This CBSE class 10 Maths Extra Practice Question is from the chapter Real Numbers. This word problem tests your understanding of computing the value of the largest divisor that divides 3 numbers leaving 3 different remainders.

Question 5 : What is the largest number that divides 170, 220, and 420 leaving remainders 8, 4, and 15 respectively?

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**Given Data**: The three numbers are 170, 220, and 420.

Let x be the largest number that divides 170, 220, and 420 leaving reaminders of 8, 4, and 15 respectively.

Remainder of \\frac{\text{170}}{\text{x}}) is 8.

=> 170 – 8 = 162 is divisible by x.

Remainder of \\frac{\text{220}}{\text{x}}) is 4.

=> 220 – 4 = 216 is divisible by x.

Remainder of \\frac{\text{420}}{\text{x}}) is 15.

=> 420 – 15 = 405 is divisible by x.

So, we can infer that x divides 162, 216, and 405 without leaving any remainder.

Or, x is a factor of 162, 216, and 405. i.e., x is a common factor of 162, 216, and 405.

Because x is the largest such number, ** x is the HCF OF 162, 216 and 405.**

**Step 1**: Apply Euclid's Division Lemma on 216 with 162 as divisor.

216 = 162 × 1 + 54

The remainder is not zero.

**Step 2**: Apply Euclid's Division Lemma on 162 with 54 as divisor.

162 = 54 × 3 + 0.

The remainder is 0.

∴ The divisor of this step, 54 is the HCF of 162 and 216.

**Step 3**: Find HCF of 54 and 405

Apply Euclid's Division Lemma on 405 with 54 as divisor.

405 = 54 × 7 + 27

The remainder is not zero.

**Step 4**: Apply Euclid's Division Lemma on 54 with 27 as divisor.

54 = 27 × 2 + 0

The remainder is zero.

∴ The divisor of this step, 27 is the HCF of 54 and 405.

Hence, 27 is the HCF of 162, 216, and 405.

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CBSE Class 10 Maths - 2021

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