ISEE Middle Level Quantitative Reasoning Quiz: Divisibility And Factors

What this quiz covers

This quiz focuses on Divisibility And Factors, giving you a quick way to practice the rules, question types, and explanations that matter most for ISEE Middle Level Quantitative Reasoning.

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. The quantity in Column A is greater.
2. The quantity in Column B is greater.
3. The two quantities are equal. (correct answer)
4. The relationship cannot be determined from the information given.

Explanation: This question tests a property of the greatest common factor (GCF). The GCF of two numbers that share a common factor is that common factor multiplied by the GCF of the remaining parts. Let G = GCF(m, n). Then m = Gx and n = Gy, where x and y are coprime. Column A is GCF(12Gx, 12Gy) = 12G × GCF(x, y) = 12G × 1 = 12G. Column B is 12 × GCF(m, n) = 12G. The two quantities are always equal.

Question 2

1. 6
2. 12 (correct answer)
3. 18
4. 60

Explanation: This question tests middle school quantitative reasoning skills: understanding and applying divisibility and factors. Divisibility means a number can be divided by another number without leaving a remainder. Factors are numbers you can multiply to get another number. In this problem, the scenario involves finding the greatest common factor (GCF) of 60 cookies and 36, requiring understanding of common factors. The correct answer, choice B (12), is correct because the factors of 60 are {1,2,3,4,5,6,10,12,15,20,30,60} and the factors of 36 are {1,2,3,4,6,9,12,18,36}, with 12 being the largest number in both lists. Choice C (18) is incorrect because while 18 is a factor of 36, it is not a factor of 60 (60 ÷ 18 = 3.33...). To help students: Encourage listing all factors of each number systematically. Teach identifying common factors by finding numbers in both lists. Emphasize that the GCF is the largest of the common factors.

Question 3

1. $\{1,2,3,4,5,6,10,12,15,20,30,60\}$ (correct answer)
2. $\{1,2,3,4,5,6,7,10,12,15,20,30,60\}$
3. $\{1,2,3,5,6,10,12,15,20,30\}$
4. $\{2,4,6,8,10,12,14,16,18,20\}$

Explanation: This question tests middle school quantitative reasoning skills: understanding and applying divisibility and factors. Divisibility means a number can be divided by another number without leaving a remainder. Factors are numbers you can multiply to get another number. In this problem, the scenario involves identifying the complete set of factors of 60 cookies, requiring systematic factor finding. The correct answer, choice A, is correct because it lists all factors of 60: 1×60, 2×30, 3×20, 4×15, 5×12, and 6×10 are all the factor pairs. Choice B incorrectly includes 7, which is not a factor (60 ÷ 7 = 8.57...). To help students: Encourage finding factors in pairs starting from 1. Teach checking completeness by verifying each number divides 60 evenly. Emphasize that the list should include all numbers from 1 up to the square root, plus their pairs.

Question 4

1. 7
2. 6 (correct answer)
3. 40
4. 0

Explanation: This question tests middle school quantitative reasoning skills: understanding and applying divisibility and factors. Divisibility means a number can be divided by another number without leaving a remainder. Factors are numbers you can multiply to get another number. In this problem, the scenario involves identifying which number is a factor of 36 chairs, requiring understanding of what makes a number a factor. The correct answer, choice B (6), is correct because 6 × 6 = 36, making 6 a factor of 36. Choice A (7) is incorrect because 36 ÷ 7 = 5.14..., which is not a whole number, so 7 is not a factor of 36. To help students: Encourage practice with factor pairs (1×36, 2×18, 3×12, 4×9, 6×6). Teach systematic checking by dividing to see if you get whole numbers. Emphasize that factors always divide evenly into the original number.

Question 5

1. 9
2. 3
3. 4
4. 8 (correct answer)

Explanation: This question tests middle school quantitative reasoning skills: understanding and applying divisibility and factors. Divisibility means a number can be divided by another number without leaving a remainder. Factors are numbers you can multiply to get another number. In this problem, the scenario involves identifying which number is NOT a factor of 36 chairs, requiring understanding of factor relationships. The correct answer, choice D (8), is correct because 36 ÷ 8 = 4.5, which is not a whole number, meaning 8 is not a factor of 36. Choices A (9), B (3), and C (4) are all factors of 36 because 36 ÷ 9 = 4, 36 ÷ 3 = 12, and 36 ÷ 4 = 9, all giving whole number results. To help students: Encourage listing all factor pairs systematically. Teach checking each option by division. Emphasize that non-whole quotients indicate non-factors.

Question 6

1. The two quantities are equal. (correct answer)
2. The quantity in Column B is greater.
3. The quantity in Column A is greater.
4. The relationship cannot be determined from the information given.

