Samacheer Kalvi Class 10 Maths Solutions Chapter 2 Numbers and Sequences Exercise 2.1

Question 1. Find all positive integers which when divided by 3 leaves remainder 2.
Answer: The question asks for all positive whole numbers that leave a remainder of 2 when divided by 3. We can use Euclid's division lemma for this. This lemma states that any integer \( a \) can be written as \( a = bq + r \), where \( b \) is the divisor, \( q \) is the quotient, and \( r \) is the remainder, with \( 0 \le r < b \).
Here, the divisor \( b \) is 3, and the remainder \( r \) is 2. So, the positive integer \( a \) can be expressed in the form \( a = 3q + 2 \).
Since \( a \) must be a positive integer, the possible values for \( q \) (the quotient) start from 0:
If \( q = 0 \), then \( a = 3(0) + 2 = 2 \).
If \( q = 1 \), then \( a = 3(1) + 2 = 5 \).
If \( q = 2 \), then \( a = 3(2) + 2 = 8 \).
If \( q = 3 \), then \( a = 3(3) + 2 = 11 \).
And so on.
The positive integers that fit this condition are 2, 5, 8, 11, ... and this pattern continues indefinitely. Each of these numbers gives a remainder of 2 when divided by 3.
In simple words: We are looking for numbers that, when divided by 3, always leave 2 as the leftover. Using a math rule, these numbers can be found by adding 2 to any multiple of 3. So, the numbers are 2, 5, 8, 11, and so on.

🎯 Exam Tip: Remember to list out a few initial terms for patterns like this to demonstrate understanding of the sequence and to ensure your general form is correct.

Question 2. A man has 532 flower pots. He wants to arrange them in rows such that each row contains 21 flower pots. Find the number of completed rows and how many flower pots are left over.
Answer: We need to arrange 532 flower pots into rows, with 21 pots in each row. To find out how many full rows can be made and how many pots will remain, we can use Euclid's division algorithm.
The algorithm is written as \( a = bq + r \), where \( a \) is the total number of items, \( b \) is the number of items per group, \( q \) is the number of groups, and \( r \) is the remainder. The remainder \( r \) must be between 0 and \( b-1 \).
In this problem, \( a = 532 \) (total flower pots) and \( b = 21 \) (flower pots per row).
Divide 532 by 21:
\( 532 = 21 \times 25 + 7 \)
From this calculation, the quotient \( q \) is 25, and the remainder \( r \) is 7.
Therefore, the man can complete 25 rows of flower pots. There will be 7 flower pots left over. This method is useful for quickly solving problems that involve division and finding a remainder.
In simple words: A man has 532 pots and wants to put 21 pots in each row. We divide 532 by 21. He can make 25 full rows, and 7 pots will be left without a row.

🎯 Exam Tip: Clearly identify the dividend, divisor, quotient, and remainder in your explanation when using the division algorithm. The remainder is always less than the divisor.

Question 3. Prove that the product of two consecutive positive integers is divisible by 2.
Answer: We need to show that if you multiply any two positive whole numbers that follow each other (consecutive), the result can always be divided by 2. Let these two consecutive positive integers be \( n \) and \( n-1 \). Their product is \( n(n-1) \).
Every positive integer \( n \) can be categorized in one of two ways: it is either an even number or an odd number. We will look at both cases.

Case 1: If \( n \) is an even integer.
An even integer can be written in the form \( n = 2q \), where \( q \) is any integer.
Now, let's find the product \( n(n-1) \):
\( n(n-1) = (2q)(2q-1) \)
Since this expression already has 2 as a factor, we can write it as \( 2 \times [q(2q-1)] \).
This shows that the product is a multiple of 2, so it is divisible by 2.

