This CBSE class 10 Maths Extra Practice Question is from Real Numbers. This word problem tests your understanding of computing HCF of two numbers that leave the same remainder.

Question 2 : Find the largest number that will divide 382 and 710 and leaves a remainder 13 in each case.

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**Given**: The two numbers are 382 and 710

Let 'x' be the largest number that divides 382 and 710, leaving 13 as remainder in both cases.

Remainder of \\frac{\text{382}}{\text{x}}) is 13

∴ 382 – 13 = 369 is divisible by x.

Similarly, remainder of \\frac{\text{710}}{\text{x}}) is 13

∴ 710 – 13 = 697 is divisible by x.

So, x divides both 369 and 697.

Or, x is a factor common of both 369 and 697.

Because x is the largest such number, **x is the HCF of 369 and 697.**

**Step 1**: Apply Euclid's Lemma on 697 with 369 as divisor

697 = 369 × 1 + 328

The remainder is not zero.

**Step 2**: Apply Euclid's Division Lemma on 369 with 328 as divisor

369 = 328 × 1 + 41

The remainder is not zero.

**Step 3**: Apply Euclid's Lemma on 328 with 41 as divisor

328 = 41 × 8 + 0

The remainder is zero.

∴ The divisor of this step, 41 is the HCF of 369 and 697

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CBSE Class 10 Maths - 2021

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