Question

VVICI
15 UIVisible by 2.
Hence, n? - n is divisible by 2 for every posi
integer n.
Hence Prove
7. Find HCF of 81 and 237 and express it as a lin
combination of 81 and 237 i.e., HCF (81, 237) =
+ 237y for some x and y.
A [Board Term-I, 2012, Set-​

Answer 1

Question 1 :

If n is divisible by 2 , then prove that n²- n is divisible by 2 .

Proof :

Given that , n is divisible by 2 .

Thus let n = 2m , m € Z .

Now ,

→ n² - n = (2m)² - 2m

→ n² - n = 4m² - 2m

→ n² - n = 2m(2m - 1)

→ n² - n = 2p , p = m(2m - 1) € N

(divisible by 2)

Clearly ,

If n is divisible by 2 , then n² - n is also divisible by 2 .

Hence proved .

Question 2 :

Find HCF of 81 and 237 and express it as a linear combination of 81 and 237 i.e., HCF (81, 237) = 81x + 237y for some x and y.

Solution :

Let's find the HCF of 81 and 237 by long division method .

$\sf 81) \overline{ \: \: \: 237 \: \:}(2 \\ \sf \dfrac{162 }{ \: \: \: \: \: \: \: \: 75) \: 81\:(1 }\\ \quad \: \: \sf \dfrac{\:\:\:\:\:\:\:\:\:\:75}{ \: \: \: \: \: \: 6) \: \: 15 \: \: (12} \\ \sf {\quad\:\: }\dfrac{12}{3) \: \: \: 6 \: \: (2} \\ {\:\:\:\: \qquad} \sf \dfrac{6}{ 0}$

Clearly ,

HCF (81, 237) = 3

Also ,

From the division , we have ;

• 237 = 2×81 + 75 -----(1)

• 81 = 1×75 + 6 ------(2)

• 75 = 12×6 + 3 ------(3)

• 6 = 2×3 + 0 ------(4)

Now ,

→ HCF(81,237) = 3

→ HCF(81,237) = 75 - 12×6

→ HCF(81,237) = 75 - 12×(81 - 1×75)

→ HCF(81,237) = 75 - 12×81 + 12×75

→ HCF(81,237) = -12×81 + 13×75

→ HCF(81,237) = -12×81 + 13×(237 - 2×81)

→ HCF(81,237) = -12×81 + 13×237 - 26×81

→ HCF(81,237) = -38×81 + 13×237

→ HCF(81,237) = 81×(-38) + 237×(13)

Clearly ,

HCF (81, 237) = 81×(-38) + 237×(13) is of the form (81x + 237y) , where x = -38 and y = 13 .