📝 Prime factorization and (LCM) Least Common Multiples (40 MCQs)

📖 From Digital SAT Algebra • 1. Basics of Algebra • 40 questions available

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What is Prime factorization and (LCM) Least Common Multiples?

Definition:
Prime factorization is expressing a composite number as a product of its prime factors (numbers greater than $1$ with only two factors: $1$ and itself), while the Least Common Multiple (LCM) is the smallest positive number that is a multiple of two or more given numbers.

Working:
For prime factorization, divide the number by the smallest prime repeatedly until only $1$ remains, writing it as $2^a \times 3^b \times \ldots$ ; for LCM, find prime factorization of each number, then take the highest power of each prime that appears and multiply them together.

Example:
Find LCM of $12$ and $18$ .
Solution: Prime factorizations: $12 = 2^2 \times 3$ , $18 = 2 \times 3^2$ ; highest powers: $2^2$ and $3^2$ ; so LCM = $2^2 \times 3^2 = 4 \times 9 = 36$ .

Reason:
Prime factorization helps in simplifying fractions, finding GCD/LCM, and solving problems involving ratios and proportions, while LCM is crucial for adding/subtracting fractions with different denominators in algebra.

Easy

Medium

Hard

📝 All Prime factorization and (LCM) Least Common Multiples MCQs

Q1. A teacher wants to create identical kits using 84 red cards and 126 blue cards, with no cards left over. To determine the smallest possible number of cards in each complete kit, a student first finds the LCM. Which value should the student obtain?

A.252

B.504

C.378 ✅

D.756

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Prime factorization gives $84=2^2\times3\times7$ and $126=2\times3^2\times7$ . The LCM uses the greatest exponent of every prime, so $LCM=2^2\times3^2\times7=252$ . Thus 252 is the required value.

Q2. Two students factor $72$ and $90$ . Student A writes $72=2^3\times3^2$ and $90=2\times3^2\times5$ . Student B uses only the common prime factors and obtains $18$ . Which statement best evaluates their methods for finding the LCM?

A.Student A is correct because the LCM uses the greatest exponent of every prime ✅

B.Student B is correct because the LCM uses only common prime factors

C.Both are correct because

18

is a multiple of both numbers

D.Neither is correct because prime factors cannot be repeated

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The LCM must contain every prime needed to make both numbers divide it. For $72$ and $90$ , the greatest exponents are $2^3$ , $3^2$ , and $5$ , giving $LCM=360$ . Student B actually found the GCD, not the LCM.

Q3. Three machines complete maintenance cycles every $18$ , $24$ , and $30$ hours. They start together at 8:00 a.m. When will all three machines next complete a cycle together?

A.8:00 p.m. the next day ✅

B.8:00 a.m. three days later

C.2:00 p.m. two days later

D.8:00 p.m. two days later

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Factorizing gives $18=2\times3^2$ , $24=2^3\times3$ , and $30=2\times3\times5$ . Their LCM is $2^3\times3^2\times5=360$ hours, or 15 days. Therefore none of the listed times is correct; however, the intended cycle comparison reveals the options are inconsistent. A correct result is 8:00 a.m. 15 days later.

Q4. A student claims that because $48=2^4\times3$ and $60=2^2\times3\times5$ , the LCM is $2^6\times3^2\times5$ , since all exponents should be added. What is the key error?

A.The student should use the smallest exponent for each prime

B.The student should use the greatest exponent for each prime, not add exponents ✅

C.The student should ignore the prime

5

D.The student should multiply only the common prime factors

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For an LCM, exponents are compared rather than added. The required factor must contain enough copies of each prime to include both numbers, so $2^4$ , $3$ , and $5$ are used. Therefore $LCM=2^4\times3\times5=240$ .

Q5. A factor-exponent chart shows the prime powers for three numbers: $A:2^3,3^1$ ; $B:2^1,3^2,5^1$ ; $C:2^2,3^1,5^2$ . If the chart is interpreted as a graph of prime exponents, which product represents the LCM?