Case 2: If \( n \) is an odd integer.
An odd integer can be written in the form \( n = 2q+1 \), where \( q \) is any integer.
If \( n \) is \( 2q+1 \), then the integer just before it, \( n-1 \), will be \( (2q+1)-1 = 2q \). So, \( n-1 \) is an even integer.
Now, let's find the product \( n(n-1) \):
\( n(n-1) = (2q+1)(2q) \)
Again, this expression has 2 as a factor, so we can write it as \( 2 \times [q(2q+1)] \).
This shows that the product is a multiple of 2, so it is divisible by 2.
In both possible cases, the product of two consecutive positive integers is always divisible by 2. This is because one of any two consecutive integers will always be an even number.
In simple words: Take any two numbers that come one after the other, like 3 and 4. Multiply them (3 x 4 = 12). The answer (12) can always be divided by 2. This works because one of the two numbers will always be an even number.

🎯 Exam Tip: When proving properties related to integers, consider classifying numbers as either even (2q) or odd (2q+1) as a common starting point for your proof.

Question 4. When the positive integers a, b and c are divided by 13, the respective remainders are 9, 7 and 10. Show that a + b + c is divisible by 13.
Answer: We are given three positive integers, \( a \), \( b \), and \( c \). We know their remainders when divided by 13.
Using the division algorithm (\( \text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder} \)):
For \( a \): When \( a \) is divided by 13, the remainder is 9. So, we can write \( a = 13q_1 + 9 \).
For \( b \): When \( b \) is divided by 13, the remainder is 7. So, we can write \( b = 13q_2 + 7 \).
For \( c \): When \( c \) is divided by 13, the remainder is 10. So, we can write \( c = 13q_3 + 10 \).
Now, let's find the sum of these integers, \( a + b + c \):
\( a + b + c = (13q_1 + 9) + (13q_2 + 7) + (13q_3 + 10) \)
Group the terms with \( 13q \) and the constant terms together:
\( a + b + c = (13q_1 + 13q_2 + 13q_3) + (9 + 7 + 10) \)
\( a + b + c = 13(q_1 + q_2 + q_3) + 26 \)
We can also take 13 as a common factor from 26 (since \( 26 = 13 \times 2 \)):
\( a + b + c = 13(q_1 + q_2 + q_3) + 13(2) \)
Now, factor out 13 from the entire expression:
\( a + b + c = 13(q_1 + q_2 + q_3 + 2) \)
Since the sum \( a+b+c \) can be written as 13 multiplied by another integer \( (q_1 + q_2 + q_3 + 2) \), it means that the sum \( a+b+c \) is divisible by 13. This shows a property of remainders under addition related to a common divisor.
In simple words: If you have three numbers and divide each by 13, you get remainders of 9, 7, and 10. If you add these three numbers together, the total sum will also be perfectly divisible by 13. This happens because the sum of the remainders (9+7+10=26) is also divisible by 13.

🎯 Exam Tip: When working with sums of numbers and their remainders, remember that if the sum of remainders is divisible by the divisor, then the sum of the original numbers will also be divisible by that divisor.

Question 5. Prove that square of any integer leaves the remainder either 0 or 1 when divided by 4.
Answer: We need to prove that when you square any whole number, and then divide that square by 4, the remainder will always be either 0 or 1. Let the integer be \( x \). Its square is \( x^2 \).
Any integer \( x \) can be classified as either an even number or an odd number. We will examine both cases.

Case 1: If \( x \) is an even integer.
An even integer can always be written in the form \( x = 2q \), where \( q \) is some integer.
Let's find the square of \( x \):
\( x^2 = (2q)^2 \)
\( x^2 = 4q^2 \)
If we divide \( 4q^2 \) by 4, the remainder is clearly 0, because \( 4q^2 \) is exactly 4 times \( q^2 \).
So, if \( x \) is an even integer, \( x^2 \) leaves a remainder of 0 when divided by 4.

