RBSE Solutions Class 10 Maths Chapter 2 Real Numbers Exercise 2.1

Question 1. Show that square of any positive odd integer is of the form 8q + 1, where q is any positive integer.
Answer: Let \(a\) be any positive odd integer. We know that any odd positive integer can be written in the form of \(4m + 1\) or \(4m + 3\), where \(m\) is an integer. Let's examine both cases:
(i) If \(a = 4m + 1\):
\( a^2 = (4m + 1)^2 \)
\( = (4m)^2 + 2(4m)(1) + 1^2 \) (Using \( (A+B)^2 = A^2+2AB+B^2 \))
\( = 16m^2 + 8m + 1 \)
\( = 8m(2m + 1) + 1 \)
Let \( q = m(2m + 1) \). This \(q\) will be a positive integer because \(m\) is a positive integer.
\( \implies a^2 = 8q + 1 \)
(ii) If \(a = 4m + 3\):
\( a^2 = (4m + 3)^2 \)
\( = (4m)^2 + 2(4m)(3) + 3^2 \)
\( = 16m^2 + 24m + 9 \)
\( = 16m^2 + 24m + 8 + 1 \)
\( = 8(2m^2 + 3m + 1) + 1 \)
Let \( q = 2m^2 + 3m + 1 \). This \(q\) will also be a positive integer.
\( \implies a^2 = 8q + 1 \)
In both possible cases, the square of a positive odd integer is of the form \(8q + 1\). This proves the statement.
In simple words: When you square any positive odd number, the result will always be one more than a multiple of 8. For example, \(3^2 = 9 = 8 \times 1 + 1\), and \(5^2 = 25 = 8 \times 3 + 1\).

🎯 Exam Tip: Remember to consider all possible forms of the integer (like \(4m+1\) and \(4m+3\)) and show the result for each case. The "let \(q = \text{expression}\)" step is crucial for clarity.

Question 2. Use Euclid's division lemma to show that the cube of any positive integer is of the form 9q or 9q + 8, where q is some integer.
Answer: Let \(a\) be any positive integer. According to Euclid's division lemma, if we divide \(a\) by 3, the remainder can be 0, 1, or 2. So, \(a\) can be written in one of three forms: \(3b\), \(3b + 1\), or \(3b + 2\), where \(b\) is an integer.
We will look at each case for \(a^3\):
**Case I: When \(a = 3b\)**
\( a^3 = (3b)^3 \)
\( = 27b^3 \)
\( = 9(3b^3) \)
Let \( q = 3b^3 \). This \(q\) is an integer.
\( \implies a^3 = 9q \)
**Case II: When \(a = 3b + 1\)**
\( a^3 = (3b + 1)^3 \)
\( = (3b)^3 + 3(3b)^2(1) + 3(3b)(1)^2 + 1^3 \) (Using \( (A+B)^3 = A^3+3A^2B+3AB^2+B^3 \))
\( = 27b^3 + 27b^2 + 9b + 1 \)
\( = 9(3b^3 + 3b^2 + b) + 1 \)
Let \( q = 3b^3 + 3b^2 + b \). This \(q\) is an integer.
\( \implies a^3 = 9q + 1 \)
**Case III: When \(a = 3b + 2\)**
\( a^3 = (3b + 2)^3 \)
\( = (3b)^3 + 3(3b)^2(2) + 3(3b)(2)^2 + 2^3 \)
\( = 27b^3 + 54b^2 + 36b + 8 \)
\( = 9(3b^3 + 6b^2 + 4b) + 8 \)
Let \( q = 3b^3 + 6b^2 + 4b \). This \(q\) is an integer.
\( \implies a^3 = 9q + 8 \)
From all three cases, the cube of any positive integer can be expressed in the form \(9q\), \(9q + 1\), or \(9q + 8\).
In simple words: If you take any whole number and cube it (multiply it by itself three times), the answer will always fit one of these patterns: it will be a perfect multiple of 9, or it will be one more than a multiple of 9, or it will be eight more than a multiple of 9.