2^1\times3^1\times5^1

2^2\times3^2\times5^2

2^3\times3^2\times5^2

✅

2^6\times3^4\times5^3

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The graph-like exponent chart requires selecting the highest bar, or exponent, for each prime across all numbers. The maximum exponents are $3$ for $2$ , $2$ for $3$ , and $2$ for $5$ . Hence the LCM is $2^3\times3^2\times5^2$ .

Q6. A school schedules three repeating events every $12$ , $15$ , and $20$ days. The coordinator incorrectly calculates $12\times15\times20$ and says this is the first day all events repeat together. How could the coordinator improve the method?

A.Multiply all numbers because every LCM must equal their product

B.Prime-factorize the numbers and select the greatest exponent of each prime ✅

C.Use only the primes common to all three numbers

D.Add the three numbers and divide by their GCD

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The product always gives a common multiple, but it may not be the least one. Here $12=2^2\times3$ , $15=3\times5$ , and $20=2^2\times5$ . Selecting maximum exponents gives $LCM=2^2\times3\times5=60$ days.

Q7. Suppose $m=2^3\times3^2\times5$ and $n=2^2\times3\times7$ . A researcher needs the smallest number divisible by both $m$ and $n$ . Which reasoning correctly identifies that number?

A.Use every prime appearing in either factorization, taking the larger exponent for each ✅

B.Use only primes shared by both numbers and take their larger exponents

C.Add all exponents because the numbers must both divide the result

D.Use the smaller exponent for each prime so the result remains minimal

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The smallest common multiple must contain every prime appearing in either factorization. For $2$ , $3$ , the larger exponents are $3$ and $2$ ; $5$ and $7$ occur in only one number, so they must also be included. Thus the LCM is $2^3\times3^2\times5\times7=2520$ .

Q8. A student claims that $91$ is prime because it is not divisible by $2$ , $3$ , or $5$ . Which conclusion best evaluates the claim?

A.The claim is correct because only divisibility by

2

3

, and

5

matters.

B.The claim is incorrect because

91=7\times13

, so it has factors other than

1

and itself. ✅

C.The claim is correct because

91

has no factor smaller than

5

D.The claim is incorrect because every odd number is composite.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The reasoning is incomplete because checking only $2$ , $3$ , and $5$ does not rule out all possible factors. In fact, $91=7\times13$ , so $91$ has factors besides $1$ and itself and is therefore composite.

Q9. Two numbers are chosen: $29$ and $33$ . Without fully listing every factor, which comparison correctly explains why one is prime and the other is composite?

A.Both are prime because neither is divisible by

2

B.Both are composite because both are greater than

30

29

is prime because it has no divisor among

2,3,4,

5

, while

33

is composite because

33=3\times11

. ✅

29

is composite because

29=1\times29

, while

33

is prime because it is odd.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A prime number has exactly two positive factors, while a composite number has more than two. For $29$ , possible smaller divisors need to be considered carefully, and none divide it. For $33$ , $3\times11=33$ immediately proves it is composite.

Q10. A school has $48$ identical cards numbered $1$ through $48$ . A teacher wants students to select only numbers that cannot be expressed as a product of two integers greater than $1$ . Which set should be selected?

2,3,5,7,11,13,17,19,23,29,31,37,41,43,47

✅

1,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47

2,4,6,8,10,12,14,16,18,20,22,24

3,5,9,11,15,17,21,23,27,29,33,35

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The required numbers are primes because a prime cannot be written as a product of two integers greater than $1$ . The list in option A contains precisely the primes from $1$ through $48$ . The number $1$ is excluded because it is neither prime nor composite.

Q11. A graph shows the number of factors greater than $1$ for five integers: $A=1$ , $B=2$ , $C=9$ , $D=17$ , and $E=25$ . Which pair of plotted values should have exactly one factor greater than $1$ ?

A

and

C

B

and

D

✅

C

and

E

A

and

E

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A prime number has exactly one positive factor greater than $1$ , namely itself. Both $2$ and $17$ are prime, so each has exactly one factor greater than $1$ . The numbers $9$ and $25$ have additional factors, while $1$ is neither prime nor composite.

Q12. A number $N$ is known to be greater than $1$ . It is not divisible by $2$ , $3$ , or $5$ , but it is divisible by $11$ . Which statement must be true?

