GMAT Quantitative Quiz: Gcf And Lcm

What this quiz covers

This quiz focuses on Gcf And Lcm, giving you a quick way to practice the rules, question types, and explanations that matter most for GMAT Quantitative.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 7
2. 14 (correct answer)
3. 21
4. 28

Explanation: When you encounter a problem about creating identical groups with no items left over, you're dealing with a greatest common divisor (GCD) problem. The key insight is that the number of bracelets must evenly divide both the red and blue bead quantities. To find the maximum number of identical bracelets, you need to calculate the GCD of 42 and 70. Using prime factorization: $42 = 2 \times 3 \times 7$ and $70 = 2 \times 5 \times 7$. The common factors are $2 \times 7 = 14$, so the GCD is 14. This means she can make 14 bracelets, each containing $42 ÷ 14 = 3$ red beads and $70 ÷ 14 = 5$ blue beads. Looking at the wrong answers: Choice (A) 7 is a factor of both numbers but not the greatest one—you'd only be using half the available beads efficiently. Choice (C) 21 doesn't divide evenly into 70 ($70 ÷ 21 = 3.33...$), so you'd have leftover blue beads. Choice (D) 28 doesn't divide evenly into either 42 or 70, making it impossible to create identical bracelets with no beads remaining. Study tip: When you see "greatest possible number" with "no items left over," immediately think GCD. Don't get distracted by factors that work for only one quantity—the answer must work for all given numbers simultaneously. Practice finding GCDs quickly using either prime factorization or the Euclidean algorithm.

Question 2

1. $18x^{2}y^{2}$
2. $6x^{3}y^{2}$
3. $12x^{2}y^{3}$
4. $6x^{2}y^{2}$ (correct answer)

Explanation: When you see a question asking for the greatest common factor (GCF) of algebraic expressions, you need to find the largest expression that divides evenly into both terms. This means taking the GCF of the numerical coefficients and the lowest power of each variable that appears in both expressions. To find the GCF of $18x^{3}y^{2}$ and $30x^{2}y^{4}$, work with each part separately: For the coefficients: The GCF of 18 and 30 is 6 (since $18 = 2 \times 3^2$ and $30 = 2 \times 3 \times 5$, sharing factors of 2 and 3). For the x terms: You have $x^3$ and $x^2$. The GCF uses the lowest power, which is $x^2$. For the y terms: You have $y^2$ and $y^4$. The GCF uses the lowest power, which is $y^2$. Therefore, the GCF is $6x^{2}y^{2}$, which is choice D. Choice A ($18x^{2}y^{2}$) incorrectly uses 18 instead of 6 as the coefficient. Choice B ($6x^{3}y^{2}$) mistakenly uses $x^3$ (the higher power) instead of $x^2$. Choice C ($12x^{2}y^{3}$) has two errors: using 12 instead of 6 for the coefficient, and $y^3$ instead of $y^2$. Strategy tip: Always take the GCF of coefficients and the minimum power for each variable. A quick check is to verify that your answer divides evenly into both original expressions.

Question 3

1. $(14,42)$
2. $(21,28)$ (correct answer)
3. $(12,49)$
4. $(28,35)$

Explanation: When you encounter problems involving both least common multiple (LCM) and greatest common factor (GCF), remember the fundamental relationship: $\text{LCM}(m,n) \times \text{GCF}(m,n) = m \times n$. This gives you a powerful constraint to check potential answers. Given that $\text{LCM}(m,n) = 84$ and $\text{GCF}(m,n) = 7$, we know that $m \times n = 84 \times 7 = 588$. Additionally, since the GCF is 7, both $m$ and $n$ must be multiples of 7. Let's verify each option by checking if $m \times n = 588$ and confirming the LCM and GCF: For choice B: $(21,28)$. First, $21 \times 28 = 588$ ✓. To find the LCM, factor: $21 = 3 \times 7$ and $28 = 4 \times 7 = 2^2 \times 7$. The LCM takes the highest power of each prime: $2^2 \times 3 \times 7 = 84$ ✓. The GCF is the common factor: $7$ ✓. Choice A: $(14,42)$ gives $14 \times 42 = 588$, but $\text{LCM}(14,42) = 42$ and $\text{GCF}(14,42) = 14$. Choice C: $(12,49)$ gives $12 \times 49 = 588$, but $\text{GCF}(12,49) = 1$ since 12 and 49 share no common factors. Choice D: $(28,35)$ gives $28 \times 35 = 980 \neq 588$. Study tip: Always use the product relationship $\text{LCM} \times \text{GCF} = m \times n$ as your first filter—it quickly eliminates impossible answers before you calculate LCM and GCF directly.

