### Chapter 1: Real Numbers

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Q
##### Real Number CBSE NCERT Solutions

Question:

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

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

Answer:

# (i) 135 and 225

As we can see, 225 is greater than 135. Therefore, by using Euclid’s division lemma, 𝒂 = 𝒃𝒒 + 𝒓,𝟎 ≤ 𝒓 < 𝒃, we have,

225 = 135 × 1 + 90

Now, remainder 90 ≠ 0,

We consider new dividend 135 and new divisor 90, and again using Euclid’s division lemma, we get,

135 = 90 × 1 + 45

Again, remainder 45 ≠ 0,

We consider new dividend 90 and new divisor 45, and again using Euclid’s division lemma, we get,

90 = 45 × 2 + 0

The remainder is now zero, and the divisor at this stage is 45.

∴ HCF (225,135) = HCF (135, 90) = HCF (90, 45) = 45.

Hence, the HCF of 225 and 135 is 45.

# (ii) 196 and 38220

As we can see, 38220 is greater than 196. Therefore, by using Euclid’s division lemma, 𝒂 = 𝒃𝒒 + 𝒓,𝟎 ≤ 𝒓 < 𝒃, we have,

38220 = 196 × 195 + 0

The remainder is zero, and the divisor at this stage is 196.

∴ HCF (38220, 196) = 196.

Hence, the HCF of 38220 and 196 is 196.

# (iii) 867 and 255

As we can see, 867 is greater than 255. Therefore, by using Euclid’s division lemma, 𝒂 = 𝒃𝒒 + 𝒓,𝟎 ≤ 𝒓 < 𝒃, we have,

867 = 255 × 3 + 102

Now, remainder 102 ≠ 0,

We consider new dividend 255 and new divisor 102, and again using Euclid’s division lemma, we get,

255 = 102 × 2 + 51

Again, 51 ≠ 0,

We consider new dividend 102 and new divisor 51, and again using Euclid’s division lemma, we get,

102 = 51 × 2 + 0

The remainder is now zero, since the divisor at this stage is 51.

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

Hence, the HCF of 867 and 255 is 51.

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