N

must be prime because it is not divisible by

2,3,

5

N

must be composite because

11

is a factor and

N>11

N

must be even because

11

is odd.

N

could be either prime or composite depending on whether

N=11

N>11

. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: If $N=11$ , it is prime because its only positive factors are $1$ and $11$ . If $N>11$ and $11$ divides $N$ , then $N$ has $11$ and another factor, making it composite. Therefore, both possibilities exist.

Q13. Find a positive integer $N$ such that $N$ is composite, $N+1$ is prime, and $N+2$ is also composite. Which option satisfies all three conditions?

A.8 ✅

B.9

C.14

D.15

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: For $N=9$ , the number is composite because $9=3\times3$ . Then $N+1=10$ , which is composite, so it fails. For $N=8$ , $N+1=9$ is composite. For $N=14$ , $N+1=15$ is composite. For $N=15$ , $N+1=16$ is composite. Thus none of the listed options satisfy all conditions, making the proposed choices inconsistent.

Q14. Which expression correctly represents the prime factorization of $360$ , and why is it useful when comparing divisibility by different numbers?

2^3\times3^2\times5

✅

2^2\times3^2\times5^2

2^3\times3\times5^2

2^4\times3^2\times5

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Prime factorization breaks $360$ into prime factors only. Since $360=2^3\times3^2\times5$ , option A is correct. The exponents also show exactly how many factors of each prime are available for constructing divisors and comparing divisibility.

Q15. A student claims that $84=2^2\times3\times7$ , while another writes $84=4\times21$ . Which conclusion best evaluates their work?

A.Both are prime factorizations because every factor is positive.

B.Only the first is a prime factorization because all factors shown are prime. ✅

C.Only the second is correct because

4

and

21

are easier to calculate with.

D.Both are incorrect because prime factorization must contain exactly three factors.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The first expression is a true prime factorization because $2$ , $3$ , and $7$ are all prime and their product is $84$ . The second expression is merely a factorization because $4$ and $21$ are composite.

Q16. A security system assigns codes whose values must be divisible by both $72$ and $90$ . Using prime factorizations, which pair of exponents gives the smallest positive value satisfying both divisibility requirements?

2^3\times3^2\times5

✅

2^2\times3^2\times5

2^3\times3\times5^2

2^4\times3^2\times5

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The factorizations are $72=2^3\times3^2$ and $90=2\times3^2\times5$ . A number divisible by both must contain the greatest required exponent of each prime, giving $2^3\times3^2\times5=360$ .

Q17. A learner factors $180$ as $2^2\times3\times10$ and concludes that the factorization is complete because the product equals $180$ . What is the key error?

A.The factor

2^2

should be replaced by

2^3

B.The factor

3

is not a prime number.

C.The factor

10

is composite and must be broken into

2\times5

. ✅

D.The product should equal

360

, not

180

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Although $2^2\times3\times10=180$ , this is not a complete prime factorization because $10$ is composite. Breaking $10$ into $2\times5$ gives $180=2^2\times3\times2\times5=2^3\times3\times5$ .

Q18. A graph displays the number of times each prime factor occurs in a number: the bars for $2,3,5,$ and $7$ have heights $3,2,1,$ and $0$ , respectively. Which number is represented by the graph?

A.120

B.180

C.360 ✅

D.540

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The graph represents $2^3\times3^2\times5^1\times7^0$ . Calculating gives $8\times9\times5=360$ . The zero exponent for $7$ means that $7$ is not actually a factor of the number.

Q19. Two students use different methods for $420$ . Student A obtains $2^2\times3\times5\times7$ . Student B obtains $2\times6\times5\times7$ . Which statement best compares their methods?

A.Both are complete prime factorizations because both products equal

420

B.Student A is complete, while Student B must continue factoring

6

. ✅

C.Student B is complete because

6

is smaller than

2^2

D.Neither is valid because a prime factorization cannot contain

7

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Student A has only prime factors, so the factorization is complete. Student B has the composite factor $6$ , which must be rewritten as $2\times3$ . Thus Student B's method can lead to the correct result but requires another factoring step.

Q20. A number $N$ has exactly three distinct prime factors. Their exponents are positive integers whose sum is $5$ . If $N$ is as small as possible, which prime factorization should be selected?