Question 4

1. 92
2. 122
3. 182 (correct answer)
4. 242

Explanation: This is a remainder problem that tests your understanding of modular arithmetic and the Chinese Remainder Theorem. When you see a question asking for a number that leaves the same remainder when divided by different values, you're looking for a pattern involving the least common multiple (LCM). Since we need a number that leaves remainder 2 when divided by 5, 9, and 12, we can write this as: $n = 5k + 2$, $n = 9j + 2$, and $n = 12i + 2$ for some integers $k$, $j$, and $i$. This means $n - 2$ must be divisible by all three numbers. First, find the LCM of 5, 9, and 12. Since $5 = 5$, $9 = 3^2$, and $12 = 2^2 \times 3$, the LCM is $2^2 \times 3^2 \times 5 = 180$. Therefore, $n - 2$ must be a multiple of 180, so $n = 180m + 2$ for some positive integer $m$. For the smallest positive value, use $m = 1$: $n = 180(1) + 2 = 182$. Let's verify: $182 ÷ 5 = 36$ remainder $2$, $182 ÷ 9 = 20$ remainder $2$, and $182 ÷ 12 = 15$ remainder $2$. ✓ Choice A (92) gives different remainders when divided by these numbers. Choice B (122) also fails to give remainder 2 for all three divisors. Choice D (242) would work since $242 = 180 + 62 = 180(1) + 62$, but this doesn't follow our pattern and isn't the smallest solution. Remember: when a number leaves the same remainder $r$ when divided by multiple values, subtract $r$ and find the LCM of those divisors.

Question 5

1. 2
2. 3
3. 4 (correct answer)
4. 5

Explanation: When you encounter LCM problems with a specific target value, you need to find all numbers that produce that exact least common multiple with the given number. To solve $\operatorname{lcm}(n,15)=60$, start by analyzing the prime factorizations. Since $15 = 3 \cdot 5$ and $60 = 2^2 \cdot 3 \cdot 5$, the LCM formula tells us that $\operatorname{lcm}(n,15)$ takes the highest power of each prime from both numbers. For $\operatorname{lcm}(n,15) = 60 = 2^2 \cdot 3 \cdot 5$, we need:

• The factor $2^2$ must come from $n$ (since 15 has no factors of 2)
• The factor $3$ can come from either $n$ or 15
• The factor $5$ can come from either $n$ or 15

Question 6

1. 150
2. 160
3. 175 (correct answer)
4. 200

Explanation: When you encounter LCM (least common multiple) problems, remember that the LCM of two numbers must be divisible by both numbers and contain all prime factors from each number at their highest powers. To find all values of $t$ where $\text{lcm}(20,t) = 100$, start by finding the prime factorizations: $20 = 2^2 \times 5$ and $100 = 2^2 \times 5^2$. Since $\text{lcm}(20,t) = 100$, the value $t$ must have prime factors that, when combined with those of 20, produce exactly $2^2 \times 5^2$. The LCM takes the highest power of each prime from both numbers, so $t$ can only contain factors of 2 and 5. For the LCM to equal 100, $t$ must contribute the factor $5^2$ (since 20 only has $5^1$), and can have at most $2^2$ (since 100 has $2^2$). Therefore, $t$ must be of the form $2^a \times 5^2$ where $0 \leq a \leq 2$. This gives us three possibilities: $t = 5^2 = 25$, $t = 2 \times 5^2 = 50$, and $t = 2^2 \times 5^2 = 100$. You can verify: $\text{lcm}(20,25) = 100$, $\text{lcm}(20,50) = 100$, and $\text{lcm}(20,100) = 100$. The sum is $25 + 50 + 100 = 175$. Answer (A) 150 likely comes from missing one of the values. Answer (B) 160 might result from incorrectly including 60. Answer (D) 200 could come from including an invalid value like 125. Strategy tip: Always find prime factorizations first in LCM problems, then systematically determine what prime factors the unknown value must contain.

