SHSAT Math Quiz: Factors And Multiples

20 questions · exam conditions

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Factors And MultiplesQuestion 1 of 20

If $p$ is the greatest prime such that $p^{2}$ is a factor of $180$ , what is the value of $p$ ?

SHSAT Math Quiz

SHSAT Math Quiz: Factors And Multiples

Practice Factors And Multiples in SHSAT Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Factors And Multiples, giving you a quick way to practice the rules, question types, and explanations that matter most for SHSAT Math.

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.

All questions

Question 1

If $p$ is the greatest prime such that $p^{2}$ is a factor of $180$ , what is the value of $p$ ?

2
3 (correct answer)
5
7

Explanation: When you see a question asking for the greatest prime whose square divides a number, you need to find the prime factorization and identify which prime appears with the highest even exponent. Let's find the prime factorization of 180. Start by dividing by the smallest primes:

180 = 4 \times 45 = 4 \times 9 \times 5 = 2^2 \times 3^2 \times 5^1

180 = 2^2 \times 3^2 \times 5^1

. For a prime

p

to have

p^2

as a factor of 180, the prime must appear with an exponent of at least 2 in the factorization. Looking at our factorization, both 2 and 3 appear with exponent 2, while 5 appears with exponent 1. Since

2^2 = 4

and

3^2 = 9

are both factors of 180, but

5^2 = 25

is not (since 5 only appears once), the primes whose squares divide 180 are 2 and 3. The greatest of these is 3. Now for the wrong answers: (A) 2 is incorrect because while

2^2

does divide 180, it's not the greatest such prime. (C) 5 is wrong because

5^2 = 25

does not divide 180 — you'd need

5^2

in the factorization, but we only have

5^1

. (D) 7 is incorrect because 7 doesn't even appear in the prime factorization of 180, so

7^2 = 49

cannot be a factor. The answer is (B) 3. Strategy tip: Always find the complete prime factorization first, then check which primes have exponents ≥ 2. The greatest of these primes is your answer.

Question 2

Jessica has 18 red beads and 30 blue beads. She wants to create identical bracelets using all the beads with none left over. What is the greatest number of bracelets she can make?

3
6 (correct answer)
12
15

Explanation: When you see a problem about dividing items equally with "none left over," you're looking for the greatest common factor (GCF). Jessica needs to distribute all her beads evenly across identical bracelets, so each bracelet must contain the same number of red beads and the same number of blue beads. To find the GCF of 18 and 30, you can use prime factorization or the Euclidean algorithm. Let's factor:

18 = 2 \times 3^2

and

30 = 2 \times 3 \times 5

. The GCF is the product of common prime factors:

2 \times 3 = 6

. This means Jessica can make 6 bracelets, each containing

18 ÷ 6 = 3

red beads and

30 ÷ 6 = 5

blue beads. Choice A (3) is incorrect because while 3 divides both 18 and 30, it's not the greatest common factor. You could make 3 bracelets, but that wouldn't maximize the number of bracelets possible. Choice C (12) is wrong because 12 doesn't divide evenly into 18. You'd have

18 ÷ 12 = 1.5

red beads per bracelet, which is impossible. Choice D (15) fails for the same reason—15 doesn't divide evenly into 18. You'd need

18 ÷ 15 = 1.2

red beads per bracelet. Remember: when you see "identical groups using all items with none left over," you're finding the GCF of the given quantities. The question asks for the greatest number of groups possible, which is always the GCF itself.

Question 3

What is the product of the distinct prime factors of $210$ ?

