# NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.1

## Chapter 1 Ex.1.1 Question 1

Use Euclid’s division algorithm to find the HCF of:

(i) \(135\) and \(225\)

(ii) \(196\) and \(38220\)

(iii) \(867\) and \(255\)

**Solution**

**Video Solution**

**What is known?**

Two different numbers

**What is unknown?**

HCF of the given numbers.

**Reasoning: **

You have to find the HCF of given integers by using Euclid’s Division Lemma. It is a technique to compute the highest common factor of two given positive integer. Recall, that the HCF of two positive integers *\(a\)* and *\(b\)* is the largest positive integer that divides both *\(a\)* and \(b.\)

To obtain the HCF of two positive integers say *\(a\)* and *\(b\)* with \(a > b,\) follow the below steps-

**Step- I.** Apply Euclid’s division lemma to *\(a\)* and \(b.\) So, we find whole numbers \(q\) and \(r\) such that

\[a = bq + r,\; 0 \le r < b\]

**Step - II. ** If \(r=0,\) *\(b\)* is the HCF of \(a \) and \(b.\) If \( r \ne 0,\) apply the division lemma to *\(b\)* and *\(r.\)*

**Step - III.** Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

**Steps:**

(i) \(135\) and \(225\)

In this case \(\begin{align} 225 > 135. \end{align}\) We apply Euclid’s division lemma to \(135\) and \( 225\) and get

\[225=(135 \times 1)+90\]

Since, the remainder \(r \ne 0,\) we apply the division lemma to \(135\) and \(90\) to get

\[135=(90 \times 1)+45\]

Now, we consider \(90\) as the divisor and \(45\) as the remainder and apply the division lemma, to get

\[90 = (45\times 2 )+ 0\]

Since, the remainder is zero and the divisor is \(45,\) therefore, the H.C.F of \(135\) and \(225\) is \(45.\)

(ii) \(196\) and \(38220\)

\(38220\) is greater than \(196, \) we apply Euclid’s division lemma to \(38220\) and \(196,\) to get

\[38220 = {{ }}(196{{ }} \times {{ }}195) + {{ }}0\]

Since, the remainder is zero and the divisor in this step is \(195,\) therefore, the H.C.F of \(38220\) and \(196\) is \(196.\)

(iii) \(867\) and \(255\)

\(867\) is greater than \(225\) and on applying Euclid’s division lemma to \(867\)and \(225,\) to get

\[867 = {{ }}(255{{ }} \times {{ }}3) + {{ }}102\]

Since, the remainder \(r{{ }} \ne {{ }}0,\) we apply the division lemma to \(225\) and \(102\) and get

\[255{ }=(102{ }\times { }2)+{ }51\]

Again, remainder is not zero, we apply Euclid’s division lemma \(102\) and \(51\) which gives

\[102=(51 \times \ 2)+0\]

Since, the remainder is zero and the divisor is \(51,\) therefore, the H.C.F of \(867\) and \(255\) is \(51.\)

## Chapter 1 Ex.1.1 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.

**Solution**

**Video Solution**

**What is known?**

Some integer \(q.\)

**What is unknown?**

That any positive odd integer is of the form \(6q+1,\) or \(6q+3,\) or \(6q+5,\) where *\(q\) *is some integer.

**Reasoning: **

To solve this question, first think about the Euclid’s division algorithm. Suppose there is any positive integer '\(a\)' and it is of the form \(6q+ r\). where \(q\) is some integer. This means that \(0 \le r < 6\) i.e. \(r=0\) or \(1\) or \(2\) or \(3\) or \(4\) or \(5\) but it can’t be \(6 \) because \(r\) is smaller than \(6.\). So, by Euclid’s division lemma possible values for '\(a\)' can be \(6q\) or \(6q+1\)or \(6q+2\) or \(6q+3\) or \( 6q+4\) or \(6q+5.\)

**Steps:**

Let '\(a\)' be any positive integer. Then according to Euclid’s algorithm,

\(a = 6q + r\) for some integer \(q \ge 0\)

and \(r= 0,1,2,3,4,5\) because \(0\le r < 6\)

Therefore,

\(a = 6q + 0\) or \(6q + 1\) or \(6q + 2\) or \(6q + 3\) or \(6q + 4\) or \(6q + 5.\)

