Name: Composite Number as Product of Primes
Uploaded: 2019-07-19
Duration: 6 min 23 s
Description: How do you write number 81? Number 81 can be written as 3x3x3x3, which cannot be divided further. What is this called? Explore more about it in this video.

Question

Use Euclid’s division algorithm to find the

H C F

of

135

and

225

.

Accepted Answer

The given numbers are $135$ and $225$ .

Since $225$ is greater than $135$ , we can apply Euclid's division lemma to $a = 225$ and $b = 135$ to find $q$ and $r$ such that $225 = 135 q + r, 0 \leq r < 135$ .

So, dividing $225$ by $135$ we get $1$ as the quotient and $90$ as remainder $i . e . 225 = (135 \times 1) + 90$ .

Remainder $r$ is $90$ and is not equal to $0$ , we apply Euclid's division lemma to $a = 135$ and $b = 90$ to find whole numbers $q$ and $r$ such that $135 = 90 \times q + r, 0 \leq r < 90$ .

So, dividing $135$ by $90$ , we get $1$ as the quotient and $45$ as remainder $i . e . 135 = (90 \times 1) + 45$ .

Again, remainder $r$ is $45$ and is not equal to $0$ , so we apply Euclid's division lemma to $a = 90$ and $b = 45$ to find $q$ and $r$ such that $90 = 45 \times q + r, 0 \leq r < 45$ .

So, dividing $90$ by $45$ , we get $2$ as the quotient and $0$ as remainder $i . e . 90 = (2 \times 45) + 0$ .

Since the remainder is $0$ , the divisor at this stage will be $HCF$ of $(135, 225)$ and the divisor at this stage is $45$ .

Therefore, the $H C F$ of $135$ and $225$ is $45$ .

Exercise 1.1

Jharkhand Board Mathematics Solutions for Exercise 1.1

Questions from Exercise 1.1 with Hints & Solutions

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