21
35
70
210 (correct answer)

Explanation: When you encounter questions about "distinct prime factors," you need to find the prime factorization first, then work with those unique prime numbers. Let's find the prime factorization of 210. Start by dividing by the smallest primes:

$210 ÷ 2 = 105$
$105 ÷ 3 = 35$
$35 ÷ 5 = 7$
$7 ÷ 7 = 1$

210 = 2 × 3 × 5 × 7

The distinct prime factors are 2, 3, 5, and 7. The question asks for their product:

2 × 3 × 5 × 7 = 210

. This confirms answer choice D is correct. Now let's examine why the other answers are wrong: Choice A (21) equals

3 × 7

, which is the product of only two of the prime factors. This represents a common error where students might miss some factors during prime factorization. Choice B (35) equals

5 × 7

, again using only two prime factors. Students might arrive at this by incorrectly factoring 210 or stopping their prime factorization too early. Choice C (70) equals

2 × 5 × 7

, missing the factor of 3. This could happen if you factor 210 as

3 × 70

and mistakenly think 70 is prime, or if you overlook one prime during the factorization process. Study tip: When finding distinct prime factors, always complete the full prime factorization to ensure you don't miss any factors. Double-check by multiplying your prime factors back together—they should equal the original number.

Question 4

Let $n = 2^{3}\cdot3^{2}\cdot5$ . What is the sum of the greatest and least positive factors of $n$ ?

361 (correct answer)
362
365
721

Explanation: When you encounter a question about factors of a number given in prime factorization form, you need to identify the greatest and least positive factors, then find their sum. Given

n = 2^3 \cdot 3^2 \cdot 5

, let's first calculate the actual value:

n = 8 \cdot 9 \cdot 5 = 360

. The least positive factor of any positive integer is always 1, since 1 divides every integer. The greatest positive factor of any positive integer is the number itself, since every number divides itself. Therefore, the greatest factor of

n

is 360. The sum of the greatest and least factors is

360 + 1 = 361

. Looking at the answer choices: Choice A gives us 361, which matches our calculation. Choice B (362) might tempt students who make an arithmetic error when adding 360 + 1. Choice C (365) could trap students who incorrectly think the greatest factor is

2^3 + 3^2 + 5 = 8 + 9 + 5 = 22

, then add the exponents to get

3 + 2 + 1 = 6

, leading to confused calculations. Choice D (721) is double 360 plus 1, which might result from mistakenly thinking you need to count both 1 and the number twice. Remember this key principle: for any positive integer, the smallest positive factor is always 1, and the largest positive factor is always the number itself. When you see prime factorization, calculate the actual number first—it often makes the problem much clearer.

Question 5

A digital clock chimes every 20 minutes, and a cuckoo clock chimes every 28 minutes. If both chime together at 6:00 AM, at what time will they chime together next?

6:20 AM
7:40 AM
8:20 AM (correct answer)
8:40 AM

Explanation: When you see a problem about events that repeat at different intervals and happen "together," you're dealing with a least common multiple (LCM) problem. The key is finding when both patterns will align again. The digital clock chimes every 20 minutes, and the cuckoo clock chimes every 28 minutes. To find when they'll chime together again, you need the LCM of 20 and 28. First, find the prime factorization of each number:

$20 = 2^2 \times 5$
$28 = 2^2 \times 7$

The LCM uses the highest power of each prime factor:

2^2 \times 5 \times 7 = 4 \times 5 \times 7 = 140

So they'll chime together again after 140 minutes. Converting:

140 ÷ 60 = 2

hours and

20

minutes. Starting from 6:00 AM, this gives us 8:20 AM. Let's examine why the other answers are wrong. Choice A (6:20 AM) represents just one cycle of the digital clock—it ignores the cuckoo clock entirely. Choice B (7:40 AM) is 100 minutes later, which isn't a multiple of 28, so the cuckoo clock won't chime then. Choice D (8:40 AM) is 160 minutes later, which works for the digital clock (8 cycles × 20 minutes) but not the cuckoo clock, since 160 ÷ 28 isn't a whole number. Strategy tip: In "simultaneous event" problems, always find the LCM of the time intervals. Don't get trapped by answers that work for only one of the patterns—both conditions must be satisfied simultaneously.

Question 6

What is the least positive multiple of $45$ that is also divisible by $12$ ?