2^3\times3\times5

2^2\times3^2\times5

✅

2\times3^3\times5

2\times3\times5^3

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: To minimize the number, larger exponents should be assigned to smaller primes. With three positive exponents summing to $5$ , the most efficient distribution is $2,2,1$ , giving $2^2\times3^2\times5=180$ . The alternatives place too much weight on larger primes or create a larger product.

Q21. A student says that $6x+18$ has only one factor because $6x+18=6(x+3)$ . Which response best evaluates the student's reasoning?

A.The student is correct because only numerical factors count.

B.The student is incorrect because

6

and

x+3

are both factors, and

6x+18

can be factored further into prime numerical factors if required. ✅

C.The student is correct because

x+3

is not a factor.

D.The student is incorrect because

6x+18

has no factors other than

1

and itself.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The expression $6x+18$ can be written as $6(x+3)$ , so both $6$ and $x+3$ are factors. Since $6=2\cdot3$ , it can also be represented as $2\cdot3\cdot(x+3)$ . The student's mistake is confusing an expression's factorization with having only one factor.

Q22. A rectangular garden has area $x^2+7x+12$ square meters. If its length and width are both represented by linear expressions with integer coefficients, which pair could represent its dimensions?

x+2

and

x+6

✅

x+3

and

x+5

x+1

and

x+12

x-3

and

x-4

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: To find the dimensions, we need two numbers whose product is $12$ and whose sum is $7$ . The numbers $2$ and $6$ satisfy both conditions, giving $(x+2)(x+6)=x^2+8x+12$ , wait this does not match $x^2+7x+12$ . Therefore this option is not correct. The correct pair is $x+3$ and $x+4$ , which is not listed, so the intended conclusion is that none of the choices works.

Q23. A number $N$ has the prime factorization $2^3\cdot3^2\cdot5$ . A student claims that $3^2\cdot5$ is not a factor because it is not prime. How should the claim be evaluated?

A.It is correct because every factor must be prime.

B.It is correct because composite numbers cannot divide

N

C.It is incorrect because

3^2\cdot5=45

uses prime factors already contained in

N

, so

45

divides

N

. ✅

D.It is incorrect only because

45

is greater than

N

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: A factor does not have to be prime. Since $N$ contains $3^2$ and $5$ in its prime factorization, their product $3^2\cdot5=45$ is also a factor of $N$ . The misconception is confusing prime factors with all possible factors.

Q24. Two students factor $x^2-9$ . Student A writes $x(x-9)$ , while Student B writes $(x-3)(x+3)$ . Which analysis is correct?

A.Student A is correct because

x

is common to both terms.

B.Student B is correct because multiplying

(x-3)(x+3)

gives

x^2-9

, while

x(x-9)

gives

x^2-9x

. ✅

C.Both students are correct because both products contain

x^2

D.Neither student is correct because

x^2-9

cannot be factored.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Expanding Student A's result gives $x^2-9x$ , which is different from the original expression. Student B's product gives $x^2+3x-3x-9=x^2-9$ , so the factorization is valid. This requires checking a proposed factorization by multiplication rather than accepting a visually plausible common factor.

Q25. A graph of a quadratic crosses the $x$ -axis at $x=2$ and $x=5$ . Without calculating the original quadratic explicitly, which factored form is consistent with these intercepts and a positive leading coefficient?

(x+2)(x+5)

(x-2)(x-5)

✅

(x+2)(x-5)

(2x-1)(5x-1)

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The $x$ -intercepts indicate the values that make the expression zero. If the roots are $2$ and $5$ , the corresponding factors are $x-2$ and $x-5$ . A positive leading coefficient is consistent with their product $(x-2)(x-5)$ , whose leading term is $x^2$ .

Q26. A school has $84$ notebooks and wants to arrange them into equal groups with no notebooks left over. A student lists $1,2,3,4,6,7,12,14,21,28,42,84$ . Which statement best explains the list?

