📝 Identify Multiples and divisibility Tests and rules (17 MCQs)

📖 From Digital SAT Algebra • 1. Basics of Algebra • 17 questions available

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What is Identify Multiples and divisibility Tests and rules?

Definition:
Multiples are the products of a given number and any integer, while divisibility rules are shortcuts that determine whether one whole number can be divided by another without a remainder, based on specific patterns in the digits of the number.

Working:
To find multiples, multiply the number by $1, 2, 3, \ldots$ , and for divisibility, apply rules such as a number divisible by $2$ if it ends in even digit, by $3$ if sum of digits is divisible by $3$ , by $5$ if it ends in $0$ or $5$ , and by $10$ if it ends in $0$ .

Example:
Check if $2,340$ is divisible by $6$ (rule: divisible by both $2$ and $3$ ).
Solution: Ends with $0$ (even, so divisible by $2$ ), and sum of digits $2+3+4+0=9$ , which is divisible by $3$ , so $2,340$ is divisible by $6$ .

Reason:
These concepts help in simplifying fractions, finding common denominators, and solving number theory problems, making arithmetic faster and more systematic, especially in algebra and factorization.

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📝 All Identify Multiples and divisibility Tests and rules MCQs

Q1. A number is divisible by both 4 and 9. Which of the following must be true about the number?

A.It is divisible by 36

B.It is divisible by 18

C.It is divisible by 6

D.All of the above ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: If a number is divisible by both 4 and 9, since 4 and 9 are coprime (their only common factor is 1), the number must be divisible by their product, 36. Divisibility by 36 automatically implies divisibility by all factors of 36, which include 18 and 6. Option D is correct. A common error is to only consider the LCM as 36 but forget that 18 and 6 are divisors of 36, hence any multiple of 36 is also a multiple of those numbers.

Q2. A teacher wrote $N = 2^3 \times 3^2 \times 5 \times 7$ on the board. She then stated, 'This number is divisible by 12, 18, and 30.' A student disagrees, saying it is not divisible by 18. Who is correct and why?

A.Teacher is correct; 18 divides N because

18 = 2 \times 3^2

, and both factors are present in N. ✅

B.Student is correct; 18 requires

3^3

, but N only has

3^2

C.Teacher is correct; 18 divides N because 18 is a factor of 72.

D.Student is correct; 18 is not a factor because N is not a multiple of 6.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The teacher is correct. For 18 = $2 \times 3^2$ , N has $2^3$ and $3^2$ , which means it has enough prime factors to form 18. The student's likely error is thinking that '18' needs a factor of 3 three times or confusing it with 27. Option B is a strong distractor because it mistakes the exponent requirement. Option C is true but does not explain the reasoning fully, making A the best answer as it directly applies prime factorization.

Q3. A student claims that the number 5,760 is divisible by 45. To verify, she checks divisibility by 5 and 9 and finds that it is divisible by both. Is her method sufficient and is her claim correct?

A.Method is sufficient and claim is correct. ✅

B.Method is sufficient but claim is incorrect.

C.Method is insufficient because 45 requires divisibility by 3 and 15, not 5 and 9.

D.Method is insufficient because 5 and 9 are not coprime.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: 45 = 5 × 9, and 5 and 9 are coprime (gcd=1). Therefore, a number divisible by both 5 and 9 is indeed divisible by 45. 5,760 ends with 0 (divisible by 5) and sum of digits = 5+7+6+0=18 (divisible by 9), so it is divisible by 45. The student's method is correct. Option C is a distractor based on the misconception that you must check factors of 45 like 3 and 15, but 3 and 15 are not coprime (gcd=3), so that would be insufficient. Option D incorrectly states they are not coprime. So A is correct.

Q4. A palindrome number reads the same forward and backward. Which of the following 5-digit palindromes is always divisible by 11?

A.Any 5-digit palindrome ✅

B.Only if the middle digit is 0

C.Only if the sum of outer digits equals the middle digit

D.None of the above

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: For a 5-digit palindrome of the form $abcba$ , the alternating sum for divisibility by 11 is: (a + c + a) - (b + b) = (2a + c - 2b). For this to be a multiple of 11, it must be 0, ±11, etc. But it is not always true. Example: 12321: (1+3+1)-(2+2)=5-4=1 not divisible by 11. So not all palindromes. Actually, the correct property: Any palindrome with an even number of digits is divisible by 11, but for odd number, not always. So the correct answer is D. Let's confirm: 12321 not divisible by 11 (12321/11=1120.09). So D is correct. This is a high-order question testing the rule's application to palindromes.

