Factors and Multiples Quiz - ICSE Class 5 Mathematics
Free practice quiz: Factors and Multiples (ICSE Class 5 Mathematics). 59 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
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Which of the following numbers is a factor of 72, but not a factor of 48?
- 6
- 8
- 12
- 9
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Answer: D. 9
Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72. Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. The number 9 is a factor of 72 but not of 48.
What is the smallest composite number?
- 1
- 2
- 3
- 4
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Answer: D. 4
A composite number is a natural number that has at least one divisor other than 1 and itself. The number 1 is neither prime nor composite. The number 2 is prime (divisors are 1 and 2). The number 3 is prime (divisors are 1 and 3). The number 4 has divisors 1, 2, and 4. Since it has a divisor other than 1 and itself (which is 2), it is a composite number. It is the smallest such number.
What is the smallest number that leaves a remainder of 1 when divided by 2, 3, 4, 5, and 6?
- 60
- 61
- 120
- 121
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Answer: B. 61
First, find the LCM of 2, 3, 4, 5, and 6. Prime factorization of 2 = 2. Prime factorization of 3 = 3. Prime factorization of 4 = 2^2. Prime factorization of 5 = 5. Prime factorization of 6 = 2 x 3. LCM(2, 3, 4, 5, 6) = 2^2 x 3 x 5 = 4 x 3 x 5 = 60. The numbers that leave a remainder of 1 when divided by these numbers will be of the form (LCM x k) + 1. The smallest such number is when k=1. So, (60 x 1) + 1 = 60 + 1 = 61.
What is the greatest 4-digit number divisible by 15?
- 9990
- 9995
- 9985
- 9975
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Answer: A. 9990
The greatest 4-digit number is 9999. For a number to be divisible by 15, it must be divisible by both 3 and 5. A number is divisible by 5 if its last digit is 0 or 5. A number is divisible by 3 if the sum of its digits is divisible by 3. Let's check numbers downwards from 9999. 9999: Sum of digits = 36 (divisible by 3). Last digit is 9 (not divisible by 5). 9998: Sum of digits = 35 (not divisible by 3). Last digit is 8 (not divisible by 5). 9997: Sum of digits = 34 (not divisible by 3). Last digit is 7 (not divisible by 5). 9996: Sum of digits = 33 (divisible by 3). Last digit is 6 (not divisible by 5). 9995: Sum of digits = 32 (not divisible by 3). Last digit is 5 (divisible by 5). 9994: Sum of digits = 31 (not divisible by 3). Last digit is 4 (not divisible by 5). 9993: Sum of digits = 30 (divisible by 3). Last digit is 3 (not divisible by 5). 9992: Sum of digits = 29 (not divisible by 3). Last digit is 2 (not divisible by 5). 9991: Sum of digits = 28 (not divisible by 3). Last digit is 1 (not divisible by 5). 9990: Sum of digits = 27 (divisible by 3). Last digit is 0 (divisible by 5). Since 9990 is divisible by both 3 and 5, it is divisible by 15. As we checked downwards, it is the greatest 4-digit number divisible by 15. Alternatively, divide 9999 by 15: 9999 / 15 = 666 with a remainder of 9. So, 9999 - 9 = 9990 is the greatest multiple of 15 less than or equal to 9999.
What is the smallest number divisible by 11 and 13?
- 11
- 13
- 143
- 273
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Answer: C. 143
To find the smallest number divisible by 11 and 13, we need to find their LCM. Since 11 and 13 are prime numbers, their LCM is their product. LCM(11, 13) = 11 x 13 = 143.