🎯 Exam Tip: When using Euclid's division lemma, choose the divisor (here, 3) carefully based on the target form (here, 9q). \(3^3 = 27\), which is a multiple of 9, making it a suitable choice for proving forms related to 9.

Question 3. Show that any positive odd integer is of the form 6q + 1, or 6q + 3 or 6q + 5, where q is some integer.
Answer: Let \(a\) be a positive odd integer. We can apply Euclid's division algorithm to divide \(a\) by 6. The form is \(a = 6q + r\), where \(q\) is the quotient and \(r\) is the remainder. The possible remainders (\(r\)) when dividing by 6 are 0, 1, 2, 3, 4, or 5.
So, \(a\) can be of the form:
1. \(a = 6q + 0 = 6q\)
2. \(a = 6q + 1\)
3. \(a = 6q + 2\)
4. \(a = 6q + 3\)
5. \(a = 6q + 4\)
6. \(a = 6q + 5\)
Since \(a\) is an odd integer, it cannot be divisible by 2. Therefore, \(a\) cannot be of the forms \(6q\), \(6q + 2\), or \(6q + 4\), because these expressions are all even numbers (as \(6q\), \(6q+2\), and \(6q+4\) can all be written as \(2 \times (\text{something})\)). For example, \(6q = 2(3q)\), \(6q+2 = 2(3q+1)\), and \(6q+4 = 2(3q+2)\).
Thus, any positive odd integer must be of the form \(6q + 1\), \(6q + 3\), or \(6q + 5\). This completes the proof.
In simple words: When you divide any positive odd number by 6, the remainder will always be 1, 3, or 5. It will never be 0, 2, or 4 because if it were, the number would be even.

🎯 Exam Tip: To prove a number is odd, ensure it cannot be expressed as \(2k\) for some integer \(k\). When showing forms like \(6q+r\), clearly eliminate the even remainder cases based on the definition of an odd integer.

Question 4. Find the HCF of the following pairs of numbers using Euclid's division lemma.
(i) 210, 55
(ii) 420, 130
(iii) 75, 243
(iv) 135, 225
(v) 196, 38220
(vi) 867, 255.
Answer:
(i) **For 210 and 55:**
We apply Euclid's division lemma to 210 and 55, since \(210 > 55\).
\( 210 = 55 \times 3 + 45 \)
Since the remainder 45 is not 0, we now apply the lemma to the divisor 55 and remainder 45.
\( 55 = 45 \times 1 + 10 \)
Since the remainder 10 is not 0, we apply the lemma to the divisor 45 and remainder 10.
\( 45 = 10 \times 4 + 5 \)
Since the remainder 5 is not 0, we apply the lemma to the divisor 10 and remainder 5.
\( 10 = 5 \times 2 + 0 \)
The remainder is now 0. Therefore, the HCF of 210 and 55 is the last non-zero divisor, which is 5.

(ii) **For 420 and 130:**
We apply Euclid's division lemma to 420 and 130, since \(420 > 130\).
\( 420 = 130 \times 3 + 30 \)
Since the remainder 30 is not 0, we now apply the lemma to the divisor 130 and remainder 30.
\( 130 = 30 \times 4 + 10 \)
Since the remainder 10 is not 0, we apply the lemma to the divisor 30 and remainder 10.
\( 30 = 10 \times 3 + 0 \)
The remainder is now 0. Therefore, the HCF of 420 and 130 is the last non-zero divisor, which is 10.

(iii) **For 75 and 243:**
We apply Euclid's division lemma to 243 and 75, since \(243 > 75\).
\( 243 = 75 \times 3 + 18 \)
Since the remainder 18 is not 0, we apply the lemma to the divisor 75 and remainder 18.
\( 75 = 18 \times 4 + 3 \)
Since the remainder 3 is not 0, we apply the lemma to the divisor 18 and remainder 3.
\( 18 = 3 \times 6 + 0 \)
The remainder is now 0. Therefore, the HCF of 75 and 243 is the last non-zero divisor, which is 3.