Q5. A student claims that the 15th multiple of 6 is 90, and the 20th multiple of 6 is 120. He then concludes that the 25th multiple of 6 must be 150 because the pattern of multiples increases by 30 every 5 steps. Is his reasoning correct for finding the 25th multiple? If not, what is the correct 25th multiple?

A.Yes, correct, 150

B.No, correct is 140

C.No, correct is 150 but reasoning is flawed ✅

D.No, correct is 180

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The student's final answer (150) happens to be correct because the 25th multiple of 6 is indeed $6 \times 25 = 150$ . However, his reasoning is flawed: he assumed a pattern based on two data points (15th to 20th) and extrapolated linearly, but for multiples of a fixed number, the increase is constant per step, not per block of 5 steps. The correct method is direct multiplication. This tests whether students understand that correlation does not imply causation in patterns, and that direct definition is always reliable.

Q6. A school is arranging students in rows for a drill. If there are 144 students and they must be arranged in rows such that each row has the same number of students, and the number of students per row must be a multiple of 8. Which of the following row counts is NOT possible?

A.8

B.12 ✅

C.16

D.24

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The number of students per row must divide 144 and be a multiple of 8. The divisors of 144 are 1,2,3,4,6,8,9,12,16,18,24,36,48,72,144. Among these, multiples of 8 are 8,16,24,48,72,144. 12 is a divisor but not a multiple of 8. Thus, 12 is not possible. This requires students to combine divisibility and multiples in a real-world context, testing both conceptual understanding and application.

Q7. If $n$ is a positive integer such that $n$ is a multiple of 4 and also a multiple of 6, then which of the following must be a multiple of 12?

n/2

n

2n

D.Both

n

and

2n

✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: If $n$ is a multiple of both 4 and 6, then $n$ is a common multiple of 4 and 6. The least common multiple (LCM) of 4 and 6 is 12, so $n$ is a multiple of 12. Therefore $n$ itself is a multiple of 12, and any multiple of $n$ (like $2n$ ) is also a multiple of 12. $n/2$ may not be an integer if $n$ is not even, but since n is multiple of 4 it is even, but $n/2$ may be 6, which is not multiple of 12. So only $n$ and $2n$ are guaranteed. This tests the fundamental relationship between common multiples and LCM.

Q8. Riya and Sohan are playing a game. Riya says: 'If a number is a multiple of 3, then it must be a multiple of 9.' Sohan says: 'No, every multiple of 9 is a multiple of 3, but not vice versa.' Who is correct and why?

A.Riya is correct because 3 divides 9

B.Sohan is correct because 9 is a multiple of 3 ✅

C.Both are correct because 3 and 9 are related

D.Neither is correct because multiples are independent

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Sohan is correct: If a number is a multiple of 9, say $9m$ , then it can be written as $3(3m)$ , so it is a multiple of 3. However, a multiple of 3 like 6 is not a multiple of 9. Riya's statement is a common misconception that divisibility in one direction implies the reverse. This tests the understanding of the subset relationship: multiples of 9 are a subset of multiples of 3, but not equal.

Q9. A number $x$ is such that when divided by 7, the quotient is a multiple of 4 and the remainder is 0. Which of the following could be the value of $x$ ?

A.56

B.84

C.112

D.All of the above ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: If $x$ is divisible by 7 and the quotient is a multiple of 4, then $x = 7 \times (4m) = 28m$ for some integer $m$ . So $x$ must be a multiple of 28. Check options: 56 = 28×2, 84 = 28×3, 112 = 28×4. All are multiples of 28. This question combines division, quotient, multiples, and requires checking multiple options, testing multi-step reasoning and the ability to generalize the condition into a single divisibility rule.

Q10. Find the least positive integer that is divisible by 6, 8, and 10, and is also a perfect square.