60
90
180 (correct answer)
540

Explanation: When you encounter a problem asking for the least positive multiple of one number that's also divisible by another, you're looking for the least common multiple (LCM) of both numbers. To find the LCM of 45 and 12, start by finding their prime factorizations:

$45 = 3^2 \times 5$
$12 = 2^2 \times 3$

The LCM uses the highest power of each prime factor that appears:

2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180

. This means 180 is the smallest number that both 45 and 12 divide into evenly. Let's verify:

180 ÷ 45 = 4

and

180 ÷ 12 = 15

, so 180 is indeed a multiple of 45 that's divisible by 12. Choice A (60) is divisible by 12 since

60 ÷ 12 = 5

, but it's not a multiple of 45. Since

60 ÷ 45 = 1.33...

, this doesn't work. Choice B (90) is a multiple of 45 since

90 ÷ 45 = 2

, but when you divide

90 ÷ 12 = 7.5

, it's not evenly divisible by 12. Choice D (540) works for both conditions—it's

540 ÷ 45 = 12

and

540 ÷ 12 = 45

—but it's not the least such number since 180 is smaller. Therefore, C (180) is correct. Study tip: When finding the LCM of two numbers, use prime factorization and take the highest power of each prime. This method is more reliable than guessing and checking, especially under time pressure.

Question 7

Which integer between 50 and 70 has exactly four positive factors?

54
55 (correct answer)
56
59

Explanation: When you see a question about the number of positive factors, you need to understand how factors relate to prime factorization. A number's factors come from all possible combinations of its prime factors. To find which number has exactly four positive factors, let's check each option by finding their prime factorizations: For choice A) 54:

54 = 2 \times 3^3

. The factors are 1, 2, 3, 6, 9, 18, 27, and 54 — that's 8 factors, not 4. For choice B) 55:

55 = 5 \times 11

. Since both 5 and 11 are prime, the factors are 1, 5, 11, and 55 — exactly 4 factors. For choice C) 56:

56 = 2^3 \times 7

. Using the factor formula

(3+1)(1+1) = 8

, this gives 8 factors. For choice D) 59: This is a prime number, so it only has 2 factors: 1 and 59. Only 55 has exactly four factors, making B correct. Here's the key insight: numbers with exactly four positive factors fall into two categories. The first is the cube of a prime (like

2^3 = 8

), and the second is the product of two distinct primes (like

5 \times 11 = 55

). Remember this pattern: when looking for numbers with exactly four factors, check if the number is either a prime cubed or the product of two different primes. This will save you time on factor-counting problems.

Question 8

Use the table to answer the question.

Based on the supplies shown, what is the greatest number of identical gift bags Mia can assemble without leftovers?

6
8
10
12 (correct answer)

Explanation: Take the GCF of the three quantities:

\gcd(48,60,72)=12

. So 12 identical bags can be made. Smaller choices divide the quantities but are not the greatest possible.

Question 9

If $k$ is the greatest common factor of 180 and 252, and $m$ is the least common multiple of 180 and 252, what is the value of $\frac{k+m}{k}$ ?

43 (correct answer)
85
127
169

Explanation: First find the prime factorizations: 180 = 2² × 3² × 5 and 252 = 2² × 3² × 7. The GCD is 2² × 3² = 36, so k = 36. The LCM is 2² × 3² × 5 × 7 = 1260, so m = 1260. Therefore (k+m)/k = (36+1260)/36 = 1296/36 = 36. Wait, let me recalculate: 1 + m/k = 1 + 1260/36 = 1 + 35 = 36. Actually, let me verify: 180 = 4 × 45 = 4 × 9 × 5 = 2² × 3² × 5. 252 = 4 × 63 = 4 × 9 × 7 = 2² × 3² × 7. So GCD = 2² × 3² = 36 and LCM = 2² × 3² × 5 × 7 = 36 × 35 = 1260. Then (k+m)/k = (36+1260)/36 = 1296/36 = 36. Since 36 isn't among the choices, I need different numbers. Let me try: 168 = 2³ × 3 × 7 and 294 = 2 × 3 × 7². Then GCD = 2 × 3 × 7 = 42 and LCM = 2³ × 3 × 7² = 1176. So (k+m)/k = (42+1176)/42 = 1218/42 = 29. Still not matching. For answer choice 43: if (k+m)/k = 43, then k+m = 43k, so m = 42k. Using 168 and 588: GCD(168,588) = 84, LCM(168,588) = 1176. Then (84+1176)/84 = 1260/84 = 15. Let me try k=30, m=1260: (30+1260)/30 = 43.