Case 2: If \( x \) is an odd integer.
An odd integer can always be written in the form \( x = 2k + 1 \), where \( k \) is some integer.
Let's find the square of \( x \):
\( x^2 = (2k + 1)^2 \)
Expanding this expression using the formula \( (a+b)^2 = a^2 + 2ab + b^2 \):
\( x^2 = (2k)^2 + 2(2k)(1) + 1^2 \)
\( x^2 = 4k^2 + 4k + 1 \)
Now, we can factor out 4 from the first two terms:
\( x^2 = 4(k^2 + k) + 1 \)
Let \( q = k^2 + k \). Since \( k \) is an integer, \( q \) will also be an integer. The expression becomes:
\( x^2 = 4q + 1 \)
If we divide \( 4q + 1 \) by 4, the remainder is 1. This is because \( 4q \) is divisible by 4, leaving only the +1.
So, if \( x \) is an odd integer, \( x^2 \) leaves a remainder of 1 when divided by 4.
In summary, for any integer \( x \), its square \( x^2 \) will always leave a remainder of either 0 or 1 when divided by 4. This property is foundational in number theory.
In simple words: If you take any whole number, multiply it by itself, and then divide the answer by 4, the leftover number will always be either 0 or 1. For example, \( 2 \times 2 = 4 \) (remainder 0 when divided by 4), and \( 3 \times 3 = 9 \) (remainder 1 when divided by 4).

🎯 Exam Tip: When dealing with proofs involving properties of integers, always cover all possible cases (e.g., even and odd numbers) to ensure the proof is complete and correct.

Question 6. Use Euclid's Division Algorithm to find the Highest Common Factor (H.C.F) of
(i) 340 and 412
(ii) 867 and 255
(iii) 10224 and 9648
(iv) 84, 90 and 120
Answer: We will use Euclid's Division Algorithm to find the Highest Common Factor (HCF) for each pair or set of numbers. This algorithm involves repeatedly dividing the larger number by the smaller number until a remainder of zero is obtained. The divisor at the step where the remainder becomes zero is the HCF.

(i) For 340 and 412:
Start with the larger number (412) and the smaller number (340):
\( 412 = 340 \times 1 + 72 \)
The remainder is \( 72 \), which is not zero.
Next, use the previous divisor (340) and the remainder (72):
\( 340 = 72 \times 4 + 52 \)
The remainder is \( 52 \), which is not zero.
Next, use 72 and 52:
\( 72 = 52 \times 1 + 20 \)
The remainder is \( 20 \), not zero.
Next, use 52 and 20:
\( 52 = 20 \times 2 + 12 \)
The remainder is \( 12 \), not zero.
Next, use 20 and 12:
\( 20 = 12 \times 1 + 8 \)
The remainder is \( 8 \), not zero.
Next, use 12 and 8:
\( 12 = 8 \times 1 + 4 \)
The remainder is \( 4 \), not zero.
Next, use 8 and 4:
\( 8 = 4 \times 2 + 0 \)
The remainder is 0. The divisor at this step is 4.
Therefore, the HCF of 340 and 412 is 4.

(ii) For 867 and 255:
Start with 867 and 255:
\( 867 = 255 \times 3 + 102 \)
The remainder is \( 102 \), not zero.
Next, use 255 and 102:
\( 255 = 102 \times 2 + 51 \)
The remainder is \( 51 \), not zero.
Next, use 102 and 51:
\( 102 = 51 \times 2 + 0 \)
The remainder is 0. The divisor at this step is 51.
Therefore, the HCF of 867 and 255 is 51.

(iii) For 10224 and 9648:
Start with 10224 and 9648:
\( 10224 = 9648 \times 1 + 576 \)
The remainder is \( 576 \), not zero.
Next, use 9648 and 576:
\( 9648 = 576 \times 16 + 432 \)
The remainder is \( 432 \), not zero.
Next, use 576 and 432:
\( 576 = 432 \times 1 + 144 \)
The remainder is \( 144 \), not zero.
Next, use 432 and 144:
\( 432 = 144 \times 3 + 0 \)
The remainder is 0. The divisor at this step is 144.
Therefore, the HCF of 10224 and 9648 is 144.