(iv) **For 135 and 225:**
We apply Euclid's division lemma to 225 and 135, since \(225 > 135\).
\( 225 = 135 \times 1 + 90 \)
Since the remainder 90 is not 0, we apply the lemma to the divisor 135 and remainder 90.
\( 135 = 90 \times 1 + 45 \)
Since the remainder 45 is not 0, we apply the lemma to the divisor 90 and remainder 45.
\( 90 = 45 \times 2 + 0 \)
The remainder is now 0. Therefore, the HCF of 135 and 225 is the last non-zero divisor, which is 45.

(v) **For 196 and 38220:**
We apply Euclid's division lemma to 38220 and 196, since \(38220 > 196\).
\( 38220 = 196 \times 195 + 0 \)
The remainder is immediately 0. Therefore, the HCF of 196 and 38220 is the divisor, which is 196.

(vi) **For 867 and 255:**
We apply Euclid's division lemma to 867 and 255, since \(867 > 255\).
\( 867 = 255 \times 3 + 102 \)
Since the remainder 102 is not 0, we apply the lemma to the divisor 255 and remainder 102.
\( 255 = 102 \times 2 + 51 \)
Since the remainder 51 is not 0, we apply the lemma to the divisor 102 and remainder 51.
\( 102 = 51 \times 2 + 0 \)
The remainder is now 0. Therefore, the HCF of 867 and 255 is the last non-zero divisor, which is 51.
In simple words: To find the HCF of two numbers using Euclid's method, you keep dividing the larger number by the smaller one. Then you replace the larger number with the smaller number and the smaller number with the remainder. You repeat this until the remainder is zero. The last number you divided by (the last non-zero divisor) is your HCF. This process helps find the greatest common factor efficiently.

🎯 Exam Tip: Always state which number is the divisor and which is the remainder at each step of the Euclid's division lemma. Clearly mark the step where the remainder becomes zero, as the divisor at that point is your HCF.

Question 5. If HCF of numbers 408 and 1032 is expressed in the form of \(1032x - 408 \times 5\), then find the value of x.
Answer: First, we need to find the HCF of 408 and 1032 using Euclid's division lemma.
We apply the lemma to 1032 and 408, since \(1032 > 408\).
\( 1032 = 408 \times 2 + 216 \)
Since the remainder 216 is not 0, we apply the lemma to the divisor 408 and remainder 216.
\( 408 = 216 \times 1 + 192 \)
Since the remainder 192 is not 0, we apply the lemma to the divisor 216 and remainder 192.
\( 216 = 192 \times 1 + 24 \)
Since the remainder 24 is not 0, we apply the lemma to the divisor 192 and remainder 24.
\( 192 = 24 \times 8 + 0 \)
The remainder is now 0. So, the HCF of 408 and 1032 is 24.
Now, we are given that the HCF is expressed as \(1032x - 408 \times 5\).
\( \text{HCF} = 1032x - 408 \times 5 \)
We found HCF = 24. So, we set up the equation:
\( 24 = 1032x - 408 \times 5 \)
\( 24 = 1032x - 2040 \)
To find \(x\), we add 2040 to both sides:
\( 24 + 2040 = 1032x \)
\( 2064 = 1032x \)
Now, divide both sides by 1032:
\( x = \frac{2064}{1032} \)
\( x = 2 \)
Therefore, the value of \(x\) is 2.
In simple words: First, we find the biggest number that divides both 408 and 1032 without leaving a remainder; this is called the HCF, which turns out to be 24. Then, we are told that this HCF can be written in a special way using \(x\). We put our HCF (24) into that equation and solve it like a puzzle to find that \(x\) is 2.

🎯 Exam Tip: For problems involving HCF in a linear combination form like \(ax + by\), first find the HCF using Euclid's algorithm. Then, substitute the HCF value into the given expression and solve the resulting linear equation for the unknown variable.