Explanation: When you see a question about divisibility and remainders, think about what the given information tells you about the number's properties. If a positive integer $N$ is divisible by 6, this means $N = 6k$ for some positive integer $k$. Since $6 = 2 \times 3$, any number divisible by 6 must also be divisible by both 2 and 3. This is a fundamental property of divisibility: if a number is divisible by a product of factors, it's divisible by each individual factor. For Column A: When $N$ is divided by 2, since $N$ is divisible by 2, the remainder is 0. For Column B: When $N$ is divided by 3, since $N$ is divisible by 3, the remainder is also 0. Therefore, both quantities equal 0, making them equal. Looking at the wrong answers: Choice B suggests the remainder when dividing by 3 is greater, but both remainders are 0. Choice C suggests the remainder when dividing by 2 is greater, which is also incorrect for the same reason. Choice D claims the relationship cannot be determined, but the divisibility by 6 gives us complete information about both remainders. Study tip: Remember that divisibility by a composite number automatically means divisibility by all its prime factors. When a number is divisible by another, the remainder is always 0. Questions testing this concept often try to confuse you by asking about remainders when the given information already tells you the remainder must be zero.

Question 7

1. The two quantities are equal. (correct answer)
2. The quantity in Column B is greater.
3. The quantity in Column A is greater.
4. The relationship cannot be determined from the information given.

Explanation: When you see questions about distinct prime factors, you need to find the prime factorization of each number and count how many different prime numbers appear in each factorization. Let's find the prime factorization of 120 first. Start by dividing by the smallest prime, 2: $120 = 2^3 \times 15 = 2^3 \times 3 \times 5$. So 120 has the distinct prime factors 2, 3, and 5 - that's 3 distinct prime factors. Now for 90: $90 = 2 \times 45 = 2 \times 9 \times 5 = 2 \times 3^2 \times 5$. So 90 also has the distinct prime factors 2, 3, and 5 - that's also 3 distinct prime factors. Both quantities equal 3, making them equal. Choice A is correct because both numbers have exactly 3 distinct prime factors. Choice B is wrong because Column B (3) is not greater than Column A (3). Choice C is wrong because Column A (3) is not greater than Column B (3). Choice D is wrong because we have sufficient information to determine that both quantities are equal - there's no ambiguity here. Remember that "distinct prime factors" means you count each prime number only once, regardless of how many times it appears in the factorization. The exponent doesn't matter for this count - whether it's $2^1$ or $2^3$, the number 2 contributes just one distinct prime factor. Practice prime factorization systematically by always starting with the smallest prime and working your way up.

Question 8

1. 2
2. 22
3. 60 (correct answer)
4. 120

Explanation: This problem requires finding the least common multiple (LCM) of 12 and 10 to determine the smallest number that is a multiple of both. The prime factorization of 12 is 2² × 3. The prime factorization of 10 is 2 × 5. The LCM is the product of the highest powers of all prime factors involved: LCM(12, 10) = 2² × 3 × 5 = 4 × 3 × 5 = 60. Therefore, the minimum number of pencils (and erasers) the store must sell to have an equal amount of each is 60.

Question 9

1. 2
2. 3
3. 6 (correct answer)
4. 12

Explanation: Let the three consecutive even integers be represented by (2n), (2n+2), and (2n+4), where (n) is an integer. Their sum is (2n + (2n+2) + (2n+4) = 6n + 6). This sum can be factored as (6(n+1)). Since the sum can always be expressed as 6 times an integer, it is always divisible by 6. To confirm 6 is the greatest such integer, we can test a few cases. For 2, 4, 6, the sum is 12 (divisible by 6). For 4, 6, 8, the sum is 18 (divisible by 6). Since the sums are not always divisible by 12 (e.g., 18), the greatest integer that must be a factor is 6.

Question 10

1. The quantity in Column A is greater.
2. The quantity in Column B is greater. (correct answer)
3. The two quantities are equal.
4. The relationship cannot be determined from the information given.

Explanation: By definition, a prime number has exactly two positive factors: 1 and itself. So, the quantity in Column A is 2. Since (p) is a prime number greater than 2, it must be an odd number. Therefore, (p+1) must be an even number greater than 3. Any even number greater than 2 is composite and has at least three factors: 1, 2, and the number itself. For example, if p=3, p+1=4, which has 3 factors (1,2,4). If p=5, p+1=6, which has 4 factors (1,2,3,6). The number of factors of (p+1) will always be greater than 2. Thus, the quantity in Column B is always greater than the quantity in Column A.

Question 11

1. 19 (correct answer)
2. 31
3. 37
4. 41

Explanation: Let the integer be N. From the first condition, N can be written as 7k + 5 for some integer k. Listing possible values of N: 5, 12, 19, 26, 33, 40... From the second condition, N can be written as 6m + 1 for some integer m. Listing possible values of N: 1, 7, 13, 19, 25, 31... We are looking for the smallest number that appears in both lists. The smallest such number is 19.

Question 12

1. 27
2. 29
3. 33
4. 37 (correct answer)

Explanation: Let S be the number of students. We are given 20 < S < 40. The first condition means S leaves a remainder of 1 when divided by 4 (S = 4k + 1). Possible values for S in the given range are: 21, 25, 29, 33, 37. The second condition means S leaves a remainder of 2 when divided by 5 (S = 5m + 2). Possible values for S in the given range are: 22, 27, 32, 37. The only number that appears in both lists is 37. So, there are 37 students in the club.

