# Ex.1.1 Q1 Real Numbers Solution - NCERT Maths Class 10

## Question

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

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

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

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

## Text 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 \(51.\)