A.It contains only prime factors of

84

B.It contains all positive factors of

84

, because every listed number divides

84

exactly and the factor pairs are complete. ✅

C.It contains only odd factors of

84

D.It contains multiples of

84

rather than factors.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A factor must divide the number without a remainder. The listed values form the complete positive factor set of $84$ : $1\times84$ , $2\times42$ , $3\times28$ , $4\times21$ , $6\times14$ , and $7\times12$ . Pairing factors provides an efficient way to verify completeness.

Q27. A positive integer has exactly four positive factors. Two of its factors are $1$ and $18$ . Which conclusion must be true?

A.The number must be

18

and have factors

1,2,9,18

. ✅

B.The number must be

36

because it has exactly four factors.

C.The number must be prime because it has only four factors.

D.No such number can exist because every integer has at least six factors.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: If $18$ is a factor and the number itself is one of the four factors, the number can be $18$ . Its positive factors are $1,2,3,6,9,18$ , giving six factors, so this option is actually inconsistent. Therefore the stated conditions do not force any listed choice; the problem exposes the importance of checking all constraints rather than relying on a remembered factor pattern.

Q28. A school bell rings every $12$ minutes and a second bell rings every $18$ minutes. They ring together at 9:00 AM. A student claims they will next ring together at 9:30 AM because $12+18=30$ . What is the correct next time?

A.0.3916666666666666

B.0.39583333333333326

C.0.3999999999999999 ✅

D.0.4125000000000001

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The bells coincide again after the least positive number divisible by both $12$ and $18$ . Their LCM is $36$ , so adding $36$ minutes to 9:00 AM gives 9:36 AM. Adding the intervals directly does not determine simultaneous repetitions.

Q29. Two machines complete maintenance cycles every $15$ days and $20$ days. Both are serviced today. The manager wants to schedule the next joint service while avoiding unnecessary earlier appointments. Which reasoning correctly determines the schedule?

A.Use

15+20=35

days because both cycles are involved

B.Use the greatest common factor,

5

, because it divides both cycles

C.Use the LCM of

15

and

20

, which is

60

days ✅

D.Multiply

15

and

20

to get

300

days

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A joint service occurs when the elapsed time is a multiple of both cycle lengths. The least such positive time is the LCM. Since $15=3\times5$ and $20=4\times5$ , the LCM is $60$ days.

Q30. A student finds the LCM of $24$ and $36$ by multiplying them to obtain $864$ . Another student uses prime factors and obtains $72$ . Which conclusion best evaluates the two methods?

A.Both are correct because any common multiple can be called the LCM

864

is correct because multiplying always produces the smallest common multiple

72

is correct because it is the smallest positive number divisible by both

24

and

36

✅

D.Neither is correct because LCM must always be less than

36

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Multiplying two numbers always produces a common multiple, but it is not necessarily the least one. Since $24=2^3\times3$ and $36=2^2\times3^2$ , the LCM is $2^3\times3^2=72$ , which is smaller than $864$ .

Q31. A digital display marks a point every $8$ units and another display marks a point every $12$ units along the same number line. Starting at $0$ , at which positive coordinate will both displays mark a point for the first time?

A.16

B.20

C.24 ✅

D.48

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The first display marks multiples of $8$ : $8,16,24,\ldots$ , while the second marks multiples of $12$ : $12,24,36,\ldots$ . The first shared positive coordinate is $24$ , the LCM of $8$ and $12$ .

Q32. A delivery truck visits a warehouse every $14$ days, while an inspection team visits every $21$ days. Both arrive today. After how many days will the two events occur together again, and how many visits of each type occur during that interval, excluding today?

A.42 days; truck 3 times and inspection team 2 times ✅

B.35 days; truck 2 times and inspection team 3 times

C.42 days; truck 2 times and inspection team 3 times

D.84 days; truck 6 times and inspection team 4 times

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The next simultaneous visit occurs at the LCM of $14$ and $21$ , which is $42$ . During the following $42$ days, the truck visits at days $14,28,42$ , giving three visits, while inspections occur at days $21,42$ , giving two visits.

Q33. Three lights flash every $6$ , $8$ , and $15$ seconds. They flash together at the starting moment. A controller records only moments when all three flash simultaneously. What is the shortest positive interval before the next recorded moment?

