before choosing.

1. The piece length must divide 72 and 120 and be as long as possible → GCF.
2. 72 = 2^3 \times 3^2 and 120 = 2^3 \times 3 \times 5. GCF = 2^3 \times 3 = 24 inches.
3. Pieces from the first roll: 72 \div 24 = 3.
4. Pieces from the second roll: 120 \div 24 = 5.
5. Total pieces: 3 + 5 = 8.

Answer: 8 pieces (each 24 inches long).

Check: 8 \times 24 = 192 = 72 + 120, so all the ribbon is used. ✓

Tip: The GCF is the length of each piece. The question asks for the number of pieces, so finish with total length \div piece length.

Look at what each situation asks for:

1. Packing 24 apples and 30 oranges into identical bags: splitting into equal groups → GCF (6 bags).
2. Two sprinklers that turn on every 15 and every 20 minutes: when do they turn on together again? Times are multiples of each number → LCM (60 minutes).
3. Cutting 18-ft and 27-ft boards into the longest equal pieces: the piece length divides both → GCF (9 ft).
4. Total time of a 15-minute and a 20-minute task: just add → 35 minutes. No GCF or LCM.

Answer: the sprinklers.

Check: 60 \div 15 = 4 and 60 \div 20 = 3, and no smaller time works (30 \div 20 and 45 \div 20 are not whole). ✓

Tip: Ask yourself: is the answer smaller than the given numbers (pieces, groups) → GCF; or larger (a later time, a bigger amount) → LCM.

Test each statement:

1. "1 is a prime number." 1 has only one factor (itself). A prime needs exactly two factors. False.
2. "Every odd number is prime." 9 = 3 \times 3 and 15 = 3 \times 5 are odd but not prime. False.
3. "2 is the only even prime number." 2 has factors 1 and 2, so it is prime. Every other even number has 2 as an extra factor, so it is composite. True.
4. "51 is prime because it is odd and does not end in 5." 5 + 1 = 6 is divisible by 3, so 51 = 3 \times 17. False.

Answer: 2 is the only even prime number.

Check: even numbers 4, 6, 8, 10, \ldots all have the factors 1, 2 and themselves — at least three factors. ✓

Tip: To disprove a statement that says "every" or "always", one counterexample is enough. To prove a number is not prime, find one factor besides 1 and itself.

1. Prime factorizations:
   • 18 = 2 \times 3^2
   • 30 = 2 \times 3 \times 5
   • 42 = 2 \times 3 \times 7
2. Primes in all three: 2 and 3.
3. Smallest power of each: 2^1 and 3^1.
4. GCF = 2 \times 3 = 6.

Answer: 6.

Check: 18 \div 6 = 3, 30 \div 6 = 5, 42 \div 6 = 7. The results 3, 5, 7 share no common factor except 1. ✓

Tip: 2 and 3 are also common factors, but they are not the greatest. Multiply all the shared primes together to get the GCF.

We need n > 50 and \text{GCF}(n, 36) = 12. Note 36 = 2^2 \times 3^2.

1. 48: 48 = 2^4 \times 3, so \text{GCF}(48, 36) = 2^2 \times 3 = 12 ✓ — but 48 < 50 ✗.
2. 54: 54 = 2 \times 3^3, so \text{GCF}(54, 36) = 2 \times 3^2 = 18 ✗.
3. 60: 60 = 2^2 \times 3 \times 5, so \text{GCF}(60, 36) = 2^2 \times 3 = 12 ✓, and 60 > 50 ✓.
4. 72: 72 = 2^3 \times 3^2, so \text{GCF}(72, 36) = 2^2 \times 3^2 = 36 ✗.

Answer: n = 60.

Check: 60 \div 12 = 5 and 36 \div 12 = 3; 5 and 3 share no factor except 1, so 12 really is the greatest. ✓

Tip: Being a multiple of 12 is not enough — 72 is a multiple of 12, but it shares the bigger factor 36 with 36. Check every condition the question gives.

1. The tile side must divide 84 and 120, and be as large as possible → find the GCF.
2. 84 = 2^2 \times 3 \times 7 and 120 = 2^3 \times 3 \times 5. GCF = 2^2 \times 3 = 12. Each tile is 12 in by 12 in.
3. Tiles along the 84-inch side: 84 \div 12 = 7.
4. Tiles along the 120-inch side: 120 \div 12 = 10.
5. Total tiles: 7 \times 10 = 70.

Answer: 70 tiles.

Check (area): floor area 84 \times 120 = 10{,}080 in²; one tile is 12^2 = 144 in²; 10{,}080 \div 144 = 70. ✓

Tip: Multi-step GED problems often stop at a value you need along the way (here, 12). Before choosing, reread the question: it asks for the number of tiles, not the tile size.

1. Fern days are multiples of 4 after June 1; cactus days are multiples of 10.
2. The first number of days on both lists is the LCM of 4 and 10.
3. 4 = 2^2 and 10 = 2 \times 5, so LCM = 2^2 \times 5 = 20 days.
4. Count forward: June 1 + 20 days = June 21.

Answer: June 21.

Check: from June 1 to June 21 is 20 days; 20 \div 4 = 5 fern waterings and 20 \div 10 = 2 cactus waterings. ✓

Tip: Add the number of days to the start date: June 1 + 20 = June 21, not June 20. Product trap: 4 \times 10 = 40 days would skip the June 21 match.