(iv) For 84, 90, and 120:
First, find the HCF of the first two numbers, 84 and 90:
\( 90 = 84 \times 1 + 6 \)
The remainder is \( 6 \), not zero.
Next, use 84 and 6:
\( 84 = 6 \times 14 + 0 \)
The remainder is 0. So, the HCF of 84 and 90 is 6.
Now, find the HCF of this result (6) and the third number (120):
\( 120 = 6 \times 20 + 0 \)
The remainder is 0. So, the HCF of 6 and 120 is 6.
Therefore, the HCF of 84, 90, and 120 is 6. This method extends naturally for finding the HCF of more than two numbers.
In simple words: To find the HCF using Euclid's method, you keep dividing the larger number by the smaller one, then replacing the larger number with the smaller one and the smaller number with the remainder. You repeat until the remainder is zero. The last number you divided by is the HCF. For three numbers, find the HCF of two, then find the HCF of that result and the third number.

🎯 Exam Tip: Clearly show each step of the Euclidean algorithm, including the remainder at each stage, and explicitly state when the remainder is zero to identify the HCF correctly.

Question 7. Find the largest number which divides 1230 and 1926 leaving remainder 12 in each case.
Answer: We are looking for the largest number that divides 1230 and 1926, and in both cases, leaves a remainder of 12.
If a number divides 1230 and leaves a remainder of 12, it means that \( (1230 - 12) \) must be perfectly divisible by that number. So, the number must exactly divide \( 1218 \).
Similarly, if the same number divides 1926 and leaves a remainder of 12, then \( (1926 - 12) \) must be perfectly divisible by that number. So, the number must exactly divide \( 1914 \).
Therefore, we need to find the Highest Common Factor (HCF) of 1218 and 1914. We will use Euclid's division algorithm for this.

Applying Euclid's division algorithm to 1914 and 1218:
\( 1914 = 1218 \times 1 + 696 \)
The remainder is \( 696 \), which is not zero.
Next, use the previous divisor (1218) and the remainder (696):
\( 1218 = 696 \times 1 + 522 \)
The remainder is \( 522 \), not zero.
Next, use 696 and 522:
\( 696 = 522 \times 1 + 174 \)
The remainder is \( 174 \), not zero.
Next, use 522 and 174:
\( 522 = 174 \times 3 + 0 \)
The remainder is 0. The divisor at this step is 174.
Thus, the HCF of 1218 and 1914 is 174.
The largest number that divides 1230 and 1926, leaving a remainder of 12 in each case, is 174. This method is crucial when dealing with problems involving common remainders across multiple numbers.
In simple words: To find a number that leaves a remainder of 12 when dividing two other numbers, first subtract 12 from both original numbers. Then, find the largest number that can divide these new numbers perfectly. This largest number is the answer.

🎯 Exam Tip: When a problem states a common remainder, always subtract that remainder from each given number before finding their HCF. This converts the problem into a standard HCF calculation.

Question 8. If d is the Highest Common Factor of 32 and 60, find x and y satisfying d = 32x + 60y.
Answer: First, we need to find \( d \), which is the Highest Common Factor (HCF) of 32 and 60, using Euclid's division algorithm. Then, we will express \( d \) in the form \( 32x + 60y \) to find the integer values of \( x \) and \( y \).

Step 1: Find the HCF of 32 and 60.
Apply Euclid's division lemma:
\( 60 = 32 \times 1 + 28 \) (Equation i)
The remainder \( 28 \neq 0 \).
Now, apply the lemma to 32 and 28:
\( 32 = 28 \times 1 + 4 \) (Equation ii)
The remainder \( 4 \neq 0 \).
Now, apply the lemma to 28 and 4:
\( 28 = 4 \times 7 + 0 \) (Equation iii)
The remainder is 0. The divisor at this step is 4. So, the HCF \( d \) is 4.