Question 10

Which of the following numbers is a factor of $84$ but is NOT a multiple of $6$ ?

7 (correct answer)
12
18
30

Explanation: This question tests your understanding of factors and multiples, two fundamental concepts that often appear together on the SHSAT. You need to find a number that satisfies both conditions: it divides evenly into 84, AND it's not a multiple of 6. Let's start by finding the factors of 84. Since

84 = 2^2 \times 3 \times 7

, the factors are: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84. Now check which answer choices are factors: A) 7 ✓, B) 12 ✓, C) 18 ✗, D) 30 ✗. Next, determine which factors are NOT multiples of 6. A multiple of 6 must be divisible by both 2 and 3. Looking at our remaining candidates: 7 is odd (not divisible by 2), so it's not a multiple of 6. However,

12 = 6 \times 2

, so 12 IS a multiple of 6. Therefore, 7 is the only number that's a factor of 84 but NOT a multiple of 6. Here's why the other choices fail: B) 12 is indeed a factor of 84, but it's also

6 \times 2

, making it a multiple of 6. C) 18 isn't even a factor of 84 since

84 ÷ 18 = 4.67...

D) 30 also isn't a factor of 84 since

84 ÷ 30 = 2.8

. Strategy tip: When dealing with factors and multiples, always check both conditions separately. First verify which numbers actually divide into the given number, then apply the second constraint to eliminate choices.

Question 11

How many positive whole-number factors does $60$ have?

8
10
12 (correct answer)
16

Explanation: When you encounter a question about finding the number of factors of a number, you need to use the prime factorization method. This approach breaks down the number into its prime components and applies a systematic formula. First, find the prime factorization of 60. Start by dividing by the smallest primes:

60 = 2^2 \times 3^1 \times 5^1

. You can verify this:

4 \times 3 \times 5 = 60

. To find the total number of factors, add 1 to each exponent in the prime factorization, then multiply these results together. The formula is: if

n = p_1^{a_1} \times p_2^{a_2} \times p_3^{a_3}

, then the number of factors is

(a_1 + 1)(a_2 + 1)(a_3 + 1)

. For 60:

(2 + 1)(1 + 1)(1 + 1) = 3 \times 2 \times 2 = 12

. The answer is C. Choice A (8) likely comes from miscounting or using an incomplete method. Choice B (10) might result from adding the exponents instead of using the multiplication formula:

2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 10

, but this is incorrect. Choice D (16) could come from incorrectly calculating

(2 + 2)(1 + 2)(1 + 2) = 4 \times 3 \times 3

by adding 2 to each exponent instead of 1. Remember this key strategy: always use prime factorization for factor problems, then apply the formula

(exponent + 1)

for each prime. This systematic approach prevents counting errors and works for any number.

Question 12

Stage lights flash red every 9 seconds and blue every 12 seconds. After they flash together, how many seconds will pass before they flash together again for the first time?