1. The number of teams must divide 45 and 60 and be as large as possible → GCF.
2. 45 = 3^2 \times 5 and 60 = 2^2 \times 3 \times 5. GCF = 3 \times 5 = 15 teams.
3. Men per team: 45 \div 15 = 3. Women per team: 60 \div 15 = 4.
4. People per team: 3 + 4 = 7.

Answer: 7 people on each team.

Check: 15 teams \times 7 people = 105 = 45 + 60. ✓

Tip: Multi-step GCF problems usually ask for something after the GCF. Here the GCF (15) is the number of teams; the question wants the team size.

1. Step 1: 2^2 \times 3 = 4 \times 3 = 12. Correct.
2. Step 2: 2 \times 3^2 = 2 \times 9 = 18. Correct.
3. Step 3: Taking the smaller powers gives the GCF, not the LCM. For the LCM, take each prime with its larger power: 2^2 \times 3^2. This is the first error.
4. Step 4: follows from Step 3. The correct LCM is 2^2 \times 3^2 = 36.

Answer: Step 3.

Check: 6 cannot be a multiple of 12 — it is smaller than 12. And 36 \div 12 = 3, 36 \div 18 = 2. ✓

Tip: Memory aid: GCF = Give up the extra powers (smaller); LCM = Larger powers. A quick sanity check: the LCM is never smaller than the larger number.

1. Use the rule \text{GCF} \times \text{LCM} = a \times b.
2. Substitute: 6 \times 72 = 18 \times b, so 432 = 18b.
3. Solve: b = 432 \div 18 = 24.
4. Verify with prime factors: 18 = 2 \times 3^2 and 24 = 2^3 \times 3.
   • GCF = 2 \times 3 = 6 ✓
   • LCM = 2^3 \times 3^2 = 72 ✓

Answer: 24.

Check: multiples of 24: 24, 48, 72 — and 72 \div 18 = 4, so 72 is the first common multiple. ✓

Tip: When a question gives you some facts and asks for a missing number, try each answer choice — plugging in is fast here. 12 gives the right GCF but the wrong LCM (36).

1. A common multiple of 6 and 15 must be a multiple of \text{LCM}(6, 15). 6 = 2 \times 3 and 15 = 3 \times 5, so LCM = 2 \times 3 \times 5 = 30.
2. Test each choice:
   • 45: 45 \div 30 = 1.5 ✗ (it is not even, so not a multiple of 6)
   • 90: 90 \div 30 = 3 ✓, but digit sum 9 → multiple of 9 ✗
   • 120: 120 \div 30 = 4 ✓, digit sum 3 → not a multiple of 9 ✓
   • 180: 180 \div 30 = 6 ✓, but digit sum 9 → multiple of 9 ✗

Answer: 120.

Check: 120 = 6 \times 20 = 15 \times 8, and 120 \div 9 = 13.\overline{3}. ✓

Tip: "Multiple of both" means "multiple of the LCM". Then check the extra condition — the GED often adds a "but not" to catch students who stop early.

1. Remove the staff items: 78 - 3 = 75 sandwiches and 108 - 3 = 105 bottles.
2. The number of tables must divide 75 and 105 and be as large as possible → GCF.
3. 75 = 3 \times 5^2 and 105 = 3 \times 5 \times 7. GCF = 3 \times 5 = 15.
4. Each table gets 75 \div 15 = 5 sandwiches and 105 \div 15 = 7 bottles.

Answer: 15 tables.

Check: 15 \times 5 + 3 = 78 sandwiches and 15 \times 7 + 3 = 108 bottles. ✓

Tip: Read every condition before computing. Using the original 78 and 108 gives a GCF of 6 — a correct calculation for the wrong numbers.

1. The side of each square must divide 150 and 210 and be as large as possible → GCF.
2. 150 = 2 \times 3 \times 5^2 and 210 = 2 \times 3 \times 5 \times 7. GCF = 2 \times 3 \times 5 = 30 ft.
3. Plots along each side: 150 \div 30 = 5 and 210 \div 30 = 7.
4. Number of plots: 5 \times 7 = 35.
5. Rent: 35 \times \$45 = \$1{,}575.

Answer: $1,575.

Check (area): field area 150 \times 210 = 31{,}500 ft²; one plot is 30^2 = 900 ft²; 31{,}500 \div 900 = 35 plots. ✓

Tip: This problem has three stages: GCF \rightarrow count \rightarrow money. Each wrong answer stops at, or skips, one of them. Write the stages down before you start.

Test each statement with examples:

1. \text{GCF}(a, b) \times \text{LCM}(a, b) = a \times b. Example 6 and 8: 2 \times 24 = 48 = 6 \times 8 ✓. Example 4 and 8: 4 \times 8 = 32 = 4 \times 8 ✓. This always holds, because the GCF takes the smaller power of each prime and the LCM takes the larger, so together they use every prime factor exactly once from each number. True.
2. \text{LCM}(a, b) = a \times b. For 6 and 8: LCM = 24 \ne 48. False.
3. The GCF of two different primes is the smaller prime. For 3 and 7: the only common factor is 1. False.
4. The LCM is always greater than both numbers. For 4 and 8: LCM = 8, which equals b. False.

Answer: \text{GCF}(a, b) \times \text{LCM}(a, b) = a \times b.

Check: a = 12, b = 18: \text{GCF} = 6, \text{LCM} = 36, and 6 \times 36 = 216 = 12 \times 18. ✓

Tip: For "always true" questions, hunt for one counterexample to knock out each wrong choice. Pairs where one number divides the other (like 4 and 8) and pairs of primes are great test cases.