# Important Questions For Class 10 Maths Chapter 1 | Q5

###### Real Numbers | Largest divisor of 3 numbers leaving 3 different remainders

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?

## NCERT Solution to Class 10 Maths

#### With Videos

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

#### Step 1 - Decoding Given Information

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 2 - Use Euclid's Division Lemma to find the HCF of the 3 Numbers

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.

##### 27 is the largest number that divides 170, 220, and 420 leaving remainders of 8, 4, and 15 respectively.

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