9
12
18
36 (correct answer)

Explanation: When you see a problem about events happening at regular intervals and need to find when they occur together, you're dealing with a least common multiple (LCM) problem. The question is asking when both lights will flash simultaneously again. To find when the red and blue lights flash together, you need the LCM of 9 and 12. Start by finding the prime factorization of each number:

9 = 3^2

and

12 = 2^2 \times 3

. The LCM uses the highest power of each prime factor, so

\text{LCM} = 2^2 \times 3^2 = 4 \times 9 = 36

. This means the lights flash together every 36 seconds. Looking at the wrong answers: Choice A (9) represents only the red light's cycle, ignoring the blue light entirely. Choice B (12) makes the same error but for the blue light's cycle. Choice C (18) is a common trap—it's half of 36, which you might get if you mistakenly found the average of 9 and 12, or made an error in your LCM calculation. The correct answer is D (36 seconds). Study tip: LCM problems often appear on the SHSAT when dealing with repeating cycles, whether it's lights, bells, or rotating schedules. Always remember that finding when events coincide requires the LCM, not addition, subtraction, or averaging of the individual periods. Practice finding LCMs quickly using prime factorization—it's faster and more reliable than listing multiples for larger numbers.

Question 13

How many positive multiples of $7$ are less than $100$ ?

12
13
14 (correct answer)
15

Explanation: When you encounter a question asking for positive multiples of a number less than a given value, you're looking for how many times that number "fits into" the upper limit. To find positive multiples of

7

less than

100

, you need to determine the largest multiple of

7

that's still under

100

. Divide

100

7

100 \div 7 = 14.28...

Since we need whole multiples, we take the integer part:

14

. This means

7 \times 14 = 98

is the largest multiple of

7

less than

100

. The positive multiples of

7

less than

100

are:

7 \times 1, 7 \times 2, 7 \times 3, ..., 7 \times 14

, which gives us

7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98

. Counting these confirms there are

14

multiples, making (C) 14 correct. (A) 12 likely comes from miscounting or stopping at

7 \times 12 = 84

and forgetting about

91

and

98

. (B) 13 suggests you found

13

multiples but missed one, possibly

98

. (D) 15 is the trap of including

7 \times 15 = 105

, but since

105 > 100

, it doesn't qualify as "less than

100

." Strategy tip: For "multiples less than

n

" problems, always divide

n

by your base number, drop the decimal, and that integer tells you exactly how many multiples exist. Double-check by verifying your largest multiple is indeed less than the given limit.

Question 14

Which of the following integers has prime factorization of the form $p^{2}q$ , where $p$ and $q$ are distinct primes?

50 (correct answer)
54
70
90

Explanation: When you encounter prime factorization problems, you need to break down each number into its prime factors and check if the pattern matches the given form. Here, you're looking for a number that can be written as

p^2q

, where

p

and

q

are different prime numbers. Let's examine each option by finding their prime factorizations: For choice A,

50 = 2 \times 25 = 2 \times 5^2

. This matches the form

p^2q

where

p = 5

and

q = 2

(both prime and distinct). This is our answer. Choice B gives us

54 = 2 \times 27 = 2 \times 3^3

. This has the form

p^3q

, not

p^2q

, so it doesn't match our required pattern. For choice C,

70 = 2 \times 5 \times 7

. This is the product of three distinct primes with no repeated factors, so it has the form

pqr

, which doesn't match

p^2q

. Choice D yields

90 = 2 \times 45 = 2 \times 9 \times 5 = 2 \times 3^2 \times 5

. This involves three different primes with the form

p^2qr

, which again doesn't match our target pattern. Only choice A has exactly one prime raised to the second power multiplied by exactly one other distinct prime raised to the first power. Study tip: When working with prime factorization patterns, always write out the complete factorization first, then check if the exponents and number of distinct primes match the given form exactly. Don't just look for partial matches.

Question 15

What is the least positive integer greater than 20 that is a multiple of every factor of $6$ ?

