# 1. Find the HCF of 81 and 237 and express it as a linear combination of 81 and 237.2. Find the HCF of 65 and 117 and express it in the form 65m + 117n.3. If the Hcf of the 210 and 55 is expressible in the form 210 X 5 + 55y, find y.

1>Among 81 and 237; 237 > 81
Since 237 > 81, we apply the division lemma to 237 and 81 to obtain
237 = 81 × 2 + 75  … Step 1
Since remainder 75 ≠ 0, we apply the division lemma to 81 and 75 to obtain
81 = 75 × 1 + 6  … Step 2
Since remainder 6 ≠ 0, we apply the division lemma to 75 and 6 to obtain
75 = 6 × 12 + 3  … Step 3
Since remainder 3 ≠ 0, we apply the division lemma to 6 and 3 to obtain
6 = 3 × 2 + 0     … Step 4
In this step the remainder is zero. Thus, the divisor i.e. 3 in this step is the H.C.F. of the given numbers
The H.C.F. of 237 and 81 is 3

From Step 3:
3 = 75 – 6 × 12    … Step 5
From Step 2:
6 = 81 – 75 × 1
Thus, from Step 5, it is obtained
3 = 75 – (81 – 75 × 1) × 12
⇒ 3 = 75 – (81× 12 – 75 × 12)
⇒ 3 = 75 × 13 – 81× 12     … Step 6
From Step 1;
75 = 237 – 81 × 2
Thus, from Step 6;
3 = (237 – 81 × 2) × 13 – 81× 12
⇒ 3 = (237 × 13 – 81 × 26) – 81× 12
⇒ 3 = 237 × 13 – 81 × 38
⇒ H.C.F. of 237 and 81 = 237 × 13 + 81 × (–38)

2>

Among 65 and 117; 117 > 65
Since 117 > 65, we apply the division lemma to 117 and 65 to obtain
117 = 65 × 1 + 52  … Step 1
Since remainder 52 ≠ 0, we apply the division lemma to 65 and 52 to obtain
65 = 52 × 1 + 13   … Step 2
Since remainder 13 ≠ 0, we apply the division lemma to 52 and 13 to obtain
52 = 4 × 13 + 0   … Step 3
In this step the remainder is zero. Thus, the divisor i.e. 13 in this step is the H.C.F. of the given numbers
The H.C.F. of 65 and 117 is 13

From Step 2:
13 = 65 – 52 × 1     … Step 4
From Step 1:
52 = 117 – 65 × 1
Thus, from Step 4, it is obtained
13 = 65 – (117 – 65 × 1) × 1
⇒13 = 65 × 2 – 117
⇒13 = 65 × 2 + 117 × (–1)
In the above relationship the H.C.F. of 65 and 117 is of the form 65m + 117 n, where m = 2 and n = –1

3>

Let us first find the HCF of 210 and 55.

Applying Euclid division lemna on 210 and 55, we get

210 = 55 × 3 + 45 ....(1)

Since the remainder 45 ≠ 0. So, again applying the Euclid division lemna on 55 and 45, we get

55 = 45 × 1 + 10 .... (2)

Again, considering the divisor 45 and remainder 10 and applying division lemna, we get

45 = 4 × 10 + 5 .... (3)

We now, consider the divisor 10 and remainder 5 and applying division lemna to get

10 = 5 × 2 + 0 .... (4)

We observe that the remainder at this stage is zero. So, the last divisor i.e., 5 is the HCF of 210 and 55.

∴ 5 = 210 × 5 + 55y

⇒ 55y = 5 - 1050 = -1045

⇒ y = -19

• 54
What are you looking for?