CBSE Class 6 Mathematics Playing With Numbers Worksheet Set 05

Fill in the blanks.

Question 1. The greatest single digit prime number is ________.
Answer: 7
In simple words: Prime numbers between 1 and 9 are 2, 3, 5, and 7, among which 7 is the largest.

Exam Tip: Remember that 1 is neither prime nor composite, and 2 is the only even prime number.

Question 2. The only even prime number is________.
Answer: 2
In simple words: All other even numbers are divisible by 2, making them composite.

Exam Tip: 2 is the smallest and the only even prime number.

Question 3. ________ is an example for a perfect number.
Answer: 6 (or 28)
In simple words: A number is called a perfect number if the sum of all its proper divisors equals the number itself. For 6, the divisors are 1, 2, and 3, and \(1 + 2 + 3 = 6\).

Exam Tip: The first two perfect numbers are 6 and 28.

Question 4. The HCF of two prime numbers is________.
Answer: 1
In simple words: Prime numbers have no common factors other than 1.

Exam Tip: HCF of any two distinct prime numbers is always 1.

Question 5. The LCM of two numbers that have no common factor is ________.
Answer: Their product
In simple words: When two numbers are co-prime, their least common multiple is simply obtained by multiplying them together.

Exam Tip: If HCF of two numbers is 1, their LCM is equal to their product.

Question 6. A number that is neither prime nor composite is ________.
Answer: 1
In simple words: 1 has only one factor (itself), so it does not fit the definition of prime (two factors) or composite (more than two factors).

Exam Tip: 1 is a unique number - neither prime nor composite.

Question 7. The greatest prime number less than 100 is ________.
Answer: 97
In simple words: Checking downwards from 99, 99 is divisible by 3 and 9, 98 is even, but 97 has no factors other than 1 and 97.

Question 8. ________ is the smallest composite number.
Answer: 4
In simple words: The numbers 1, 2, 3 are not composite. 4 has factors 1, 2, and 4, making it the first composite number.

Exam Tip: Smallest prime is 2, smallest composite is 4.

Question 9. A number is divisible by 6 if it is divisible by both ________ and ________.
Answer: 2, 3
In simple words: Since \(6 = 2 \times 3\) and 2 and 3 are co-prime, any number divisible by both must be divisible by 6.

Exam Tip: Check for divisibility by 2 (even number) and 3 (sum of digits divisible by 3).

Question 10. The LCM of 5 and 20 is ________.
Answer: 20
In simple words: Since 20 is a multiple of 5, 20 is the smallest common multiple.

Question 11. Two numbers having only 1 as common factor are called ________.
Answer: Co-prime numbers (or relatively prime numbers)
In simple words: Numbers that share no common factors other than 1 are co-prime.

Exam Tip: Co-prime numbers do not need to be prime numbers themselves (e.g., 8 and 9).

Question 12. The sum of any two odd numbers is ________.
Answer: Even
In simple words: Adding two odd quantities always results in pairs, making the sum even (e.g., \(3 + 5 = 8\)).

Exam Tip: Odd + Odd = Even; Odd + Even = Odd.

Question 13. ________ is a factor of every number.
Answer: 1
In simple words: Every integer can be divided evenly by 1.

Exam Tip: 1 is the universal factor and every number is a multiple of itself.

Question 14. The numbers that are multiples of two are called ________.
Answer: Even numbers
In simple words: Numbers ending in 0, 2, 4, 6, 8 are multiples of 2.

Exam Tip: Even numbers are divisible by 2 without any remainder.

Question 15. The HCF of two consecutive even numbers is ________.
Answer: 2
In simple words: Consecutive even numbers differ by 2 and both are divisible by 2 (e.g., 4 and 6 have an HCF of 2).

Exam Tip: The HCF of any two consecutive even numbers is always 2.

Write True or False. If false give the correct answer.

Question 16. 10 is a perfect number.
Answer: False. Correct statement: 10 is not a perfect number (Proper divisors of 10 are 1, 2, 5, and their sum is \(1 + 2 + 5 = 8 \neq 10\)).
In simple words: A perfect number equals the sum of its proper divisors.

Exam Tip: Always verify divisors before declaring a number perfect.

