Question

without actually performing the long division ,state​

Answer 1

Answer:

Answer

To Find, The decimal expansion of the given number without actual division. Solution, ...

So,

Now, Multiply ans divide the number by 2³. Then,

Here 5³ × 2³ = (5×2)³ = 10³ ( By using the identity) So,

Therefore the value of the given number is

Answer 2

Answer:

Theorem: Let x=qp​ be a rational number, such that the prime factorisation of q is of the form 2n5m, where n, m are non-negative integers. Then, x has a decimal expansion which terminates.

(i) 312513​

Factorise the denominator, we get

3125=5×5×5×5×5=55

So, denominator is in form of 5m so, 312513​ is terminating.

(ii) 817​

Factorise the denominator, we get

8=2×2×2=23

So, denominator is in form of 2n so, 817​ is terminating.

(iii) 45564​

Factorise the denominator, we get

455=5×7×13

So, denominator is not in form of 2n5m so, 45564​ is not terminating.

(iv) 160015​

Factorise the denominator, we get

1600=2×2×2×2×2×2×5×5=2652

So, denominator is in form of 2n5m so, 160015​ is terminating.

(v) 34329​

Factorise the denominator, we get

343=7×7×7=73

So, denominator is not in form of 2n5m so, 34329​ is not terminating.

(vi) 235223​

Here, the denominator is in form of 2n5m so, 235223​ is terminating.

(vii) 225775129​

Here, the denominator is not in form of 2n5m so, 225775129​ is not terminating.

(viii) 156​

Divide nominator and denominator both by 3 we get 153​

So, denominator is in form of 5m so, 156​ is terminating.

(ix) 5035​

Divide nominator and denominator both by 5 we get 107​

Factorise the denominator, we get

10=2×5

So, denominator is in form of 2n5m so, 5035​ is terminating.

(x) 21077​

Divide nominator and denominator both by 7 we get 3011​

Factorise the denominator, we get

30=2×3×5

So, denominator is not in the form of 2n5m so 156​ is not terminating.

verified.

Step-by-step explanation: