This CBSE class 10 Maths Extra Practice Question from the topic Real Numbers. This word problem tests your understanding of computing Highest Common Factor (HCF) of 3 Numbers using Euclid's Lemma.

Question 3 : What is the largest number that divides 437, 732, and 1263 leaving remainder of 24 in each case?

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**Given**: The three numbers are 437 , 732, and 1263.

Let 'x' be the largest number that leaves a remainder of 24 when dividing these 3 numbers.

Remainder of \\frac{\text{437}}{\text{x}}) is 24.

∴ 437 – 24 = 413 is divisible by x.

Remainder of \\frac{\text{732}}{\text{x}}) is 24.

∴ 732 – 24 = 708 is divisible by x.

Remainder of \\frac{\text{1263}}{\text{x}}) is 24

∴ 1263 – 24 = 1239 is divisible by x.

So, x divides 413, 708, and 1239 without leaving any remainder.

So, x is a common factor of 413, 708, and 1239.

Because x is the largest such number, ** x is the HCF of 413, 708, and 1239.**

**Step 1**: Apply Euclid's Lemma on 1239 with 708 as divisor.

1239 = 708 × 1 + 531

The remainder is not zero.

**Step 2**: Apply Euclid's Lemma on 708 with 531 as divisor.

708 = 531 × 1 + 177

The remainder is not zero.

**Step 3**: Apply Euclid's Lemma on 531 with 177 as divisor.

531 = 177 × 3 + 0

The remainder is 0.

So, the divisor of this step, 177 is HCF of 1239 and 708.

**Step 4**: Find the HCF of 177 and 413. Apply Euclid's Lemma on 413 with 177 as divisor.

413 = 177 × 2 + 59

The remainder is not zero.

**Step 5**: Apply Euclid's Division Lemma on 177 with 59 as divisor.

177 = 59 × 3 + 0

The remainder is 0.

∴ The divisor of this step, 59 is the HCF of 413 and 177.

59 is the HCF of 413, 708, and 1239

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CBSE Class 10 Maths - 2021

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1. HCF Word Problem | CBSE Class 10 Math Extra Practice Question **Video Solution **▶

2. Highest Common Factor of 2 Numbers | CBSE 10th Math Extra Practice Question **Video Solution** ▶

3. HCF of 3 numbers | CBSE Class 10 Math Extra Practice Question **Video Solution** ▶

4. Common divisor - different remainders - 2 numbers | CBSE 10th Math Extra Practice Question **Video Solution** ▶

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