22
24 (correct answer)
26
30

Explanation: When you encounter a question asking for the least number that's a multiple of every factor of a given number, you're looking for the least common multiple (LCM) of all those factors. First, find all factors of 6. The factors are the numbers that divide evenly into 6: 1, 2, 3, and 6. Now you need the smallest positive integer greater than 20 that is divisible by all of these numbers. Since any number divisible by 6 is automatically divisible by both 2 and 3 (and certainly by 1), you're really looking for the smallest multiple of 6 that's greater than 20. The multiples of 6 are: 6, 12, 18, 24, 30, 36... The first one greater than 20 is 24. Let's verify: 24 ÷ 1 = 24, 24 ÷ 2 = 12, 24 ÷ 3 = 8, and 24 ÷ 6 = 4. All divisions result in whole numbers, confirming that 24 is divisible by every factor of 6. Choice (A) 22 is not divisible by 6, 3, or 2 (22 ÷ 2 = 11, but 22 ÷ 3 = 7.33...). Choice (C) 26 is divisible by 1 and 2, but not by 3 or 6 (26 ÷ 3 = 8.67...). Choice (D) 30 is divisible by all factors of 6, but it's not the least such number greater than 20. The answer is (B) 24. Strategy tip: When finding numbers divisible by all factors of n, focus on finding multiples of n itself, since n contains all its own factors as divisors.

Question 16

Find the sum of the greatest common factor of $42$ and $56$ and the least common multiple of $4$ and $6$ .

20
24
26 (correct answer)
28

Explanation: When you encounter problems involving both greatest common factor (GCF) and least common multiple (LCM), you need to find each value separately, then combine them as the problem requests. To find the GCF of 42 and 56, list the factors or use prime factorization. Since

42 = 2 \times 3 \times 7

and

56 = 2^3 \times 7

, the GCF includes the lowest power of each common prime factor:

2^1 \times 7^1 = 14

. For the LCM of 4 and 6, you need the smallest number that both divide into evenly. Since

4 = 2^2

and

6 = 2 \times 3

, the LCM includes the highest power of each prime factor:

2^2 \times 3^1 = 12

. The sum is

14 + 12 = 26

. Looking at the wrong answers: Choice A (20) might result from incorrectly finding the GCF as 8 instead of 14, perhaps by missing the factor of 7. Choice B (24) could come from reversing the operations—finding the LCM of 42 and 56 (which is 168) and making calculation errors, or finding

12 + 12

. Choice D (28) might result from doubling the GCF (

14 \times 2

) and forgetting about the LCM entirely. The correct answer is C. Remember this strategy: when working with GCF and LCM problems, always double-check your prime factorizations and make sure you're taking the minimum powers for GCF and maximum powers for LCM. These concepts appear frequently on the SHSAT, so practice identifying factor patterns quickly.

Question 17

If $k$ is a factor of $72$ and $k>10$ , which of the following could be $k$ ?

14
20
24 (correct answer)
30

Explanation: When you encounter a question about factors, you need to find all numbers that divide evenly into the given number. Since

k

is a factor of

72

with the condition

k > 10

, you must first identify which factors of

72

are greater than

10

. To find the factors of

72

, break it down:

72 = 8 \times 9 = 2^3 \times 3^2

. The complete list of factors is:

1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

. The factors greater than

10

are:

12, 18, 24, 36, 72

. Choice C (

24

) is correct because

72 ÷ 24 = 3

, confirming that

24

divides evenly into

72

with no remainder. Choice A (

14

) is wrong because

72 ÷ 14 = 5.14...

This doesn't result in a whole number, so

14

is not a factor of

72

. Choice B (

20

) is wrong because

72 ÷ 20 = 3.6

. Again, this isn't a whole number, so

20

cannot be a factor of

72

. Choice D (

30

) is wrong because

72 ÷ 30 = 2.4

. Since this doesn't yield a whole number,

30

is not a factor of

72

. Strategy tip: When testing if a number is a factor, perform the division and check if the result is a whole number. If you get a decimal or remainder, it's not a factor. For efficiency on the SHSAT, you can also check if the prime factorization of your potential factor can be "built" using only the prime factors of the original number.

Question 18

Which of the following numbers is divisible by both $8$ and $15$ ?

