Question

please Solve the above question( advanced maths) class 10​

Answer 1

Solution

Given , a be any odd positive integer

Applying the division algorithm on ‘a’ we have :

a = bq + r , 0 ≤ r < b

When , b = 4 we get,

a = 4q + r , 0 ≤ r < 4

Since , a is an odd positive integer so

→ a = 4q + 1 and 4q + 3

Now we have a(a² - 1)

When , a = 4q + 1 we have from a²(a - 1)

➝ (4q + 1){ (4q+1)² – 1}

➝ (4q + 1){ 16q² + 8q +1-1}

➝ (4q + 1)(16q² + 8q)

➝ 4(4q + 1)(4q² + 2q) , which is divisible by 4

Similarly , for a = 4q + 3 , it can be shown that a(a² - 1) is divisible by 4

Therefore ,

$\therefore \sf 4 |a( {a}^{2} - 1) \forall \ \: a \in Z \longrightarrow(1)$

Since , 4 divides a(a² - 1) and 2 is a factor of 4 so ,

$\therefore \sf 4 |a( {a}^{2} - 1) \forall \ \: a \in Z \longrightarrow(2)$

Again when b = 3 then

a = 3q + r , 0 ≤ r < 3

since , a is odd positive integer so

a = 3q + 1

Therefore putting a = 3q + 1from a(a² - 1) we have

➝ (3q + 1){(3q + 1)² - 1}

➝ (3q + 1){ 9q² + 6q + 1 - 1}

➝ (3q + 1){ 9q² + 6q }

➝ 3(3q + 1)(3q² + 2q) , which is divisible by 3

$\therefore \sf 3 | a( {a}^{2} - 1) \forall \: a \in Z \longrightarrow(3)$

But G.C.D ( 4 , 3 , 2) = 1

From (1) , (2) and (3) we have ,

4×3×2 | a(a² – 1) [by theorem]

☞ 24 | a(a² – 1)

Proved

Answer 2

Answer:

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