This CBSE class 10 Maths Extra Practice Question is from Chapter 1 - Real Numbers. This is an importnat question and is a word problem that tests your understanding of divisors, remainders and idea of using HCF when a divisor divides 2 numbers leaving 2 different remainders.

Question 4 : What is the largest number that divides 967 and 1767 leaving remainders of 71 and 103 respectively?

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**Given**: The two numbers are 967 and 1767.

Let x be the largest number that divides both 967 and 1767 and leaves remainders of 71 and 103 respectively.

Remainder of \\frac{\text{967}}{\text{x}}) is 71.

So, 967 – 71 = 896 is divisible by x.

Remainder of \\frac{\text{1767}}{\text{x}}) is 103.

So, 1767 – 103 = 1664 is divisible by x.

So, x divides both 896 and 1664.

In other words, x is a factor of both 896 and 1664. i.e., x is a common factor of 896 and 1664.

Because x is the largest such number, **x is the HCF of 896 and 1664.**

**Step 1**: Apply Euclid's Lemma on 1664 with 896 as divisor.

1664 = 896 × 1 + 768

The remainder is not zero.

**Step 2**: Apply Euclid's Lemma on 896 with 768 as divisor.

896 = 768 × 1 + 128

The remainder is not zero.

**Step 3**: Apply Euclid's Lemma on 768 with 128 as divisor.

768 = 128 × 6 + 0

The remainder is zero.

∴ The divisor of this step, 128 is the HCF of 896 and 1664.

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CBSE Class 10 Maths - 2021

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