A.120

B.360

C.3600 ✅

D.7200

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: First, find the LCM of 6, 8, and 10. Prime factorizations: $6 = 2 \times 3$ , $8 = 2^3$ , $10 = 2 \times 5$ . LCM = $2^3 \times 3 \times 5 = 120$ . For a number to be a perfect square, each prime exponent must be even. $120 = 2^3 \times 3^1 \times 5^1$ . To make exponents even, we need to multiply by at least $2^1 \times 3^1 \times 5^1 = 30$ . So the smallest such number is $120 \times 30 = 3600$ .

Q11. A student claims that (4,536) is divisible by both (2) and (3), so it must also be divisible by (6). Which reasoning best evaluates the claim?

A.The claim is correct because divisibility by either (2) or (3) is enough

B.The claim is correct because the last digit is even

C.The claim is correct because (4+5+3+6=18)

D.The claim is incorrect because divisibility by (6) requires divisibility by both (2) and (3) ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Divisibility by (6) requires both conditions simultaneously. The last digit (6) shows divisibility by (2), while the digit sum (18) shows divisibility by (3). Therefore (4,536) is divisible by (6).

Q12. A five-digit number has the form (47x25). Without calculating the whole number, which value of (x) makes the number divisible by both (3) and (5)?

A.0 ✅

B.2

C.4

D.8

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The final digit is already (5), so divisibility by (5) is guaranteed. For divisibility by (3), the digit sum must be divisible by (3). The fixed digits give (4+7+2+5=18), so (18+x) must be divisible by (3). Among the choices, only (x=0) satisfies this.

Q13. A factory labels boxes with the numbers shown on a number line: (120, 121, 122, 123, 124, 125, 126). A box must satisfy divisibility by both (2) and (3). Which labeled position should be selected?

A.121

B.122

C.123

D.126 ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The number must satisfy both conditions. (126) is even, so it is divisible by (2), and its digit sum (1+2+6=9) is divisible by (3). The other listed candidates fail at least one of the two requirements.

Q14. A student tests (7,245) for divisibility by (2) by checking whether its digit sum is even. The student concludes that the number is divisible by (2). What is the error?

A.Digit sums are used for testing divisibility by (5)

B.Divisibility by (2) depends on the last digit ✅

C.The number must have an even number of digits

D.Divisibility by (2) depends on the first digit

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The student's method confuses two different tests. A whole number is divisible by (2) when its last digit is even. Since (7,245) ends in (5), it is not divisible by (2), regardless of its digit sum.

Q15. A delivery system accepts an ID only if it is divisible by (10). Four proposed IDs are (3,450), (3,455), (3,460), and (3,465). If the system rejects every ID whose last digit is not (0), which IDs will be accepted?

A.Only (3,450)

B.Only (3,460)

C.(3,450) and (3,460) ✅

D.All four IDs

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Divisibility by (10) requires the last digit to be (0). Both (3,450) and (3,460) end in (0), while (3,455) and (3,465) end in (5). Therefore exactly two IDs are accepted.

Q16. Two students analyze (8,132). Student A says it is divisible by (2) because it ends in (2). Student B says it is divisible by (3) because (8+1+3+2=14). Which conclusion is correct?

A.Only Student A is correct ✅

B.Only Student B is correct

C.Both students are correct

D.Neither student is correct

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Student A uses the correct test because (8,132) ends in the even digit (2). Student B incorrectly concludes divisibility by (3): the digit sum is (14), which is not divisible by (3). Thus only Student A is correct.

Q17. A number is divisible by (6) and (10). Which statement must be true about the number?

A.It must be odd and divisible by (3)

B.It must end in (0) and be divisible by (3) ✅

C.It must end in (5) and be divisible by (2)

D.It must have a digit sum of (10)

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Divisibility by (10) guarantees that the number ends in (0), while divisibility by (6) guarantees divisibility by both (2) and (3). Therefore the number must end in (0) and its digit sum must be divisible by (3). Among the options, only B states a necessary condition.

🔗 Related Topics (MCQs)

📝 Place value of whole numbers MCQs 📝 How to round whole numbers MCQs 📝 Prime factorization and (LCM) Least Common Multiples MCQs 📝 Variables and algebraic symbols MCQs 📝 Order of operations PEMDAS MCQs

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