CBSE Class 6 Mathematics Playing With Numbers Worksheet Set 08

Chapter 3. Playing With Numbers

I. Fill in the blanks:

Question 1. 1 is a factor of every number.
Answer: 1
In simple words: Number 1 divides every number completely without leaving any remainder.

Exam Tip: Remember that 1 is a universal factor and is neither prime nor composite.

Question 2. Every number is a factor of itself.
Answer: itself
In simple words: Any number can be divided by itself to give a quotient of 1.

Exam Tip: Every number has at least two trivial factors: 1 and itself (except for the number 1).

Question 3. Every factor is less than or equal to the given no.
Answer: less than or equal to
In simple words: Factors of a number are always smaller than or equal to the number itself.

Exam Tip: The largest factor of a number is the number itself.

Question 4. Every multiple is greater than or equal to the given no.
Answer: greater than or equal to
In simple words: Multiples of a number are obtained by multiplying by 1, 2, 3, etc., so they are always greater than or equal to the number.

Exam Tip: The smallest multiple of a number is the number itself.

Question 5. 1 is the smallest factor of a no.
Answer: 1
In simple words: 1 is the starting point when listing the factors of any counting number.

Question 6. The number itself is the greatest factor of a no.
Answer: The number itself
In simple words: You cannot have a factor larger than the number itself.

Exam Tip: Factors are finite in number for any given natural number.

Question 7. The number itself is the smallest multiple of a no.
Answer: The number itself
In simple words: The first multiple is obtained by multiplying the number by 1.

Exam Tip: Multiples are infinite, but the smallest multiple is fixed.

Question 8. A no. for which sum of all its factors is equal to twice the no. is called a perfect number.
Answer: perfect number
In simple words: If the sum of all factors equals \(2 \times \text{number}\), it is a perfect number (e.g., 6 and 28).

Exam Tip: For 6, factors are 1, 2, 3, 6. Sum = \(1 + 2 + 3 + 6 = 12 = 2 \times 6\).

Question 9. The smallest perfect no. is 6.
Answer: 6
In simple words: 6 is the first number whose proper divisors add up to the number itself (\(1 + 2 + 3 = 6\)).

Question 10. Numbers having only two factors are called prime numbers.
Answer: prime numbers
In simple words: Prime numbers are divisible only by 1 and themselves.

Exam Tip: 2 is the only even prime number.

Question 11. Numbers having more than two factors are called composite numbers.
Answer: composite numbers
In simple words: Composite numbers have at least one factor other than 1 and themselves.

Exam Tip: 1 is neither prime nor composite because it has only one factor.

Question 12. Smallest prime no. is 2.
Answer: 2
In simple words: 2 has exactly two factors: 1 and 2.

Question 13. Smallest composite no. is 4.
Answer: 4
In simple words: 4 has factors 1, 2, and 4 (more than two factors).

Exam Tip: Remember 1 is neither prime nor composite, 2 and 3 are prime, and 4 is the first composite number.

Question 14. Smallest two digit prime no. is 11.
Answer: 11
In simple words: 10 is composite, while 11 is divisible only by 1 and 11.

Exam Tip: Numbers 10, 12, 14, 15, 16, 18 are composite.

Question 15. Greatest two digit prime no. is 97.
Answer: 97
In simple words: 97 cannot be divided evenly by any number other than 1 and 97.

Question 16. Greatest one digit prime no. is 7.
Answer: 7
In simple words: Single-digit primes are 2, 3, 5, and 7, among which 7 is the largest.

Exam Tip: 8 and 9 are composite numbers.

Question 17. Smallest two digit composite no. is 10.
Answer: 10
In simple words: 10 has factors 1, 2, 5, and 10.

Exam Tip: 10 is the smallest number formed with two digits that is not prime.

Question 18. Greatest two digit composite no. is 99.
Answer: 99
In simple words: 99 is divisible by 1, 3, 9, 11, 33, and 99.

Exam Tip: 100 is the smallest three-digit number.

Question 19. Greatest one digit composite no. is 9.
Answer: 9
In simple words: Single-digit numbers are 1 to 9. 1 is neither, 2, 3, 5, 7 are prime, and 4, 6, 8, 9 are composite.

Exam Tip: 9 has factors 1, 3, and 9.

Question 20. 1 is neither prime nor composite.
Answer: 1
In simple words: 1 has only one factor (itself), so it fails the definition of both prime and composite numbers.

Exam Tip: This is a very common trick question in exams.

Question 21. The smallest even prime no. is 2.
Answer: 2
In simple words: All other even numbers are divisible by 2, making them composite. Thus, 2 is the only even prime.

Question 22. The sum of any two odd nos. is even.
Answer: even
In simple words: Adding two odd quantities always results in an even total (e.g., \(3 + 5 = 8\)).

