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Divisibility Rules Quiz 1 (10 MCQs)

This set of multiple-choice questions evaluates understanding of divisibility rules, least common multiple (LCM) calculation, prime factorization, and numerical computation. It covers concepts such as divisibility by 2, 3, 5, 7, and 9, sum of digits, arithmetic sequences, and digit-reversed numbers. The questions test the ability to identify even numbers, check last digits, and apply divisibility rules to three-digit numbers.

Quiz Instructions

Select an option to see the correct answer instantly.

1. Which number is divisible by 9 (without using a calculator)?

Correct Answer: Correct answer is: (B) 9459.

Exam Relevance: SAT, GRE, GMAT, ACT

Difficulty: Moderate

Concept notes: A number is divisible by 9 if the sum of its digits is divisible by 9.

Common Mistakes: A common mistake is to confuse the divisibility rule for 9 with the rule for 3, which only requires the sum of the digits to be divisible by 3.

Explanations: To determine if 9459 is divisible by 9, we sum its digits: 9 + 4 + 5 + 9 = 27. Since 27 is divisible by 9, 9459 is also divisible by 9.

Option Analysis:

Option A: 3459: Sum of digits = 3 + 4 + 5 + 9 = 21, which is not divisible by 9.
Option B: 9459: Sum of digits = 9 + 4 + 5 + 9 = 27, which is divisible by 9.
Option C: 5379: Sum of digits = 5 + 3 + 7 + 9 = 24, which is not divisible by 9.
Option D: 2792: Sum of digits = 2 + 7 + 9 + 2 = 20, which is not divisible by 9.

Mnemonic: Divisible by 9: Sum of digits, 9 or 18, or 27, or 36, etc.

2. What number is divisible by two?

Correct Answer: Correct answer is: (C) 752.

Exam Relevance: SAT, GRE, GMAT, ACT

Difficulty: Easy

Concept notes: A number is divisible by 2 if its last digit is even (0, 2, 4, 6, or 8).

Common Mistakes: A common mistake is to overlook the last digit and instead focus on other digits in the number.

Explanations: To determine if a number is divisible by 2, we need to check the last digit of the number. The last digit of 752 is 2, which is an even number. Therefore, 752 is divisible by 2.

Option Analysis:

Option A: The last digit of 345 is 5, which is odd. Therefore, 345 is not divisible by 2.
Option B: The last digit of 537 is 7, which is odd. Therefore, 537 is not divisible by 2.
Option C: The last digit of 752 is 2, which is even. Therefore, 752 is divisible by 2.
Option D: The last digit of 327 is 7, which is odd. Therefore, 327 is not divisible by 2.

Mnemonic: Even last digit, divisible by 2.

3. $42\equiv6\ \left(mod\ 5\right)$

Correct Answer: Correct answer is: (B) False.

Exam Relevance: GRE, GMAT, SAT, ACT

Difficulty: Moderate

Concept notes: The concept of congruence in modular arithmetic states that two integers are congruent modulo n if their difference is divisible by n. In this case, we need to check if 42 and 6 are congruent modulo 5.

Common Mistakes: A common mistake is to assume that if two numbers have the same remainder when divided by the modulus, they are congruent. However, the difference between the numbers must be divisible by the modulus.

Explanations: To determine if $42 \equiv 6 \pmod{5}$, we need to check if the difference between 42 and 6 is divisible by 5. 1. Calculate the difference: $42 - 6 = 36$. 2. Check if 36 is divisible by 5: $36 \div 5 = 7.2$, which is not an integer. Since 36 is not divisible by 5, $42 \not\equiv 6 \pmod{5}$.

Option Analysis:

Option A: This option is incorrect because 42 and 6 are not congruent modulo 5.
Option B: This option is correct because 42 and 6 are not congruent modulo 5.

4. Which number is a multiple of 6?

Correct Answer: Correct answer is: (C) 12.

Exam Relevance: SAT, ACT, GRE, GMAT

Difficulty: Easy

Concept notes: A number is a multiple of 6 if it is divisible by both 2 and 3.

Common Mistakes: A common mistake is to think that a number is a multiple of 6 if it is divisible by either 2 or 3, but it must be divisible by both.

Explanations: To determine if a number is a multiple of 6, it must be divisible by both 2 and 3. - 12 is divisible by 2 (12 ÷ 2 = 6) and by 3 (12 ÷ 3 = 4). - Therefore, 12 is a multiple of 6.

