Samacheer Kalvi Class 6 Maths Solutions Term 2 Chapter 1 Numbers Exercise 1.1

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Ex 1.1

Question 1. Fill in the blanks.
(i) The number of prime numbers between 11 and 60 is ________.
(ii) The numbers 29 and ________ are twin primes.
(iii) 3753 is divisible by 9 and hence divisible by ________.
(iv) The number of distinct prime factors of the smallest 4 digit number is ________.
(v) The sum of distinct prime factors of 30 is ________.
Answer:
(i) The number of prime numbers between 11 and 60 is 12.
(ii) The numbers 29 and 31 are twin primes.
(iii) 3753 is divisible by 9 and hence divisible by 3.
(iv) The number of distinct prime factors of the smallest 4-digit number is 2.
(v) The sum of distinct prime factors of 30 is 10.
In simple words: This question asks us to fill in the missing parts about prime numbers, twin primes, divisibility rules, and prime factors. We need to count primes, find prime pairs, apply divisibility rules, and identify prime factors.

🎯 Exam Tip: Remember key definitions like prime numbers (only 2 factors), twin primes (prime pairs with a difference of 2), and divisibility rules to quickly solve such questions. Always test a few small examples to be sure.

Question 2. Say True or False.
(i) The sum of any number of odd numbers is always even.
(ii) Every natural number is either prime or composite.
(iii) If a number is divisible by 6, then it must be divisible by 3.
(iv) 16254 is divisible by 2, 3, 6, and 9.
(v) The number of distinct prime factors of 105 is 3.
Answer:
(i) False. For example, the sum of two odd numbers (1+3=4) is even, but the sum of three odd numbers (1+3+5=9) is odd. So, it is not always even. An odd number of odd numbers will always give an odd sum.
(ii) False. The number 1 is a natural number, but it is neither prime nor composite. Prime numbers have exactly two distinct factors, and composite numbers have more than two factors. The number 1 has only one factor (itself).
(iii) True. If a number is divisible by 6, it means it can be divided by both 2 and 3. So, it must be divisible by 3. This is because 6 is a product of 2 and 3.
(iv) True.
\( 16254 \) is even, so it's divisible by 2.
Sum of digits \( = 1+6+2+5+4 = 18 \). Since 18 is divisible by 3 and 9, \( 16254 \) is divisible by 3 and 9.
Since \( 16254 \) is divisible by both 2 and 3, it is also divisible by 6.
(v) True. The prime factors of 105 are \( 3 \times 5 \times 7 \). These are 3 distinct prime factors.
In simple words: We check if each statement is correct. Some statements are true based on number rules, while others are false because of exceptions like the number 1 or specific sums of odd numbers.

🎯 Exam Tip: For True/False questions, if a statement is false, try to give a counterexample (a case where it's not true). For statements claiming "always," even one counterexample makes it false. Remember the special status of number 1.

Question 3. Write the smallest and the biggest two-digit prime number.
Answer: The smallest two-digit prime number is 11.
The biggest two-digit prime number is 97.
In simple words: The smallest prime number with two digits is 11, and the largest prime number with two digits is 97.

🎯 Exam Tip: To find prime numbers, remember they can only be divided by 1 and themselves. Start checking numbers from 10 onwards for the smallest two-digit prime, and work backwards from 99 for the largest.

Question 4. Write the smallest and the biggest three-digit composite number.
Answer: The smallest three-digit composite number is 100.
The biggest three-digit composite number is 999.
In simple words: The smallest number with three digits that is composite (meaning it has more than two factors) is 100. The biggest number with three digits that is composite is 999.

🎯 Exam Tip: Remember that composite numbers have more than two factors. The smallest three-digit number is 100, which is \( 10 \times 10 \), making it composite. The largest three-digit number is 999, which is \( 9 \times 111 \), also composite.

Question 5. The sum of any three odd natural number is odd. Justify this statement with an example.
Answer: This statement is True.
Example: \( 1 + 3 + 5 = 9 \). Here, 9 is an odd number. If you add an odd number to another odd number, the result is always even. If you then add a third odd number to that even sum, the final result will be odd.
In simple words: When you add three odd numbers together, the answer will always be an odd number. For example, \( 1+3+5 \) equals 9, which is odd.

🎯 Exam Tip: Remember the rules for adding odd and even numbers: Odd + Odd = Even, Even + Even = Even, Odd + Even = Odd. Use these rules to quickly verify the sum of multiple numbers.

Question 6. The digits of the prime number 13 can be reversed to get another prime number 31. Find if any such pair exists up to 100.
Answer: Yes, there are other such pairs of prime numbers up to 100:
\( (17, 71) \)
\( (37, 73) \)
\( (79, 97) \) These pairs are called 'emirp' numbers when they are both prime and their reverse is also prime.
In simple words: Just like 13 and 31, there are other special pairs of numbers. When you flip the digits of the first number, you get the second number, and both numbers are prime.

🎯 Exam Tip: To solve this, list prime numbers up to 100 and then check if their reversed digits also form a prime number. Being familiar with primes up to 100 helps a lot.

Question 7. Your friend says that every odd number is prime. Give an example to prove him/her wrong.
Answer: This statement is False.
An example to prove your friend wrong is the number 15. The number 15 is an odd number, but it is not a prime number. Its factors are 1, 3, 5, and 15. Since it has more than two factors, it is a composite number. Another example is 9, which is odd but has factors 1, 3, 9.
In simple words: Your friend is wrong. For example, 15 is an odd number, but it is not prime because you can divide it by 3 and 5, not just 1 and 15.

🎯 Exam Tip: Remember that a prime number has exactly two factors: 1 and itself. Any number with more than two factors is composite. Odd numbers like 9, 15, 21, 25 are common counterexamples.

Question 8. Each of the composite numbers has at least three factors. Justify this statement with an example.
Answer: This statement is True.
By definition, a composite number is a positive integer that has at least one divisor other than 1 and itself. This means it must have 1, itself, and at least one other factor. Example: Consider the number 4. It is a composite number.
The factors of 4 are 1, 2, and 4. It has three factors. Consider 6. Its factors are 1, 2, 3, and 6. It has four factors. Thus, all composite numbers indeed have at least three factors.
In simple words: A composite number is one that can be divided by more than just 1 and itself. So, it will always have at least three things that can divide it evenly. For example, the number 4 can be divided by 1, 2, and 4.

🎯 Exam Tip: Always remember the definition: prime numbers have exactly two factors (1 and itself), while composite numbers have more than two factors. This implies a minimum of three factors for any composite number.

Question 9. Find the dates of any month in a calendar which are divisible by both 2 and 3.
Answer: If a date is divisible by both 2 and 3, it means it must be divisible by their least common multiple, which is 6. So, the dates in any month that are divisible by both 2 and 3 are the multiples of 6. These dates are: 6, 12, 18, 24, and 30. (Excluding February, which has a maximum of 29 days).
In simple words: To find dates that can be divided by both 2 and 3, we look for numbers that can be divided by 6. These dates are 6, 12, 18, 24, and 30 in a month.

🎯 Exam Tip: A number divisible by both A and B is also divisible by their Least Common Multiple (LCM). For 2 and 3, the LCM is 6. List multiples of 6 within the calendar date range (1-31).

Question 10. I am a two-digit prime number and the sum of my digits is 10. I am also one of the factors of 57. Who am I?
Answer: Let's break down the clues:
1. I am a two-digit prime number.
2. The sum of my digits is 10.
- Two-digit numbers whose digits sum to 10 are: 19 (1+9=10), 28, 37 (3+7=10), 46, 55, 64, 73 (7+3=10), 82, 91.
- Out of these, the prime numbers are 19, 37, and 73. 3. I am also one of the factors of 57.
- The factors of 57 are: 1, 3, 19, 57.
Comparing the list of prime numbers from clue 2 (19, 37, 73) with the factors of 57 (1, 3, 19, 57), the common number is 19.
Therefore, the number is 19.
In simple words: The number is a two-digit prime number whose digits add up to 10, and it is also one of the numbers that can divide 57 evenly. The number that fits all these clues is 19.

🎯 Exam Tip: When solving riddles with multiple conditions, list out the possibilities for each condition separately. Then, find the common elements that satisfy all the conditions. This systematic approach helps avoid mistakes.

Question 11. Find the prime factorisation of each number by factor tree method and division method.
(i) 60
(ii) 128
(iii) 144
(iv) 198
(v) 420
(vi) 999
Answer:
(i) 60
The factor tree for 60 is:Prime factorization of 60 is \( 2 \times 2 \times 3 \times 5 \).
(ii) 128
The factor tree for 128 is:Prime factorization of 128 is \( 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \) or \( 2^7 \).
(iii) 144
The factor tree for 144 is:Prime factorization of 144 is \( 2 \times 2 \times 2 \times 2 \times 3 \times 3 \) or \( 2^4 \times 3^2 \).
(iv) 198
The factor tree for 198 is:Prime factorization of 198 is \( 2 \times 3 \times 3 \times 11 \) or \( 2 \times 3^2 \times 11 \).
(v) 420
The factor tree for 420 is:Prime factorization of 420 is \( 2 \times 2 \times 3 \times 5 \times 7 \) or \( 2^2 \times 3 \times 5 \times 7 \).
(vi) 999
The factor tree for 999 is:Prime factorization of 999 is \( 3 \times 3 \times 3 \times 37 \) or \( 3^3 \times 37 \).
In simple words: For each number, we break it down into its prime factors using a tree diagram. This helps us see all the prime numbers that multiply together to make the original number. We list the prime factors for each part.

🎯 Exam Tip: Always break down numbers into their smallest prime factors first (2, 3, 5, 7, etc.). Double-check your multiplication to ensure the product of your prime factors equals the original number. Both factor tree and division methods should yield the same result.

Question 12. If there are 143 math books to be arranged in equal numbers in all the stacks, then find the number of books in each stack and also the number of stacks.
Answer: To arrange 143 math books into equal stacks, we need to find the factors of 143. We can use prime factorization to find these factors. The factor tree for 143 is:From the factor tree, we see that \( 143 = 11 \times 13 \). So, the possible ways to arrange the books are: 1. There can be 11 stacks with 13 books in each stack. 2. There can be 13 stacks with 11 books in each stack. These are the only ways to arrange the books into equal stacks other than 1 stack of 143 books or 143 stacks of 1 book each. Books are physical objects, so having 1 stack or 143 stacks are trivial arrangements, and typically a "stack" implies more than one book per stack and more than one stack.
In simple words: To arrange 143 math books equally, we find what numbers multiply to make 143. The numbers are 11 and 13. So, you can have 11 stacks with 13 books each, or 13 stacks with 11 books each.

🎯 Exam Tip: When a question asks for arrangements in "equal numbers," think about factors. Prime factorization helps you find all possible pairs of factors easily, which represent the number of stacks and books per stack.

Question 13. The difference between two successive odd numbers is
(a) 1
(b) 2
(c) 3
(d) 0
Answer: (b) 2
In simple words: If you take two odd numbers that come right after each other, like 3 and 5, and subtract them, the answer is always 2.

🎯 Exam Tip: Test with examples: \( 5 - 3 = 2 \), \( 11 - 9 = 2 \). This simple pattern holds true for all successive odd (and even) numbers.

Question 14. The only even prime number is
(a) 4
(b) 6
(c) 2
(d) 0
Answer: (c) 2
In simple words: The number 2 is special because it is the only even number that is also a prime number. All other even numbers can be divided by 2.

🎯 Exam Tip: A prime number has exactly two factors: 1 and itself. All other even numbers are divisible by 2 (and other numbers), so they have more than two factors and are thus composite.

Question 15. Which of the following numbers is not a prime?
(a) 53
(b) 92
(c) 97
(d) 71
Answer: (b) 92
In simple words: We are looking for the number that is NOT prime. 92 can be divided by 2 (and other numbers), so it is not a prime number.

🎯 Exam Tip: To identify if a number is prime, check for divisibility by small prime numbers (2, 3, 5, 7, 11, etc.). Any even number greater than 2 is immediately composite. 92 is an even number greater than 2, so it's composite.

Question 16. The sum of the factors of 27 is
(a) 28
(b) 37
(c) 40
(d) 31
Answer: (c) 40
In simple words: First, list all the numbers that can divide 27 evenly. Then, add those numbers together to get the total sum.

🎯 Exam Tip: To find all factors of a number, start from 1 and go up to its square root. For 27, factors are 1, 3, 9, 27. Their sum is \( 1 + 3 + 9 + 27 = 40 \).

Question 17. The factors of the number are 1, 2, 4, 5, 8, 10, 16, 20, 40, and 80. What is the number?
(a) 80
(b) 100
(c) 128
(d) 160
Answer: (a) 80
In simple words: The factors of a number are all the numbers that can divide it evenly. The biggest factor in the list is always the number itself. Here, the largest factor is 80, so the number is 80.

🎯 Exam Tip: The largest factor of any number is always the number itself. Look for the largest number in the list of factors, and that will be your answer.

Question 18. The prime factorisation of 60 is \( 2 \times 2 \times 3 \times 5 \). Any other number which has the same prime factorisation as 60 is
(a) 30
(b) 120
(c) 90
(d) Impossible
Answer: (d) Impossible
In simple words: Every single number has its own special and unique prime factors. If two numbers had the exact same prime factors, they would actually be the same number. So, it's impossible for another number to have the same prime factors as 60.

🎯 Exam Tip: This question relates to the Fundamental Theorem of Arithmetic, which states that every integer greater than 1 is either a prime number itself or can be represented as a product of prime numbers, and this representation is unique (up to the order of the factors). Therefore, no two different numbers can have the exact same prime factorization.

Question 19. If the number \( 6354*97 \) is divisible by 9 then the value of * is
(a) 2
(b) 4
(c) 6
(d) 7
Answer: (a) 2
In simple words: To check if a number can be divided by 9, you add up all its digits. If this sum can be divided by 9, then the original number can also be divided by 9. We need to find the missing digit that makes the sum divisible by 9.

🎯 Exam Tip: The divisibility rule for 9 is: A number is divisible by 9 if the sum of its digits is divisible by 9. Sum the given digits: \( 6+3+5+4+*+9+7 = 34+* \). The next multiple of 9 after 34 is 36. So, \( 34+* = 36 \), which means \( * = 2 \).

Question 20. The number 87846 is divisible by
(a) 2 only
(b) 3 only
(c) 11 only
(d) all of the options
Answer: (d) all of the options
In simple words: We need to check if 87846 can be divided by 2, 3, and 11. Since it can be divided by all of them, the answer is "all of the options".

🎯 Exam Tip: Apply the divisibility rules:
- For 2: The last digit is even (6), so it's divisible by 2.
- For 3: Sum of digits \( (8+7+8+4+6 = 33) \). Since 33 is divisible by 3, the number is divisible by 3.
- For 11: Alternating sum of digits \( (6-4+8-7+8 = 11) \). Since 11 is divisible by 11, the number is divisible by 11.
Since it's divisible by 2, 3, and 11, the answer is all of the options.