Question 13

1. 18
2. 43 (correct answer)
3. 83
4. Cannot be determined.

Explanation: We need to find two prime numbers that multiply to 82. This means we need to find the prime factors of 82. Since 82 is an even number, one of its prime factors is 2. Dividing 82 by 2 gives 41. We must check if 41 is a prime number. It is not divisible by 2, 3, or 5. The square root of 41 is between 6 and 7, so we only need to check for divisibility by primes up to 5. Thus, 41 is prime. The two distinct prime numbers are 2 and 41. Their sum is 2 + 41 = 43.

Question 14

1. 3
2. 6
3. 7 (correct answer)
4. 8

Explanation: First, find the prime factorization of 720. We can break it down: 720 = 72 × 10 = (8 × 9) × (2 × 5) = (2³ × 3²) × (2 × 5). Combining the prime factors, we get 2³⁺¹ × 3² × 5¹ = 2⁴ × 3² × 5¹. Comparing this to (2^x \cdot 3^y \cdot 5^z), we have x = 4, y = 2, and z = 1. The question asks for the sum x + y + z, which is 4 + 2 + 1 = 7.

Question 15

1. 7
2. 8
3. 12
4. 56 (correct answer)

Explanation: The side length of the largest possible square plots must be the greatest common factor (GCF) of the field's dimensions, 84 and 96. First, find the prime factorization of each number: 84 = 2² × 3 × 7 and 96 = 2⁵ × 3. The GCF is the product of the lowest powers of common prime factors: GCF(84, 96) = 2² × 3 = 12. So, each square plot has a side length of 12 meters. The total number of plots is the area of the field divided by the area of one plot. Area of field = 84 × 96. Area of plot = 12 × 12. Number of plots = (84 × 96) / (12 × 12) = (84/12) × (96/12) = 7 × 8 = 56. There will be 56 square plots.

Question 16

1. 12
2. 24 (correct answer)
3. 36
4. 48

Explanation: For any two positive integers a and b, the product of the numbers is equal to the product of their greatest common factor (GCF) and least common multiple (LCM). The formula is a × b = GCF(a, b) × LCM(a, b). We are given one number (a = 18), the GCF (6), and the LCM (72). We can set up the equation: 18 × b = 6 × 72. So, 18 × b = 432. To find b, we divide 432 by 18. 432 ÷ 18 = 24. The other number is 24.

Question 17

1. 8:01:00 PM
2. 8:01:30 PM (correct answer)
3. 8:03:00 PM
4. 8:04:30 PM

Explanation: To find when they will next flash together, we need to find the least common multiple (LCM) of 15 and 18. The prime factorization of 15 is 3 × 5. The prime factorization of 18 is 2 × 3². The LCM is the product of the highest powers of all prime factors present: LCM(15, 18) = 2 × 3² × 5 = 2 × 9 × 5 = 90. They will flash together every 90 seconds. Since 90 seconds is equal to 1 minute and 30 seconds, the next time they flash together will be 1 minute and 30 seconds after 8:00 PM, which is 8:01:30 PM.

Question 18

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

Explanation: This is a two-step problem involving a function-like notation. First, we must find the value of the inner function, (F(27)). The factors of 27 are 1, 3, 9, and 27. There are 4 factors, so (F(27) = 4). Next, we must find the value of the outer function, (F(F(27))), which is now (F(4)). The factors of 4 are 1, 2, and 4. There are 3 factors, so (F(4) = 3). Therefore, (F(F(27)) = 3).

Question 19

1. 1
2. 4
3. 7 (correct answer)
4. 9

Explanation: A number is divisible by 6 if it is divisible by both 2 and 3. The number 4,5_2 is already divisible by 2 because its last digit is 2, which is an even number. For the number to be divisible by 3, the sum of its digits must be divisible by 3. The sum of the digits is 4 + 5 + _ + 2 = 11 + _. The possible values for 11 + _ that are multiples of 3 are 12, 15, and 18. If 11 + _ = 12, _ = 1. If 11 + _ = 15, _ = 4. If 11 + _ = 18, _ = 7. The possible digits are 1, 4, and 7. The largest of these is 7.

Question 20

1. 7
2. 8 (correct answer)
3. 9
4. 81

Explanation: This question tests middle school quantitative reasoning skills: understanding and applying divisibility and factors. Divisibility means a number can be divided by another number without leaving a remainder. Factors are numbers you can multiply to get another number. In this problem, the scenario involves finding the missing factor in 9 × __ = 72, requiring understanding of inverse operations. The correct answer, choice B (8), is correct because 9 × 8 = 72, making 8 the missing factor. This can be found by dividing: 72 ÷ 9 = 8. Choice C (9) would give 9 × 9 = 81, not 72. To help students: Encourage using division to find missing factors (72 ÷ 9 = ?). Teach checking answers by multiplying back. Emphasize the relationship between multiplication and division as inverse operations.