Option Analysis:

Option A: 2 is not a multiple of 6 because it is not divisible by 3.
Option B: 3 is not a multiple of 6 because it is not divisible by 2.
Option C: 12 is a multiple of 6 because it is divisible by both 2 and 3.
Option D: 16 is not a multiple of 6 because it is not divisible by 3.

Mnemonic: Divisible by 2 and 3, it's a 6.

5. A three-digit number subtracted from its digit-reversed version, when divided by 9, leaves what remainder?

Correct Answer: Correct answer is: (A) 0.

Exam Relevance: SAT, GRE, GMAT, Competitive Math Exams

Difficulty: Moderate

Concept notes: The problem involves the divisibility rule for 9 and the properties of digit-reversed numbers.

Common Mistakes: A common mistake is to assume that the remainder could be any number without considering the properties of digit-reversed numbers and the divisibility rule for 9.

Explanations: Let the three-digit number be represented as $ \overline{abc} $, where $ a, b, $ and $ c $ are its digits. The number can be expressed as $ 100a + 10b + c $. The digit-reversed version of this number is $ \overline{cba} $, which can be expressed as $ 100c + 10b + a $. The difference between the original number and its digit-reversed version is: \[ (100a + 10b + c) - (100c + 10b + a) = 100a + 10b + c - 100c - 10b - a = 99a - 99c = 99(a - c) \] Since $ 99(a - c) $ is a multiple of 99, and 99 is a multiple of 9, the difference is always divisible by 9. Therefore, when this difference is divided by 9, the remainder is 0.

Option Analysis:

Option A: Correct. The remainder is 0.
Option B: Incorrect. The remainder cannot be 1.
Option C: Incorrect. The remainder cannot be 2.
Option D: Incorrect. The remainder can be determined.

6. The numerator can be divided by the denominator.

Correct Answer: Correct answer is: (A) True.

Exam Relevance: SAT, ACT, GRE, GMAT

Difficulty: Easy

Concept notes: The question is about the concept of divisibility, which is the ability of one number (the numerator) to be divided by another number (the denominator) without leaving a remainder.

Common Mistakes: A common misunderstanding is that divisibility only applies to whole numbers, but it can also apply to fractions and other numerical expressions.

Explanations: The statement "The numerator can be divided by the denominator" is true because divisibility rules apply to the relationship between the numerator and the denominator. If the numerator is divisible by the denominator, the division will result in a whole number or a simplified fraction without a remainder.

Option Analysis:

Option A: This option is correct because the numerator can indeed be divided by the denominator if the division results in a whole number or a simplified fraction without a remainder.
Option B: This option is incorrect because it contradicts the basic principle of divisibility.

Mnemonic: Divide, No Remainder, True

7. Which of these numbers is divisible by 5?

Correct Answer: Correct answer is: (C) 45.

Exam Relevance: SAT, ACT, GRE, GMAT

Difficulty: Easy

Concept notes: A number is divisible by 5 if its last digit is 0 or 5.

Common Mistakes: A common mistake is to assume that any number ending in 5 is divisible by 5, but it is important to check the last digit of the number.

Explanations: The number 45 ends in 5, which means it is divisible by 5. The other numbers (33, 51, and 18) do not end in 0 or 5, so they are not divisible by 5.

Option Analysis:

Option A: 33 does not end in 0 or 5, so it is not divisible by 5.
Option B: 51 does not end in 0 or 5, so it is not divisible by 5.
Option C: 45 ends in 5, so it is divisible by 5.
Option D: 18 does not end in 0 or 5, so it is not divisible by 5.

Mnemonic: Divisible by 5: Ends in 0 or 5.

8. When 52 is divided by 5, what is the remainder?

Correct Answer: Correct answer is: (A) 2.

Exam Relevance: SAT, GRE, GMAT, ACT

Difficulty: Easy

Concept notes: The concept of divisibility rules involves understanding how to determine the remainder when a number is divided by another number.

Common Mistakes: A common mistake is to confuse the remainder with the quotient or to perform the division incorrectly.

Explanations: To find the remainder when 52 is divided by 5, we perform the division: 52 ÷ 5 = 10 with a remainder of 2. This can be verified by the equation: 52 = 5 × 10 + 2. Thus, the remainder is 2.

Option Analysis:

Option A: Correct. The remainder when 52 is divided by 5 is 2.
Option B: Incorrect. 7 is not the remainder when 52 is divided by 5.
Option C: Incorrect. 10 is the quotient, not the remainder.
Option D: Incorrect. 10.4 is not a valid remainder as remainders must be integers.