Question 7

1. 420
2. 455
3. 525 (correct answer)
4. 560

Explanation: When you see a problem asking for a positive integer divisible by two numbers, you need to find their least common multiple (LCM). The smallest value of $n$ that's divisible by both 45 and 70 will be $\text{LCM}(45, 70)$. To find the LCM, first determine the prime factorizations: $45 = 3^2 \times 5$ and $70 = 2 \times 5 \times 7$. The LCM takes the highest power of each prime factor that appears: $\text{LCM}(45, 70) = 2^1 \times 3^2 \times 5^1 \times 7^1 = 2 \times 9 \times 5 \times 7 = 630$. Therefore, the least possible value of $n$ is 630, making $n - 105 = 630 - 105 = 525$. This confirms answer choice C. Let's examine why the other options are incorrect. Choice A (420) would mean $n = 525$. However, $525 = 3 \times 5^2 \times 7$ is not divisible by 45 since it lacks sufficient factors of 3. Choice B (455) gives $n = 560 = 2^4 \times 5 \times 7$, which isn't divisible by 45 because it has no factor of 3. Choice D (560) means $n = 665 = 5 \times 7 \times 19$, which is divisible by neither 45 nor 70. Strategy tip: When finding numbers divisible by multiple values, always calculate the LCM using prime factorization. Take the highest power of each prime factor that appears in any of the numbers. This guarantees the smallest number divisible by all given values.

Question 8

1. 48
2. 60 (correct answer)
3. 72
4. 120

Explanation: When you encounter problems involving greatest common factor (GCF) and least common multiple (LCM) with given product constraints, remember the fundamental relationship: for any two positive integers $a$ and $b$, we have $a \times b = \text{GCF}(a,b) \times \text{LCM}(a,b)$. Given that the two integers have a product of 720 and GCF of 12, you can substitute directly into this formula: $720 = 12 \times \text{LCM}$. Solving for the LCM: $\text{LCM} = \frac{720}{12} = 60$. To verify this makes sense, if two numbers have GCF 12, they can be written as $12m$ and $12n$ where $m$ and $n$ are relatively prime. Their product is $(12m)(12n) = 144mn = 720$, so $mn = 5$. Since 5 is prime, we must have $m = 1, n = 5$ (or vice versa), giving us the numbers 12 and 60. Indeed, $\text{LCM}(12, 60) = 60$. Looking at the wrong answers: (A) 48 would give a product of $12 \times 48 = 576 \neq 720$. (C) 72 would yield $12 \times 72 = 864 \neq 720$. (D) 120 would produce $12 \times 120 = 1440 \neq 720$. Strategy tip: Memorize the relationship $a \times b = \text{GCF}(a,b) \times \text{LCM}(a,b)$. This formula appears frequently on the GMAT and provides a direct path to the solution when you know three of the four values.

Question 9

1. 90
2. 120
3. 180 (correct answer)
4. 270

Explanation: When you encounter a problem about cycles repeating simultaneously, you're looking for the least common multiple (LCM) of the cycle lengths. This tells you when all cycles will align and restart together. To find the LCM of 9, 12, and 15, start by finding the prime factorization of each number:

• $9 = 3^2$
• $12 = 2^2 \times 3$
• $15 = 3 \times 5$

Question 10

1. 420
2. 840 (correct answer)
3. 1,260
4. 1,680

Explanation: When you encounter least common multiple (LCM) problems on the GMAT, the most efficient approach is prime factorization. The LCM is found by taking the highest power of each prime factor that appears in either number. First, let's find the prime factorization of each number:

• $84 = 4 \times 21 = 4 \times 3 \times 7 = 2^2 \times 3 \times 7$
• $120 = 8 \times 15 = 8 \times 3 \times 5 = 2^3 \times 3 \times 5$

• Highest power of 2: $2^3 = 8$
• Highest power of 3: $3^1 = 3$
• Highest power of 5: $5^1 = 5$
• Highest power of 7: $7^1 = 7$

Question 11

1. 162
2. 324
3. 486 (correct answer)
4. 648

Explanation: When you encounter GCD and LCM problems, remember the fundamental relationship: for any two integers, $\text{GCD} \times \text{LCM} = p \times q$. This equation is your key to solving these problems systematically. Given that $\text{GCD}(p,q) = 18$ and $q = 162$, you can write $p = 18a$ and $q = 18b$ where $a$ and $b$ are coprime (share no common factors other than 1). Since $q = 162 = 18 \times 9$, we have $b = 9$. Using the fundamental relationship: $18 \times \text{LCM} = p \times 162$, so $\text{LCM} = \frac{p \times 162}{18} = 9p$. Since $p = 18a$, we get $\text{LCM} = 9 \times 18a = 162a$. For this to work, $a$ must be coprime to $9$. Since $9 = 3^2$, the value $a$ cannot be divisible by $3$. Now check each answer:

• A) 162: If $\text{LCM} = 162$, then $a = 1$. Since $\gcd(1,9) = 1$, this works.
• B) 324: If $\text{LCM} = 324$, then $a = 2$. Since $\gcd(2,9) = 1$, this works.
• C) 486: If $\text{LCM} = 486$, then $a = 3$. Since $\gcd(3,9) = 3 \neq 1$, this violates our coprime requirement.
• D) 648: If $\text{LCM} = 648$, then $a = 4$. Since $\gcd(4,9) = 1$, this works.

Question 12

1. 24
2. 36
3. 48
4. 72 (correct answer)

Explanation: When you encounter problems involving both LCM (least common multiple) and GCD (greatest common divisor), remember the fundamental relationship: for any two positive integers, $\text{lcm}(a,b) \times \gcd(a,b) = a \times b$. Given that $\text{lcm}(r,s) = 144$ and $r = 2^3 \cdot 3^2 = 8 \cdot 9 = 72$, you can use this relationship to find $\gcd(r,s)$. Rearranging the formula: $\gcd(r,s) = \frac{r \times s}{\text{lcm}(r,s)} = \frac{72s}{144} = \frac{s}{2}$. Since $\text{lcm}(r,s) = 144$, you need $s$ such that the LCM works out correctly. First, factor 144: $144 = 2^4 \cdot 3^2$. For the LCM to equal $2^4 \cdot 3^2$, and knowing $r = 2^3 \cdot 3^2$, the value $s$ must contribute the factor $2^4$ (since $\max(3,4) = 4$ for the power of 2). The simplest case is when $s = 2^4 \cdot 3^a$ where $a \leq 2$. If $s = 144$, then $\gcd(72, 144) = 72$. Looking at the answer choices: (A) 24 would require a different factorization that doesn't align with our LCM constraint. (B) 36 similarly doesn't work with the given LCM relationship. (C) 48 also fails to satisfy the fundamental relationship between the given values. The answer is (D) 72, which satisfies all the given conditions. Strategy tip: Always use the relationship $\text{lcm}(a,b) \times \gcd(a,b) = a \times b$ when you have three of these four values and need to find the fourth.

Question 13

1. 1260 (correct answer)
2. 1890
3. 2520
4. 3780

Explanation: We need the LCM of 84, 90, and 126. Prime factorizations: 84 = 2² × 3 × 7, 90 = 2 × 3² × 5, 126 = 2 × 3² × 7. LCM = 2² × 3² × 5 × 7 = 4 × 9 × 5 × 7 = 1260. Verification: 1260 ÷ 84 = 15, 1260 ÷ 90 = 14, 1260 ÷ 126 = 10 (all integers). Choice B represents 1.5 × LCM. Choice C represents 2 × LCM. Choice D represents 3 × LCM.