Now, \(6q + 1 = 2 \times 3q + 1 = 2 {k_1}+1\)

(where \(k_1\) is a positive integer)

\(\begin{align}6q+3 &= 6q+2 +1\\&= 2 (3q + 1)+1 \\&=2{k_2}+1\end{align}\)

(where \(k_2\) is a positive integer)

\(\begin{align}6q+5&=6q+4+1\\&=2 (3q+2)+1\\&=2{k_3}+1\end{align}\)

(where \(k_3\) is a positive integer)

Clearly, \(6q+1,6q+3\) and \(6q+5\) are of the form \(2k+1,\) where \(k\) is an integer.

Therefore, \(6q+1, 6q+3\) and \(6q+5\) are not exactly divisible by \(2.\)

Hence, these expressions of numbers are odd numbers and therefore any odd integers can be expressed in the form \(6q+1\) or \(6q+3\) or \(6q+5.\)

## Chapter 1 Ex.1.1 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?

**Solution**

**Video Solution**

**What is known?**

We are told that there is an army contingent of \(616\) members and an army band of \(32\) members. The two groups are to march in the same number of columns

**What is unknown?**

The maximum number of columns in which they can march.

**Reasoning: **

Here, we have to pay attention to the point that the army band members and army contingent members have to march in the **same number of columns** and that the number of columns must be the **maximum** possible. The definition of HCF states – HCF of two positive integers *\(a\) *and *\(b\)* is the largest positive integer *\(d\)* that divides both *\(a\)* and *\(b\)*. In other words, HCF of two numbers is the highest number (maximum) that divides both the numbers. Thus, we have the find the HCF of the members in the army band and the army contingent.

**Steps:**

HCF (\(616, 32\)) will give the maximum number of columns in which they can march.

We use Euclid’s algorithm to find the H.C.F

\[\begin{align} 616&=(32\,\times \,19)\,+\,8 \\32&=(8\,\times \,4)+0\end{align}\]

The HCF (\(616, 32\)) is \(8.\) Therefore, they can march in \(8\) columns each.

## Chapter 1 Ex.1.1 Question 4

Use Euclid’s division lemma to show that the square of any positive integer is either of the form \(3\,m \) or \(3\,m + 1\) for some integer \(m. \)

[Hint: Let \(x \) be any positive integer then it is of the form \(3q,\, 3q + 1 \) or \(3q + 2.\) Now square each of these and show that they can be rewritten in the form \(3\,m\) or \(3\,m + 1.\)]

**Solution**

**Video Solution**

**To Prove:**

The square of any positive integer is either of the form \(3\,m \) or \(3\,m + 1\) for some integer \(m\) (using the Euclid’s division lemma).

**Reasoning: **

Suppose that there is a positive integer ‘\(a\)’. By Euclid’s lemma, we know that for positive integers \(a\) and \(b,\) there exist unique integers \(q\) and \( r,\) such that \(a=bq+r,\,0\le r\lt b\)

If we keep the value of \(b = 3,\) then \(0 ≤ r < 3\) i.e. \(r = 0\) or \(1\) or \(2\) but it can’t be \(3\) because \(r\) is smaller than \(3.\) So, the possible values for \(a = 3q\) or \(3q + 1 \) or \(3q + 2.\) Now, find the square of all the possible values of \(a.\) If \(q\) is any positive integer then its square (let’s call it as “\(m\)”) will also be a positive integer. Now, observe carefully that the square of all the positive integers is either of the form \(3\,m \) or \(3\,m + 1\) for some integer \(m. \)

**Steps:**

Let \(“a” \) be any positive integer and \(b=3.\)

Then, \(a=3q+r\) for some integer \(q\ge 0\) and \(r = 0, 1 , 2\) because \(0\le r <3.\)