Question 17. If an even number is divided by 2, the quotient is always odd.
Answer: False. Correct statement: If an even number is divided by 2, the quotient can be either even or odd (e.g., \(4 \div 2 = 2\) which is even, and \(6 \div 2 = 3\) which is odd).
In simple words: Test with counterexamples like 4 or 8.

Exam Tip: Never assume parity of quotients without testing.

Question 18. All even numbers are composite.
Answer: False. Correct statement: 2 is an even number, but it is a prime number.
In simple words: 2 is even yet prime, making the statement false.

Exam Tip: Watch out for exceptions like 2 in number properties.

Question 19. 572 is divisible by 4.
Answer: True.
In simple words: A number is divisible by 4 if the number formed by its last two digits is divisible by 4. Here, 72 is divisible by 4 (\(72 \div 4 = 18\)).

Exam Tip: Focus only on the last two digits for divisibility by 4.

Question 20. If a number is divisible by 14, it must be divisible by 7.
Answer: True.
In simple words: Since 7 is a factor of 14, any multiple of 14 is also a multiple of 7.

Exam Tip: If a number is divisible by \(n\), it is divisible by all factors of \(n\).

Do as directed.

Question 21. Write all prime numbers less than 15.
Answer: 2, 3, 5, 7, 11, 13
In simple words: List all numbers less than 15 that have exactly two distinct factors (1 and themselves).

Exam Tip: 1 is excluded from prime numbers.

Question 22. Write all the factors and first 5 multiples of 20.
Answer:
Factors of 20: 1, 2, 4, 5, 10, 20
First 5 multiples of 20: 20, 40, 60, 80, 100
In simple words: Factors divide 20 completely; multiples are products of 20 with 1, 2, 3, 4, and 5.

Exam Tip: Factors are finite, while multiples are infinite.

Question 23. Express 48 as sum of two odd primes.
Answer: \(5 + 43 = 48\) (or \(7 + 41 = 48\), or \(11 + 37 = 48\), or \(13 + 35\) invalid since 35 not prime, or \(17 + 31 = 48\), or \(19 + 29 = 48\))
In simple words: Find two prime numbers other than 2 that add up to 48.

Exam Tip: Check that both numbers chosen are prime and odd.

Question 24. Using divisibility tests determine whether 218750 is divisible by 6, by 8 and by 9.
Answer:
Divisibility by 6: The number ends in 0 (even, so divisible by 2). Sum of digits = \(2 + 1 + 8 + 7 + 5 + 0 = 23\). Since 23 is not divisible by 3, the number is NOT divisible by 3, and hence NOT divisible by 6.
Divisibility by 8: Last three digits form the number 750. \(750 \div 8 = 93\) remainder 6. Since it does not divide evenly, it is NOT divisible by 8.
Divisibility by 9: Sum of digits = 23. Since 23 is not divisible by 9, the number is NOT divisible by 9.
In simple words: Apply divisibility rules for 2, 3, 8, and 9 step by step.

Exam Tip: Divisibility by 6 requires both 2 and 3 test to pass.

Question 25. Write the digit in the blank space so that the number 701__9308 is divisible by 11.
Answer: Let the missing digit be \(x\). The number is \(701x9308\).
Sum of digits at odd places (from right): \(8 + 3 + x + 0 = 11 + x\)
Sum of digits at even places (from right): \(0 + 9 + 1 + 7 = 17\)
Difference = \((17) - (11 + x) = 6 - x\)
For the number to be divisible by 11, the difference must be 0 or a multiple of 11.
Setting \(6 - x = 0\) gives \(x = 6\).
In simple words: Find the difference between alternate digit sums and make it a multiple of 11.

Exam Tip: Sum alternate digits carefully starting from the units place.

Question 26. Are 16 and 81 co-prime?
Answer: Yes.
Factors of 16: 1, 2, 4, 8, 16
Factors of 81: 1, 3, 9, 27, 81
Common factor: Only 1.
In simple words: Since their only common factor is 1, they are co-prime.

Exam Tip: Numbers do not need to be prime themselves to be co-prime.

Question 27. Write the smallest four digit number and express it in terms of prime factors.
Answer:
Smallest four-digit number = 1000
Prime factorization of 1000:
\(1000 = 2 \times 500\)
\(= 2 \times 2 \times 250\)
\(= 2 \times 2 \times 2 \times 125\)
\(= 2 \times 2 \times 2 \times 5 \times 5 \times 5\)
\(= 2^3 \times 5^3\)
In simple words: Break down 1000 using prime division until only primes remain.

Exam Tip: Express final prime factors in exponent form for clarity.

Question 28. Is it correct to say that the HCF of 8 and 15 is 0? If not why?
Answer: No, it is not correct. The HCF of 8 and 15 is 1 because they are co-prime and share no common factors other than 1. HCF can never be 0 for natural numbers.
In simple words: The lowest possible common factor for any set of numbers is 1.

Exam Tip: HCF of co-prime numbers is always 1, never 0.

Question 29. Find the greatest 3 digit number which is divisible by 12, 18 and 24.
Answer:
First, find the LCM of 12, 18, and 24.
Prime factorizations:
\(12 = 2^2 \times 3\)
\(18 = 2 \times 3^2\)
\(24 = 2^3 \times 3\)
\(\text{LCM} = 2^3 \times 3^2 = 8 \times 9 = 72\)
Greatest 3-digit number = 999.
Divide 999 by 72: \(999 \div 72 = 13\) with a remainder of 63.
Greatest 3-digit number divisible by 72 = \(999 - 63 = 936\).
In simple words: Find the LCM of the divisors, then find the largest 3-digit multiple of that LCM.

Exam Tip: Subtract the remainder from the 3-digit number to find the required multiple.

Question 30. Find the least number which when divided by 8, 10 and 16 and leave remainder 3 in each case.
Answer:
First, find the LCM of 8, 10, and 16.
\(8 = 2^3\)
\(10 = 2 \times 5\)
\(16 = 2^4\)
\(\text{LCM} = 2^4 \times 5 = 16 \times 5 = 80\)
Required number = \(\text{LCM} + \text{remainder} = 80 + 3 = 83\).
In simple words: Find the LCM of divisors and add the common remainder.

Exam Tip: "Leave remainder 3 in each case" means you add 3 to the LCM.

Question 31. Two tankers contain 580 litres and 860 litres of oil respectively. Find the maximum capacity of the container which can measure the oil of both the tankers when used an exact number of times.
Answer:
Maximum capacity = HCF of 580 and 860.
Prime factorization of 580: \(580 = 2^2 \times 5 \times 29\)
Prime factorization of 860: \(860 = 2^2 \times 5 \times 43\)
Common prime factors = \(2^2 \times 5 = 4 \times 5 = 20\).
Maximum capacity = 20 litres.
In simple words: To find the largest measuring container, calculate the HCF of the two quantities.

Exam Tip: Maximum capacity / greatest measure problems always require HCF.

Question 32. The length, breadth and width of a room are 80m, 85m and 90 m respectively. Find the length of the largest tape that can measure the three dimensions of the room exactly.
Answer:
Length of the largest tape = HCF of 80, 85, and 90.
\(80 = 2^4 \times 5\)
\(85 = 5 \times 17\)
\(90 = 2 \times 3^2 \times 5\)
Common factor in all three numbers = 5.
Length of the largest tape = 5 m.
In simple words: Find the HCF of 80, 85, and 90 to get the maximum exact measuring length.

Exam Tip: Look for keywords like "largest tape" or "exact measurement" to apply HCF.

Question 33. Find the smallest 3 digit number that is exactly divisible by 6, 15 and 18.
Answer:
First, find the LCM of 6, 15, and 18.
\(6 = 2 \times 3\)
\(15 = 3 \times 5\)
\(18 = 2 \times 3^2\)
\(\text{LCM} = 2 \times 3^2 \times 5 = 2 \times 9 \times 5 = 90\)
Multiples of 90: 90, 180, 270, ...
The smallest 3-digit multiple of 90 is 180.
In simple words: Find the LCM and take its smallest 3-digit multiple.

Exam Tip: First positive multiple of LCM that falls in the 3-digit range is 180.

Question 34. Three bells commence tolling simultaneously. If they ring at intervals of 40sec, 48sec and 50 sec respectively when will they ring together again?
Answer:
Time after which they ring together = LCM of 40, 48, and 50.
\(40 = 2^3 \times 5\)
\(48 = 2^4 \times 3\)
\(50 = 2 \times 5^2\)
\(\text{LCM} = 2^4 \times 3 \times 5^2 = 16 \times 3 \times 25 = 1200 \text{ seconds}\)
Converting to minutes: \(1200 \div 60 = 20 \text{ minutes}\).
In simple words: Simultaneous events recurring at different intervals are solved by finding the LCM.

Question 35. Two wires of length 448 cm and 616 cm are to be cut in to small pieces of equal length without wasting the wire. What is the maximum length of each piece?
Answer:
Maximum length of each piece = HCF of 448 and 616.
\(448 = 2^6 \times 7 = 64 \times 7\)
\(616 = 2^3 \times 7 \times 11 = 8 \times 7 \times 11 = 56 \times 11\)
Common factors = \(2^3 \times 7 = 8 \times 7 = 56\).
Maximum length = 56 cm.
In simple words: Cutting wires into equal maximum pieces requires finding the HCF of their lengths.

Exam Tip: Maximum length without waste = HCF.

Question 36. On a track for remote controlled racing cars, racing car A completes the track in 28 seconds and racing car B completes the track in 24 second. If they both start at the same time, after how many seconds will they be side by side again?
Answer:
Time when they meet again = LCM of 28 and 24.
\(28 = 2^2 \times 7\)
\(24 = 2^3 \times 3\)
\(\text{LCM} = 2^3 \times 3 \times 7 = 8 \times 3 \times 7 = 168 \text{ seconds}\).
In simple words: Meeting again at the starting point means finding the LCM of lap times.

Exam Tip: Repetitive cyclical events at starting points use LCM.

Question 37. 294 blue balls, 252 pink balls and 210 yellow balls are distributed equally among some students with none left over. What is the biggest possible number of students?
Answer:
Biggest possible number of students = HCF of 294, 252, and 210.
\(294 = 2 \times 3 \times 7^2\)
\(252 = 2^2 \times 3^2 \times 7\)
\(210 = 2 \times 3 \times 5 \times 7\)
Common prime factors = \(2 \times 3 \times 7 = 42\).
Biggest number of students = 42.
In simple words: Equal distribution among maximum students without leftovers requires finding the HCF.

Exam Tip: "Biggest possible number" dividing items equally points to HCF.

Question 38. Jesslyn goes to the market every 64 days. Christine goes to the same market every 72 days. They met each other one day. How many days later will they meet each other again?
Answer:
Days after which they meet again = LCM of 64 and 72.
\(64 = 2^6\)
\(72 = 2^3 \times 3^2 = 8 \times 9\)
\(\text{LCM} = 2^6 \times 3^2 = 64 \times 9 = 576 \text{ days}\).
In simple words: Future meeting dates for periodic schedules are calculated using LCM.

Exam Tip: Recurring events happening at different intervals require LCM.

Question 39. Leonard wants to cut identical square as big as he can from a piece of paper 168 mm by 196 mm. What is the length of each square?
Answer:
Length of each square = HCF of 168 and 196.
\(168 = 2^3 \times 3 \times 7 = 8 \times 21\)
\(196 = 2^2 \times 7^2 = 4 \times 49\)
Common prime factors = \(2^2 \times 7 = 4 \times 7 = 28\).
Length of each square = 28 mm.
In simple words: Cutting largest possible identical squares from a rectangle requires finding the HCF of the dimensions.

Exam Tip: "As big as he can" cutting problems always use HCF.

Question 40. As a humanitarian effort, food ration is distributed to each refugee in a refugee camp. If a day's ration is 284 packets of biscuits, 426 packets of instant noodles and 710 bottles of water, how many refugees are there in the camp?
Answer:
Number of refugees = HCF of 284, 426, and 710.
\(284 = 2^2 \times 71\)
\(426 = 2 \times 3 \times 71\)
\(710 = 2 \times 5 \times 71\)
Common prime factor in all three numbers = \(2 \times 71 = 142\).
Number of refugees = 142.
In simple words: Maximum number of people receiving equal shares without leftovers is found using HCF.

Exam Tip: Equal sharing among maximum recipients is a standard HCF word problem.