Question 14

1. 2160 (correct answer)
2. 1440
3. 2880
4. 3600

Explanation: For any two positive integers, GCF(a,b) × LCM(a,b) = ab. Therefore, ab = 12 × 180 = 2160. Choice B incorrectly calculates 12 × 120 instead of 12 × 180. Choice C represents 2 × 12 × 180, incorrectly doubling the product. Choice D represents 12² × 25, a common error from misapplying the relationship.

Question 15

1. 3.6 square meters
2. 36.0 square meters
3. 22.5 square meters
4. 9.0 square meters (correct answer)

Explanation: This problem tests your understanding of least common multiples (LCM) and unit conversions. When a floor can be tiled with different square sizes without cutting, the floor's dimensions must be common multiples of all the tile side lengths. To find the minimum floor area, you need the LCM of 15, 20, and 25 cm. Start by finding the prime factorization of each number: 15 = 3 × 5, 20 = 2² × 5, and 25 = 5². The LCM takes the highest power of each prime factor: 2² × 3 × 5² = 4 × 3 × 25 = 300 cm. The smallest possible floor is a 300 cm × 300 cm square. Converting to meters: 300 cm = 3 meters, so the area is 3 × 3 = 9 square meters. Let's examine why the other answers are wrong. Answer A (3.6 square meters) might result from incorrectly calculating 300 × 300 = 90,000 cm² and converting poorly to square meters. Answer B (36.0 square meters) could come from finding the LCM correctly but making an error like 300 cm = 30 meters instead of 3 meters. Answer C (22.5 square meters) might result from finding an incorrect LCM or making calculation errors during the conversion process. Study tip: When dealing with tiling problems, always look for the LCM of the given dimensions. Remember that the minimum area question asks for the smallest possible dimensions, which will be LCM × LCM for a square floor. Double-check your unit conversions—this is a common source of errors on the GMAT.

Question 16

1. 3 revolutions
2. 12 revolutions
3. 4 revolutions (correct answer)
4. 144 revolutions

Explanation: When you encounter gear problems on the GMAT, you're dealing with cyclical patterns that repeat at regular intervals. The key insight is finding when both gears return to their starting configuration simultaneously. To solve this, you need to determine the least common multiple (LCM) of the number of teeth on each gear. The smaller gear has 36 teeth and the larger has 48 teeth. First, find the prime factorization: 36 = 2² × 3² and 48 = 2⁴ × 3¹. The LCM is 2⁴ × 3² = 144. This means the same teeth will align again after a total of 144 teeth have passed the alignment point on either gear. Since the smaller gear has 36 teeth, it completes $\frac{144}{36} = 4$ full revolutions when this happens. Looking at the wrong answers: Choice A (3 revolutions) would mean only 108 teeth have passed (3 × 36), which isn't enough for both gears to return to their starting positions. Choice B (12 revolutions) represents 432 teeth passing (12 × 36), which is three complete cycles beyond the first realignment. Choice D (144 revolutions) confuses the total number of teeth that must pass (144) with the number of revolutions of the smaller gear. The correct answer is C: 4 revolutions. Strategy tip: In gear problems, always find the LCM of the teeth counts first, then divide by the number of teeth on the gear you're asked about. Don't confuse the LCM itself with the number of revolutions.

Question 17

1. $2^4 \times 3^3 \times 5^3 \times 7^2$
2. $2 \times 3^3 \times 5^3 \times 7^2$ (correct answer)
3. $2^3 \times 3^2 \times 5^3 \times 7^2$
4. $2^5 \times 3^4 \times 5^3 \times 7^2$

Explanation: GCF(m,n) = 2³ × 3² (minimum powers of common factors). LCM(m,n) = 2⁴ × 3⁴ × 5³ × 7² (maximum powers of all factors). Therefore, LCM/GCF = (2⁴ × 3⁴ × 5³ × 7²)/(2³ × 3²) = 2¹ × 3² × 5³ × 7² = 2 × 3³ × 5³ × 7². Choice A uses maximum powers throughout. Choice C incorrectly uses minimum powers for some factors. Choice D adds exponents instead of finding the ratio.