Math, asked by anujshinghyadav, 1 month ago

1. Use Euclid's division algorithm to find the HCF of:
(0) 135 and 225
(i) 196 and 38220
Show that any pocit
(ii) 867 and 255​

Answers

Answered by ansht515
0

Answer:

a) 45 b)196 c)51

Step-by-step explanation:

pls check by lemma

Answered by llTheUnkownStarll
1

\large \fbox  \red{Correct Question:}

1. Use Euclid's division algorithm to find the HCF of:

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

 \large \fbox \red {Solution:}

 \bold{(i) 135  \: and  \: 225}

As you can see, from the question 225 is greater than 135.

Therefore, by Euclid’s division algorithm, we have,

 \mapsto{\textrm{{{\color{navy}{225 = 135 × 1 + 90}}}}}

Now, remainder 90 ≠ 0, thus again using division lemma for 90, we get,

 \mapsto{\textrm{{{\color{navy} {135= 90 × 1 + 45}}}}}

Again, 45 ≠ 0, repeating the above step for 45, we get,

 \mapsto{\textrm{{{\color{navy} {90 = 45 × 2 + 0}}}}}

The remainder is now zero, so our method stops here.

Since, in the last step, the divisor is 45, therefore, HCF (225,135) = HCF (135, 90) = HCF (90, 45) = 45.

 \fbox \orange{Hence, the HCF of 225 and 135 is 45.}

 \bold{(ii) 196 \:  and \:  38220}

In this given question, 38220 >196, therefore the by applying Euclid’s division algorithm and taking 38220 as divisor, we get,

 \mapsto{\textrm{{{\color{navy} {38220 = 196 × 195 + 0}}}}}

We have already got the remainder as 0 here. Therefore, HCF(196, 38220) = 196.

 \fbox \orange{Hence, the HCF of 196 and 38220 is 196.}

 \bold{(iii) 867 \:  and \:  255}

As we know, 867 is greater than 255. Let us apply now Euclid’s division algorithm on 867, to get,

 \mapsto{\textrm{{{\color{navy} {867 = 255 × 3 + 102}}}}}

Remainder 102 ≠ 0, therefore taking 255 as divisor and applying the division lemma method, we get,

 \mapsto{\textrm{{{\color{navy} {255 = 102 × 2 + 51}}}}}

Again, 51 ≠ 0. Now 102 is the new divisor, so repeating the same step we get,

 \mapsto{\textrm{{{\color{navy} {102 = 51 × 2 + 0}}}}}

The remainder is now zero, so our procedure stops here. Since, in the last step, the divisor is 51.

Therefore, HCF (867,255) = HCF(255,102) = HCF(102,51) = 51.

 \fbox \orange{Hence, the HCF of 867 and 255 is 51.}

Thank you!

@itzshivani

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