Important Question | Class 10 Math Chapter 1 Q3

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?

Video Solution

Explanatory Answer | Real Numbers Important Question

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.

Step 1 - Decoding the given information

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 2 - Use Euclid's Division Lemma to find the HCF

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

∴ 59 is the largest number that leaves remainder 24 when dividing each of 437, 732, and 1263