Exam Tip: Odd + Odd = Even.

Question 23. The sum of any two even nos. is even.
Answer: even
In simple words: Adding two even quantities always results in an even total (e.g., \(4 + 6 = 10\)).

Exam Tip: Even + Even = Even.

Question 24. The greatest prime no. between 30 and 40 is 37.
Answer: 37
In simple words: The primes between 30 and 40 are 31 and 37, and 37 is the larger one.

Exam Tip: Check odd numbers ending in 1, 3, 7, 9 when searching for primes.

II. Answer the following:

Question 1. Find all the multiples of 13 upto 100.
Answer: Multiples of 13 up to 100 are obtained by multiplying 13 by 1, 2, 3, up to 7:
\(13 \times 1 = 13\)
\(13 \times 2 = 26\)
\(13 \times 3 = 39\)
\(13 \times 4 = 52\)
\(13 \times 5 = 65\)
\(13 \times 6 = 78\)
\(13 \times 7 = 91\)
In simple words: We list the table of 13 until the product exceeds 100.

Exam Tip: Stop when the next multiple exceeds 100 (\(13 \times 8 = 104\)).

Question 2. Write 3 pairs of prime nos. whose difference is 2.
Answer: Pairs of twin primes (prime numbers with a difference of 2) are:
1. (3, 5)
2. (11, 13)
3. (17, 19)
In simple words: Twin primes are pairs of prime numbers that have a difference of 2.

Exam Tip: Other valid pairs include (41, 43), (59, 61), and (71, 73).

Question 3. Write all the composite nos. between 50 and 70.
Answer: First, identify the prime numbers between 50 and 70, which are 53, 59, 61, and 67. All other numbers in this range are composite:
51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69
In simple words: Composite numbers are all integers between 50 and 70 that are not prime.

Exam Tip: Always eliminate primes first to avoid missing any composite numbers.

Question 4. The product of three consecutive nos. is always divisible by 6.
Answer: Yes, the product of three consecutive numbers is always divisible by 6 because among any three consecutive numbers, at least one is a multiple of 2 (even) and one is a multiple of 3. Since 2 and 3 are coprime, their product (which is 6) must divide the product of the three numbers.
In simple words: Three numbers in a row always contain an even number and a multiple of 3, so their product always divides by 6.

Exam Tip: Example: \(2 \times 3 \times 4 = 24\), and 24 is divisible by 6.

Question 5. Express 51 as the sum of three odd primes.
Answer: \(51 = 3 + 17 + 31\)
Check: 3, 17, and 31 are all odd prime numbers, and their sum is \(3 + 17 + 31 = 51\).
In simple words: Find three prime numbers (other than 2) that add up to 51.

Exam Tip: Trial and error with smaller odd primes helps solve such addition problems quickly.

Question 6. Write all the twin primes below 40.
Answer: Twin prime pairs below 40 are:
1. (3, 5)
2. (5, 7)
3. (11, 13)
4. (17, 19)
5. (29, 31)
In simple words: These are pairs of prime numbers with a difference of 2, all lying under 40.

Question 7. Determine if 25110 is divisible by 45.
Answer: Prime factorization of 45 is \(5 \times 9\) (since 5 and 9 are coprime).
1. Divisibility by 5: The number 25110 ends in 0, so it is divisible by 5.
2. Divisibility by 9: Sum of digits of 25110 = \(2 + 5 + 1 + 1 + 0 = 9\). Since 9 is divisible by 9, the number is divisible by 9.
Since 25110 is divisible by both 5 and 9, it is divisible by 45.
In simple words: If a number is divisible by both 5 and 9, it is divisible by their product, 45.

Exam Tip: Always break composite divisors into coprime factors like 5 and 9 for 45, or 3 and 4 for 12.

Question 8. Determine if 1375 is divisible by 55.
Answer: Prime factorization of 55 is \(5 \times 11\) (5 and 11 are coprime).
1. Divisibility by 5: The number 1375 ends in 5, so it is divisible by 5.
2. Divisibility by 11: Difference between the sum of digits at odd places and even places = \((5 + 3) - (7 + 1) = 8 - 8 = 0\). Since 0 is divisible by 11, the number is divisible by 11.
Since 1375 is divisible by both 5 and 11, it is divisible by 55.
In simple words: Check divisibility by 5 and 11 separately. Since both tests pass, 1375 is divisible by 55.

Exam Tip: The alternating sum rule for 11 is essential for larger numbers.

