GSEB Class 10 Maths Solutions Chapter 1 Real Numbers Exercise 1.1

Question 1. Use Euclid's division algorithm to find the HCF of
1. 135 and 225
2. 196 and 38220
3. 867 and 255
Answer:
1. To find the HCF of 225 and 135, we apply Euclid's division algorithm since \( 225 > 135 \).
\( 225 = 135 \times 1 + 90 \)
Since the remainder 90 is not zero, we apply the division lemma again to the divisor 135 and the remainder 90.
\( 135 = 90 \times 1 + 45 \)
The remainder 45 is still not zero, so we apply the division lemma once more to the divisor 90 and the remainder 45.
\( 90 = 45 \times 2 + 0 \)
Since the remainder has now become zero, the process stops. The divisor at this final stage is 45.
Therefore, the HCF of 135 and 225 is 45.

2. To find the HCF of 38220 and 196, we apply Euclid's division algorithm since \( 38220 > 196 \).
\( 38220 = 196 \times 195 + 0 \)
Since the remainder is zero at this step, the process ends. The divisor at this stage is 196.
Therefore, the HCF of 196 and 38220 is 196.

3. To find the HCF of 867 and 255, we apply Euclid's division algorithm since \( 867 > 255 \).
\( 867 = 255 \times 3 + 102 \)
Since the remainder 102 is not zero, we apply the division lemma again to the divisor 255 and the remainder 102.
\( 255 = 102 \times 2 + 51 \)
The remainder 51 is still not zero, so we apply the division lemma once more to the divisor 102 and the remainder 51.
\( 102 = 51 \times 2 + 0 \)
Since the remainder has now become zero, the process stops. The divisor at this final stage is 51.
Therefore, the HCF of 867 and 255 is 51.
In simple words: We keep dividing the bigger number by the smaller number and replace the numbers with the divisor and the remainder until the remainder is zero. The last divisor is the Highest Common Factor (HCF).

Exam Tip: Always clearly show each step of the Euclidean algorithm, indicating the dividend, divisor, quotient, and remainder for full marks. The last non-zero remainder's divisor is the HCF.

Question 2. Show that any positive odd integer is of the form \( 6q + 1 \), or \( 6q + 3 \) or \( 6q + 5 \) where \( q \) is some integer.(CBSE)
Answer: Let \( a \) be any positive integer, and let \( b = 6 \).
By using Euclid's division algorithm, we can write \( a = 6q + r \), where \( 0 \leq r < 6 \).
This means the possible values for the remainder \( r \) are \( 0, 1, 2, 3, 4, 5 \).
Now, we check each possibility for \( a \):
When \( r = 0 \), \( a = 6q \). This can be written as \( 2(3q) \), which is an even number.
When \( r = 1 \), \( a = 6q + 1 \). This cannot be divided by 2, so it is an odd number.
When \( r = 2 \), \( a = 6q + 2 \). This can be written as \( 2(3q + 1) \), which is an even number.
When \( r = 3 \), \( a = 6q + 3 \). This cannot be divided by 2, so it is an odd number.
When \( r = 4 \), \( a = 6q + 4 \). This can be written as \( 2(3q + 2) \), which is an even number.
When \( r = 5 \), \( a = 6q + 5 \). This cannot be divided by 2, so it is an odd number.
From these cases, we can see that positive odd integers only happen when the remainder is 1, 3, or 5.
Therefore, any positive odd integer must be in the form of \( 6q + 1 \), \( 6q + 3 \), or \( 6q + 5 \).
In simple words: If you divide any positive whole number by 6, the leftover part (remainder) can be 0, 1, 2, 3, 4, or 5. We showed that if the remainder is 1, 3, or 5, the number is odd. This proves that odd numbers always look like \( 6q+1 \), \( 6q+3 \), or \( 6q+5 \).

Exam Tip: Remember to consider all possible remainders when applying Euclid's division lemma. Clearly state which forms result in odd integers and which in even integers.

Question 3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Answer: To find the maximum number of columns in which both groups can march, we need to find the Highest Common Factor (HCF) of 616 and 32.
We will use Euclid's Algorithm for this purpose.
Let \( a = 616 \) and \( b = 32 \).
Applying Euclid's division lemma:
\( 616 = 32 \times 19 + 8 \)
Since the remainder 8 is not zero, we apply the division algorithm again to the divisor 32 and the remainder 8.
\( 32 = 8 \times 4 + 0 \)
Now, the remainder has become zero. The divisor at this step is 8.
Hence, the HCF of 616 and 32 is 8.
Therefore, the army contingent can march in a maximum of 8 columns.
In simple words: To have both groups march in rows with the same number of people in each, we must find the biggest number that divides both 616 and 32. This is called the HCF, and it comes out to be 8. So, they can march in 8 columns.

Exam Tip: Problems asking for the "maximum number of rows/columns/groups" always require finding the HCF, while problems asking for "least/minimum number" usually involve LCM.