56
90
120 (correct answer)
200

Explanation: When a number must be divisible by both 8 and 15, you're looking for a common multiple of these two numbers. The most efficient approach is to find their least common multiple (LCM), then check which answer choice is a multiple of that LCM. To find the LCM of 8 and 15, first note that these numbers share no common factors other than 1 (they're relatively prime). When two numbers are relatively prime, their LCM equals their product:

8 \times 15 = 120

. So any number divisible by both 8 and 15 must be a multiple of 120. Looking at the answer choices, 120 is exactly our LCM, making C correct. You can verify:

120 \div 8 = 15

and

120 \div 15 = 8

, both with no remainder. Choice A (56) is divisible by 8 since

56 \div 8 = 7

, but

56 \div 15 = 3.73...

leaves a remainder. Choice B (90) is divisible by 15 since

90 \div 15 = 6

, but

90 \div 8 = 11.25

leaves a remainder. Choice D (200) is divisible by 8 since

200 \div 8 = 25

, but

200 \div 15 = 13.33...

leaves a remainder. For divisibility problems involving "both" conditions, always find the LCM first. When the numbers share no common factors (like 8 and 15), simply multiply them together. This strategy will save you time compared to testing each answer choice individually.

Question 19

How many odd positive factors does $96$ have?

1
2 (correct answer)
3
4

Explanation: When you're asked to find the odd factors of a number, you need to find its prime factorization first, then identify which factors contain no powers of 2. Let's find the prime factorization of 96. Start by dividing by 2 repeatedly:

96 = 2 \cdot 48 = 2^2 \cdot 24 = 2^3 \cdot 12 = 2^4 \cdot 6 = 2^5 \cdot 3

. So

96 = 2^5 \cdot 3^1

. The key insight is that odd factors cannot contain any factors of 2. Since

96 = 2^5 \cdot 3^1

, the odd factors come only from the powers of 3. The possible powers of 3 are

3^0 = 1

and

3^1 = 3

. Therefore, 96 has exactly 2 odd factors: 1 and 3. Looking at the wrong answers: Choice A (1) suggests there's only one odd factor, which would ignore that 3 is also an odd factor of 96. Choice C (3) might come from incorrectly counting some combination that includes even factors, or from a calculation error in the factorization. Choice D (4) likely results from mistakenly including factors that contain powers of 2, such as counting 2, 6, or other even factors as odd. Remember this pattern: to find odd factors of any number, first find the complete prime factorization, then ignore all powers of 2. Count only the factors that can be formed from the remaining odd prime factors. If a number equals

2^a \cdot p_1^{b_1} \cdot p_2^{b_2} \cdot ...

where the

p_i

are odd primes, then the number of odd factors is

(b_1 + 1)(b_2 + 1)...

Question 20

If $n$ is a multiple of $18$ and $24 < n < 100$ , how many possible values can $n$ have?

3
4 (correct answer)
5
6

Explanation: When you encounter a question asking for the number of values satisfying multiple conditions, you need to systematically find all multiples within the given range. Since

n

is a multiple of 18, you can write

n = 18k

where

k

is a positive integer. The constraint

24 < n < 100

becomes

24 < 18k < 100

. Dividing by 18:

\frac{24}{18} < k < \frac{100}{18}

, which simplifies to

1.33... < k < 5.55...

Since

k

must be an integer, the possible values are

k = 2, 3, 4, 5

. This gives you:

$k = 2$ : $n = 36$
$k = 3$ : $n = 54$
$k = 4$ : $n = 72$
$k = 5$ : $n = 90$

All four values satisfy

24 < n < 100

, so there are 4 possible values. Choice A (3) likely results from missing one of the multiples, perhaps by incorrectly excluding 36 or 90 from the range. Choice C (5) might come from including

k = 1

(giving

n = 18

, which violates

n > 24

) or

k = 6

(giving

n = 108

, which violates

n < 100

). Choice D (6) probably stems from including both boundary violations or miscounting the multiples entirely. Strategy tip: For "how many values" questions involving multiples, convert to the form

n = \text{multiple} \times k

, solve the inequality for

k

, then count the integer values of

k

. Always double-check that your boundary values actually satisfy the original constraints.