Question 9. Determine the smallest 3-digit no. which is exactly divisible by 4, 12, 36.
Answer: First, find the LCM of 4, 12, and 36.
Multiples of 36: 36, 72, 108, ...
Since 36 is already a multiple of 4 and 12, the LCM of 4, 12, and 36 is 36.
We need the smallest 3-digit multiple of 36.
Multiples of 36 are: \(36 \times 1 = 36\), \(36 \times 2 = 72\), \(36 \times 3 = 108\).
Thus, the smallest 3-digit number is 108.
In simple words: Find the LCM of the numbers, then find the smallest multiple of that LCM which has 3 digits.

Exam Tip: If one number is a multiple of the others, it is automatically the LCM.

Question 10. Find all the factors of 86.
Answer: Factors of 86 are found by testing division:
\(1 \times 86 = 86\)
\(2 \times 43 = 86\)
Thus, the factors of 86 are 1, 2, 43, and 86.
In simple words: Find all pairs of numbers that multiply to give 86.

Exam Tip: Since 43 is a prime number, there are no further factors.

Question 11. Find the least no. which when divided by 8, 10 and 12 leave remainder 7 in each case.
Answer: First, find the LCM of 8, 10, and 12.
Prime factorization:
\(8 = 2^3\)
\(10 = 2 \times 5\)
\(12 = 2^2 \times 3\)
LCM = \(2^3 \times 3 \times 5 = 8 \times 3 \times 5 = 120\).
The required number leaves a remainder of 7 in each case, so we add 7 to the LCM:
Number = \(120 + 7 = 127\).
In simple words: Find the LCM of the divisors and add the constant remainder to it.

Exam Tip: Whenever a common remainder is required, compute \(\text{LCM} + \text{remainder}\).

Question 12. Write the smallest 4-digit no. and express it in the form of its prime factors.
Answer: The smallest 4-digit number is 1000.
Prime factorization of 1000:
\(1000 = 10 \times 10 \times 10\)
\(1000 = (2 \times 5) \times (2 \times 5) \times (2 \times 5)\)
\(1000 = 2^3 \times 5^3\) (or \(2 \times 2 \times 2 \times 5 \times 5 \times 5\))
In simple words: Break 1000 down by dividing by prime numbers 2 and 5 repeatedly.

Exam Tip: Use factor trees or repeated division by prime numbers for accuracy.

Question 13. Find the missing digit so that the no. is divisible by 3. (572__6) (smallest and greatest digits)
Answer: Let the missing digit be \(x\). The number is \(572x6\).
Sum of digits = \(5 + 7 + 2 + x + 6 = 20 + x\).
For the number to be divisible by 3, the sum of its digits must be a multiple of 3.
Multiples of 3 near 20 are 21, 24, 27, etc.
- If \(20 + x = 21 \implies x = 1\) (smallest digit)
- If \(20 + x = 24 \implies x = 4\)
- If \(20 + x = 27 \implies x = 7\) (greatest single digit since \(20 + 10 = 30\) where 10 is not a single digit)
Thus, the smallest digit is 1 and the greatest digit is 7.
In simple words: Find single digits that make the sum of all digits a multiple of 3.

Question 14. Using divisibility tests, determine 406839 is divisible by 11.
Answer: Sum of digits at odd places (from the right: 9, 8, 0) = \(9 + 8 + 0 = 17\).
Sum of digits at even places (from the right: 3, 6, 4) = \(3 + 6 + 4 = 13\).
Difference = \(17 - 13 = 4\).
Since 4 is not divisible by 11 (nor is it 0), the number 406839 is not divisible by 11.
In simple words: Subtract the sum of even-position digits from the sum of odd-position digits. If the result is 0 or a multiple of 11, it is divisible.

Exam Tip: Always count digit positions starting from the rightmost digit as position 1 (odd).

Question 15. Using divisibility test, determine 639240 is divisible by 8.
Answer: A number is divisible by 8 if the number formed by its last three digits is divisible by 8.
The last three digits of 639240 form the number 240.
\(240 \div 8 = 30\) (exact division with no remainder).
Therefore, 639240 is divisible by 8.
In simple words: You only need to check if the last three digits can be divided by 8.

Exam Tip: This rule saves time compared to performing long division on the entire 6-digit number.

Question 16. Using divisibility test, determine 297144 is divisible by 6.
Answer: A number is divisible by 6 if it is divisible by both 2 and 3.
1. Divisibility by 2: The number 297144 ends in 4 (an even number), so it is divisible by 2.
2. Divisibility by 3: Sum of digits = \(2 + 9 + 7 + 1 + 4 + 4 = 27\). Since 27 is divisible by 3, the number is divisible by 3.
Since 297144 is divisible by both 2 and 3, it is divisible by 6.
In simple words: Check if the number is even and if the sum of its digits is a multiple of 3.

Exam Tip: Divisibility by 6 always requires checking both 2 and 3 simultaneously.