9. How many natural numbers are there between 200 and 500, which are divisible by 7?

Correct Answer: Correct answer is: (B) 43.

Exam Relevance: SAT, GRE, GMAT, Competitive Exams

Difficulty: Moderate

Concept notes: To find the number of natural numbers between 200 and 500 that are divisible by 7, we need to identify the smallest and largest multiples of 7 within this range and then count the multiples.

Common Mistakes: A common mistake is to include the endpoints 200 and 500 in the count, even if they are not divisible by 7.

Explanations: 1. Identify the smallest multiple of 7 greater than or equal to 200: - $ 200 \div 7 \approx 28.57 $ - The smallest integer greater than 28.57 is 29. - Therefore, the smallest multiple of 7 is $ 29 \times 7 = 203 $. 2. Identify the largest multiple of 7 less than or equal to 500: - $ 500 \div 7 \approx 71.43 $ - The largest integer less than 71.43 is 71. - Therefore, the largest multiple of 7 is $ 71 \times 7 = 497 $. 3. Count the multiples of 7 from 203 to 497: - The sequence of multiples of 7 is an arithmetic sequence where the first term $ a = 203 $ and the common difference $ d = 7 $. - The general term of the sequence is given by $ a_n = a + (n-1)d $. - Set $ a_n = 497 $: \[ 497 = 203 + (n-1) \times 7 \] \[ 497 - 203 = (n-1) \times 7 \] \[ 294 = (n-1) \times 7 \] \[ n-1 = \frac{294}{7} = 42 \] \[ n = 42 + 1 = 43 \] Therefore, there are 43 natural numbers between 200 and 500 that are divisible by 7.

Option Analysis:

Option A: 42. This is incorrect because the count includes 43 multiples of 7.
Option B: 43. This is correct as calculated.
Option C: 44. This is incorrect because the count includes 43 multiples of 7.
Option D: 45. This is incorrect because the count includes 43 multiples of 7.

10. All numbers that are evenly divisible by both 6 and 14 are also divisible by which of the following numbers?

Correct Answer: Correct answer is: (C) 21.

Exam Relevance: SAT, GRE, GMAT, ACT

Difficulty: Moderate

Concept notes: To determine which number all numbers divisible by both 6 and 14 are also divisible by, we need to find the least common multiple (LCM) of 6 and 14. The LCM of two numbers is the smallest number that is a multiple of both.

Common Mistakes: A common mistake is to assume that the LCM of 6 and 14 is simply the product of the two numbers, which is incorrect. The LCM is the smallest number that both 6 and 14 can divide evenly.

Explanations: First, we find the prime factorizations of 6 and 14: - $6 = 2 \times 3$ - $14 = 2 \times 7$ The LCM is found by taking the highest power of each prime that appears in the factorizations: - The highest power of 2 is $2^1$ - The highest power of 3 is $3^1$ - The highest power of 7 is $7^1$ Thus, the LCM of 6 and 14 is: \[ \text{LCM} = 2^1 \times 3^1 \times 7^1 = 2 \times 3 \times 7 = 42 \] Now, we need to check which of the given options (8, 12, 21, 28) is a divisor of 42: - 42 is not divisible by 8. - 42 is not divisible by 12. - 42 is divisible by 21. - 42 is not divisible by 28. Therefore, the correct answer is 21.

Option Analysis:

Option A: 42 is not divisible by 8.
Option B: 42 is not divisible by 12.
Option C: 42 is divisible by 21.
Option D: 42 is not divisible by 28.

Mnemonic: LCM: Least Common Multiple

Frequently Asked Questions

What are divisibility rules?

Divisibility rules are shortcuts to determine if a number can be divided by another number without performing the actual division.

How can divisibility rules be applied in arithmetic?

Divisibility rules can simplify arithmetic operations by quickly identifying factors and multiples, making calculations more efficient.

What is the divisibility rule for 9?

A number is divisible by 9 if the sum of its digits is divisible by 9.

How do divisibility rules relate to prime factorization?

Divisibility rules help identify factors of a number, which can be used to find its prime factorization more easily.

What is the significance of the last digit in divisibility rules?

The last digit of a number is crucial for determining divisibility by 2, 5, and 10, as these rules depend on the last digit being even, 0 or 5, or 0, respectively.

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