Step 2: Express \( d \) (which is 4) in the form \( 32x + 60y \).
We will work backwards from the steps of the Euclidean algorithm.
From Equation (ii), we can isolate 4:
\( 4 = 32 - (28 \times 1) \)
From Equation (i), we can isolate 28:
\( 28 = 60 - (32 \times 1) \)
Now, substitute the expression for 28 back into the equation for 4:
\( 4 = 32 - (60 - 32 \times 1) \times 1 \)
Distribute the \(-1\) and simplify:
\( 4 = 32 - 60 + 32 \times 1 \)
Combine the terms with 32:
\( 4 = 32 \times 1 + 32 \times 1 - 60 \times 1 \)
\( 4 = 32 \times (1+1) - 60 \times 1 \)
\( 4 = 32 \times 2 - 60 \times 1 \)
This can be written in the form \( d = 32x + 60y \) as:
\( 4 = 32 \times 2 + 60 \times (-1) \)
By comparing this equation with \( d = 32x + 60y \), we can see that \( x = 2 \) and \( y = -1 \). This process is known as Bézout's identity and is a key application of the Euclidean algorithm beyond just finding the HCF.
In simple words: First, find the biggest number that divides both 32 and 60 without a leftover; this is 4. Then, we need to write 4 by multiplying 32 by some number (x) and 60 by some other number (y) and adding them. By working backwards through our division steps, we find that \( x = 2 \) and \( y = -1 \) make the equation true.

🎯 Exam Tip: When expressing HCF as a linear combination (Bézout's identity), carefully substitute expressions back into previous steps to avoid errors in calculation or signs. Always double-check the values of x and y at the end.

Question 9. A positive integer, when divided by 88, gives the remainder 61. What will be the remainder when the same number is divided by 11?
Answer: Let the positive integer be \( x \).
According to the problem, when \( x \) is divided by 88, the remainder is 61. Using the division algorithm, we can write this as:
\( x = 88 \times y + 61 \), where \( y \) is the quotient.
We need to find the remainder when this same number \( x \) is divided by 11.
We know that 88 is a multiple of 11, because \( 88 = 11 \times 8 \).
This means that the term \( 88 \times y \) is also a multiple of 11. Any multiple of 11 will leave a remainder of 0 when divided by 11.
So, to find the remainder of \( x \) when divided by 11, we only need to consider the remainder of 61 when divided by 11.
Let's divide 61 by 11:
\( 61 = 11 \times 5 + 6 \)
The remainder when 61 is divided by 11 is 6.
Since \( x = (\text{a multiple of } 11) + 61 \), and 61 leaves a remainder of 6 when divided by 11, then \( x \) will also leave a remainder of 6 when divided by 11.
The remainder when the number is divided by 11 is 6. This principle is very useful when dealing with nested divisibility.
In simple words: We have a number that leaves 61 as a leftover when divided by 88. Since 88 can be divided by 11 exactly, we only need to check what 61 leaves as a leftover when divided by 11. When 61 is divided by 11, the leftover is 6. So the original number also leaves 6 as a leftover when divided by 11.

🎯 Exam Tip: If the original divisor is a multiple of the new divisor, you only need to find the remainder of the original remainder when divided by the new divisor.

Question 10. Prove that two consecutive positive integers are always coprime.
Answer: We need to prove that any two positive integers that come immediately after each other (consecutive) are always coprime. Two numbers are coprime if their Highest Common Factor (HCF) is 1, meaning they share no common factors other than 1.
Let the two consecutive positive integers be \( n \) and \( n+1 \). We will use Euclid's division algorithm to find their HCF.

Divide \( n+1 \) by \( n \):
\( n+1 = n \times 1 + 1 \)
The remainder is 1, which is not zero.
Now, according to the algorithm, we take the previous divisor (\( n \)) and the remainder (1) and repeat the division:
\( n = 1 \times n + 0 \)
The remainder is 0. The divisor at this step, which is the last non-zero remainder from the previous step, is 1.
Therefore, the Highest Common Factor (HCF) of \( n \) and \( n+1 \) is 1.
Since their HCF is 1, the two consecutive positive integers are always coprime. This is a fundamental property of integers, showing that numbers right next to each other have no common prime factors.
In simple words: Take any two numbers that are next to each other, like 5 and 6. The only number that can divide both of them without a leftover is 1. This means they are "coprime." This always happens because one number is always odd and the other is always even, meaning they won't share any other factors besides 1.

🎯 Exam Tip: For proving numbers are coprime, using Euclid's algorithm to show their HCF is 1 is a direct and widely accepted method. Understanding the definition of coprime is crucial.