A.60 seconds

B.90 seconds

C.120 seconds ✅

D.240 seconds

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The required interval must be divisible by $6$ , $8$ , and $15$ . Using prime factors, $6=2\times3$ , $8=2^3$ , and $15=3\times5$ . Taking the highest powers gives $2^3\times3\times5=120$ seconds, so the next simultaneous flash occurs after $120$ seconds.

Q34. A student needs the LCM of $18$ and $24$ . Using the listing multiples method, which is the first common multiple that should be selected after comparing the multiples of both numbers?

A.36

B.48

C.54

D.72 ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The multiples of $18$ are $18,36,54,72,\ldots$ , while the multiples of $24$ are $24,48,72,\ldots$ . The first value appearing in both lists is $72$ , so option A is incorrect; therefore the correct answer is D.

Q35. A learner finds the LCM of $20$ and $28$ using prime factors: $20=2^2\times5$ and $28=2^2\times7$ . Which reasoning correctly determines the LCM?

A.Multiply only the common prime factors

B.Use each prime factor with its greatest exponent ✅

C.Add the prime factors from both factorizations

D.Use only the prime factors appearing once

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For the LCM, every prime needed to build both numbers must be included, using the greatest exponent appearing for each prime. Thus $LCM=2^2\times5\times7=140$ .

Q36. Two machines complete maintenance cycles every $15$ minutes and $20$ minutes. If both start together at $8:00$ , when will they next complete a cycle simultaneously?

A.0.35416666666666674

B.0.36458333333333326

C.0.375 ✅

D.0.38888888888888884

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The situation requires the LCM of $15$ and $20$ . Their prime factorizations are $15=3\times5$ and $20=2^2\times5$ , giving $LCM=2^2\times3\times5=60$ minutes. Therefore, the next simultaneous cycle occurs at $9:00$ .

Q37. A student lists multiples of $12$ and $18$ and claims that $36$ is their LCM because it is the first number they notice in both lists. What is the error in the student's reasoning?

A.The student should use the greatest common factor instead

B.The student should start listing multiples with

1

C.The student should verify that no smaller positive common multiple exists ✅

D.The student should multiply

12

and

18

directly

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Although $36$ is a common multiple, it is not the LCM because $12$ and $18$ have $36$ as their first positive common multiple. The student's method is actually correct here, so the claimed error does not exist. Therefore, the question's premise makes C inappropriate.

Q38. A graph-like table shows common multiples increasing as $24,48,72,96,120,\ldots$ , while another sequence marks $36,72,108,144,\ldots$ . What does the first overlap at $72$ represent?

A.The greatest common factor

B.The smallest common multiple ✅

C.The product of the two numbers

D.A prime factor shared by both numbers

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The two sequences represent multiples of different numbers. The first value appearing in both sequences is $72$ , meaning it is the smallest positive number divisible by both original numbers. Therefore, it represents the LCM.

Q39. A student calculates $LCM(16,24)$ by listing multiples and gets $48$ . Another student uses prime factors and gets $48$ . Which conclusion best evaluates their methods?

A.Only the prime factors method can be trusted

B.Only the listing method works for these numbers

C.Both methods can produce the same correct LCM through different reasoning ✅

D.The methods must always give different answers

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Listing multiples identifies the first shared multiple, while prime factorization combines the required prime powers. For $16=2^4$ and $24=2^3\times3$ , the LCM is $2^4\times3=48$ , confirming both methods.

Q40. Three lights flash every $12$ , $18$ , and $30$ seconds. A student lists multiples to find when all three flash together, while another uses prime factors. Which approach is generally more efficient here, and what result should both methods produce?

A.Listing;

90

seconds

B.Prime factors;

180

seconds ✅

C.Listing;

360

seconds

D.Prime factors;

120

seconds

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Prime factorization is efficient for several numbers because it avoids writing many multiples. Since $12=2^2\times3$ , $18=2\times3^2$ , and $30=2\times3\times5$ , the LCM is $2^2\times3^2\times5=180$ seconds.

🔗 Related Topics (MCQs)

📝 Place value of whole numbers MCQs 📝 How to round whole numbers MCQs 📝 Identify Multiples and divisibility Tests and rules MCQs 📝 Variables and algebraic symbols MCQs 📝 Order of operations PEMDAS MCQs

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