Therefore,

\[\begin{align} a&=3q \;\text{ or }\;3q+1\;\text{ or }\;3q+2\text{ or }\;\\ {{a}^{2}}&\!=\!{{\left( 3q \right)}^{2}}\;\!\text{ or }\!\;{{\left(3q+1 \right)}^{2}}\;\!\!\text{ or }\!\!\;{{\left( 3q{ }+{ }2 \right)}^{2}} \\ {{a}^{2}}&\!\!=\!\!3{{(3q^{2})}}\;\!\!\text { or }\;\!\!\!(9{{q}^{2}}\!\!+\!\!6q\!+\!1)\;\!\!\\&\text{ or }\!\!\!\;(9{{q}^{2}}\!+\!12q\!+\!{{4}^{}}) \\{{a}^{2}}&=3\left( 3{{q}^{2}} \right)\;\text{ or }3\left( 3{{q}^{2}}+2q \right)+1\\&\qquad \text{ or }\;3\left( 3{{q}^{2}}+{ }4q+1 \right)+1 \\&=m \; \text{ or } \;3 m +1 \\\end{align}\]

Where \(m \) is any positive integer.

Hence it can be said that the square of any positive integer is either of the form \(\begin{align}{3\,m \;\rm{or}\; 3\,m + 1.} \end{align}\)

## Chapter 1 Ex.1.1 Question 5

Use Euclid’s division lemma to show that the cube of any positive integer is of the form \(9\,m , 9\, m + 1\) or \(9\,m + 8.\)

**Solution**

**Video Solution**

**To Prove:**

The cube of any positive integer is of the form \( {9\,m, 9\,m + 1} \) or \( {9\,m + 8.} \)

**Reasoning: **

Suppose that there is a positive integer ‘\(a\)’ . By Euclid’s lemma, we know that for positive integers \(a\) and \(b,\) there exist unique integers \(q\) and \(r,\) such that \( a=b q+r, 0 \leq r \lt b\)

If we keep the value of \(b = 3\), then \(0 ≤ r < 3\) i.e. \(r = 0\) or \(1\) or \(2\) but it can’t be \(3\) because \(r\) is smaller than \(3.\) So, the possible values for \( a = 3q\) or \(3q + 1\) or \(3q + 2.\) Now, find the cube of all the possible values of \(a.\) If \(q\) is any positive integer then its cube (let’s call it as “\(m\)”) will also be a positive integer. Now, observe carefully that the cube of all the positive integers is either of the form \(9\,m\) or \(9\,m+ 1\) or \(9\,m + 1\) for some integer \(m.\)

**Steps:**

Let \(\begin{align}{“a”} \end{align}\) be any positive integer and \(b=3.\)

Then, \(a=3q+r\) for some integer \(q\ge 0\) and \(0\le r<3\)

Therefore, \(a=3q\) or \(3q+1\) or \(3q+2\)

**Case – I.** When \(a=3q\)

\[{{\left( a \right)}^{3}}\!=\!{{\left( 3q \right)}^{3~}}\!=\!27{{q}^{3}}\!=\!9\left( 3{{q}^{3}} \right)\!=\!9\,m\]

Where \(m\) is an integer such that \(m=3{{q}^{3}}\)

**Case – II.** When \(a=3q+1\)

\[\begin{align}{{\left( a \right)}^{3}}&={ }{{\left( 3q{ }+{ }1 \right)}^{3}} \\{{\left( a \right)}^{3}}&={ }27{{q}^{3}}+{ }27{{q}^{2}}+9q{ }+1 \\ {{\left( a \right)}^{3}}&={ }9\left( 3{{q}^{3}}+{ }3{{q}^{2}}+q \right){ }+1 \\ {{\left( a \right)}^{3}}&={ }9 m +1 \\\end{align}\]

Where \(m\) is an integer such that \(m=3{{q}^{3}}+{ }3{{q}^{2}}+{ }q\)

**Case – III. **When \(a=3q+2\)

\[\begin{align}\,{{\left( a \right)}^{3}}&={ }{{\left( 3q+2 \right)}^{3}} \\ {{\left( a \right)}^{3}}&={ }27{{q}^{3}}+{ }54{{q}^{2}}+36q{ }+{ }8 \\ {{\left( a \right)}^{3}}&={ }9\left( 3{{q}^{3}}+{ }6{{q}^{2}}+{ }4q \right){ }+{ }8 \\ {{\left( a \right)}^{3}}&={ }9 m+8 \\\end{align}\]

Where \(m\) is an integer such that \(m=3{{q}^{3}}+{ }6{{q}^{2}}+{ }4q\)

Thus, we can see that the cube of any positive integer is of the form \(9 m,\; 9{m}+ 1\) or \(9 \, m +8.\)