CBSE Class 6 Mathematics Playing with Numbers Assignments Set 04

Write True (T) or False (F) against the following Statements.

Question 1. If a number is divisible by 3, it must be divisible by 9.
Answer: (F) False
For example, the number 6 is divisible by 3 (\(6 \div 3 = 2\)), but 6 is not divisible by 9.
In simple words: Just because a number goes in the 3 times table, it does not mean it goes in the 9 times table.

Exam Tip: Test divisibility using a counter-example like 6 or 12 to easily disprove false statements.

Question 2. If a number is divisible by 9, it must be divisible by 3.
Answer: (T) True
Since 9 is a multiple of 3 (\(9 = 3 \times 3\)), any number that is a multiple of 9 is also automatically a multiple of 3.
In simple words: Any number that divides by 9 will always divide by 3 because 3 is a factor of 9.

Exam Tip: Remember that if \(a\) divides \(b\), any factor of \(a\) also divides \(b\).

Question 3. If a number is divisible by 4, it must be divisible by 8.
Answer: (F) False
For example, the number 12 is divisible by 4 (\(12 \div 4 = 3\)), but 12 is not divisible by 8.
In simple words: A smaller divisor does not guarantee divisibility by a larger multiple.

Exam Tip: Be careful not to confuse divisibility rules of lower numbers with higher multiples.

Question 4. If a number is divisible by 18, it must be divisible by 3 and 6.
Answer: (T) True
Since 3 and 6 are factors of 18, any number divisible by 18 is also divisible by its factors.
In simple words: If a number is a multiple of 18, it can be evenly divided by 3 and 6.

Exam Tip: Factors of a divisor always divide any number divisible by that divisor.

Question 5. If a number is divisible by 9 and 10, then it is divisible by 90.
Answer: (T) True
Since 9 and 10 are co-prime numbers (their HCF is 1), if a number is divisible by both, it must be divisible by their product (\(9 \times 10 = 90\)).
In simple words: If two numbers share no common factors other than 1, their product also divides the same number.

Exam Tip: Always check if the two divisors are co-prime before multiplying them to test divisibility.

Question 6. If a number exactly divides the sum of two numbers, it must exactly divide the numbers separately.
Answer: (F) False
For example, 3 divides the sum of 4 and 5 (since \(4 + 5 = 9\) and 9 is divisible by 3), but 3 does not divide 4 nor does it divide 5 separately.
In simple words: The sum being divisible does not mean the individual parts are divisible.

Exam Tip: Use simple numbers to test properties involving sums and differences.

Question 7. The sum of two consecutive odd numbers is always divisible by 4.
Answer: (T) True
Let two consecutive odd numbers be represented as \(2n + 1\) and \(2n + 3\).
Their sum is \((2n + 1) + (2n + 3) = 4n + 4 = 4(n + 1)\), which is clearly divisible by 4.
In simple words: Adding any two odd numbers next to each other always gives a multiple of 4.

Exam Tip: Represent odd numbers algebraically as \(2n + 1\) to prove divisibility properties easily.

Question 8. If a number is divisible by 8, it must be divisible by 4.
Answer: (T) True
Since 4 is a factor of 8, any multiple of 8 is also a multiple of 4.
In simple words: A larger divisor is a multiple of the smaller one, so divisibility carries over downwards.

Exam Tip: Every number divisible by 8 is also divisible by 1, 2, and 4.

Question 9. If a number is divisible by 3 and 7, it must be divisible by 21.
Answer: (T) True
Since 3 and 7 are co-prime numbers, their product is \(3 \times 7 = 21\). Thus, any number divisible by both must be divisible by 21.
In simple words: If a number goes in the 3 times table and 7 times table, it goes in the 21 times table.

Question 10. If a number is divisible by 8, it must be divisible by 6.
Answer: (F) False
For example, 8 is divisible by 8, but 8 is not divisible by 6.
In simple words: Not all multiples of 8 can be divided evenly by 6.

Exam Tip: Do not assume divisibility translates between numbers that are not factors or multiples.

Question 11. If a number divides three numbers exactly, it must divide their sum exactly.
Answer: (T) True
If a number divides each of the numbers individually, it is a common factor and therefore divides their sum.
In simple words: If a number divides separate parts, it also divides the total sum.

Exam Tip: This is a standard fundamental property of factors and division.

Question 12. The product of three nos. always equals the product of their H.C.F. and L.C.M.
Answer: (F) False
The property \(\text{Product of two numbers} = \text{HCF} \times \text{LCM}\) holds true only for two numbers, not for three or more numbers.
In simple words: The HCF and LCM product rule only works for pairs of numbers.

Exam Tip: Never apply the HCF-LCM product formula to three numbers.

Question 13. The H.C.F. of three or more counting nos. is always a factor of their L.C.M.
Answer: (T) True
The HCF of a set of numbers always divides every number in the set, and every number in the set always divides their LCM. Therefore, the HCF is a factor of the LCM.
In simple words: The highest common factor always divides evenly into the lowest common multiple.

Exam Tip: Remember that HCF is always a factor of LCM for any given set of numbers.

Question 14. The L.C.M. of two different nos. must be greater than their H.C.F.
Answer: (T) True
For two different numbers, the LCM is either greater than or equal to the larger number, while the HCF is less than or equal to the smaller number. Thus, LCM is greater than HCF.
In simple words: The lowest multiple is always larger than or equal to the highest factor.

Exam Tip: For distinct numbers, LCM is strictly greater than HCF.

Question 15. Each of the two counting nos. 'a' and 'b' is a multiple of their H.C.F.
Answer: (T) True
Since HCF is a common factor of both numbers \(a\) and \(b\), each number can be expressed as a multiple of their HCF.
In simple words: Every number is a multiple of its own factors, including its HCF.

Exam Tip: The HCF is a building block (factor) of the numbers, meaning the numbers are multiples of the HCF.