Question 4. Use Euclid division lemma to show that the square of any positive integer is either of form \( 3m \) or \( 3m + 1 \) for some integer \( m \).(CBSE)
Answer: Let \( a \) be any positive integer, and let \( b = 3 \).
By applying Euclid's division lemma, we can write \( a = 3q + r \), where \( 0 \leq r < 3 \).
This means the possible values for the remainder \( r \) are \( 0, 1, 2 \).
We will now consider each case for \( r \) and find the square of \( a \).

Case 1: When \( r = 0 \)
\( a = 3q + 0 = 3q \)
Squaring both sides:
\( a^2 = (3q)^2 = 9q^2 \)
We can rewrite this as \( a^2 = 3(3q^2) \). Let \( m = 3q^2 \).
So, \( a^2 = 3m \).

Case 2: When \( r = 1 \)
\( a = 3q + 1 \)
Squaring both sides:
\( a^2 = (3q + 1)^2 \)
\( a^2 = (3q)^2 + 2(3q)(1) + 1^2 \)
\( a^2 = 9q^2 + 6q + 1 \)
We can factor out 3 from the first two terms: \( a^2 = 3(3q^2 + 2q) + 1 \). Let \( m = 3q^2 + 2q \).
So, \( a^2 = 3m + 1 \).

Case 3: When \( r = 2 \)
\( a = 3q + 2 \)
Squaring both sides:
\( a^2 = (3q + 2)^2 \)
\( a^2 = (3q)^2 + 2(3q)(2) + 2^2 \)
\( a^2 = 9q^2 + 12q + 4 \)
We can rewrite 4 as \( 3 + 1 \): \( a^2 = 9q^2 + 12q + 3 + 1 \)
Now, factor out 3 from the first three terms: \( a^2 = 3(3q^2 + 4q + 1) + 1 \). Let \( m = 3q^2 + 4q + 1 \).
So, \( a^2 = 3m + 1 \).

In all three possible cases, the square of any positive integer is either of the form \( 3m \) or \( 3m + 1 \).
In simple words: We took any whole number, \( a \), and divided it by 3. The remainder can be 0, 1, or 2. Then we squared \( a \) in each case. We found that the squared number always turns out to be either a multiple of 3 (like \( 3m \)) or a multiple of 3 plus 1 (like \( 3m+1 \)).

Exam Tip: Remember to use the general form \( (a+b)^2 = a^2 + 2ab + b^2 \) for squaring binomials. When rewriting the expression, ensure that you factor out the required number (in this case, 3) correctly to match the target form.

Question 5. Use Euclid's division lemma to show that the cube of any positive integer is of the form \( 9m, 9m + 1 \) or \( 9m + 8 \).(CBSE)
Answer: Let \( a \) be any positive integer, and let \( b = 3 \).
By applying Euclid's division lemma, we can write \( a = 3q + r \), where \( 0 \leq r < 3 \).
This means the possible values for the remainder \( r \) are \( 0, 1, 2 \).
We will now consider each case for \( r \) and find the cube of \( a \).

Case 1: When \( r = 0 \)
\( a = 3q + 0 = 3q \)
Cubing both sides:
\( a^3 = (3q)^3 = 27q^3 \)
We can rewrite this as \( a^3 = 9(3q^3) \). Let \( m = 3q^3 \).
So, \( a^3 = 9m \).

Case 2: When \( r = 1 \)
\( a = 3q + 1 \)
Cubing both sides:
\( a^3 = (3q + 1)^3 \)
Using the identity \( (x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 \):
\( a^3 = (3q)^3 + 3(3q)^2(1) + 3(3q)(1)^2 + 1^3 \)
\( a^3 = 27q^3 + 27q^2 + 9q + 1 \)
We can factor out 9 from the first three terms: \( a^3 = 9(3q^3 + 3q^2 + q) + 1 \). Let \( m = 3q^3 + 3q^2 + q \).
So, \( a^3 = 9m + 1 \).

Case 3: When \( r = 2 \)
\( a = 3q + 2 \)
Cubing both sides:
\( a^3 = (3q + 2)^3 \)
Using the identity \( (x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 \):
\( a^3 = (3q)^3 + 3(3q)^2(2) + 3(3q)(2)^2 + 2^3 \)
\( a^3 = 27q^3 + 54q^2 + 36q + 8 \)
We can factor out 9 from the first three terms: \( a^3 = 9(3q^3 + 6q^2 + 4q) + 8 \). Let \( m = 3q^3 + 6q^2 + 4q \).
So, \( a^3 = 9m + 8 \).

In all three possible cases, the cube of any positive integer is either of the form \( 9m \), \( 9m + 1 \), or \( 9m + 8 \).
In simple words: We took any whole number, \( a \), and divided it by 3. The leftover part can be 0, 1, or 2. Then we found the cube of \( a \) in each case. We saw that the cubed number always turns out to be either a multiple of 9 (like \( 9m \)), a multiple of 9 plus 1 (like \( 9m+1 \)), or a multiple of 9 plus 8 (like \( 9m+8 \)).

Exam Tip: Be careful when expanding \( (a+b)^3 \). Make sure to include all terms from the binomial expansion formula. Factoring out 9 correctly is crucial for showing the final forms.