Question

If
$\sf\dfrac{241}{4000} = \dfrac{241}{2^m5^n}$
the find the value of m and n
Hint = They are non negative integer

Answer 1

Answer:

Since the numerators are equal the denominator must be equal too...

i.e

4000= $2^m*5^n$

The prime factorization of 4000 is

4000=2*2*2*2*2*5*5*5

         = $2^5*5^3$

Its in the form of $2^m*5^n$

$2^m*5^n$ = $2^5*5^3$

Replace values of m and n with 5 and 3

∴, m=5 and n=3

Answer 2

Solution:

$\sf{ \dfrac{241}{4000} }$

We are supposed to represent 4000 in the form 2ᵐ × 5ⁿ

For solving this we need to factorise 4000

2 |4000

2 |2000

2 |1000

2 |500

2 |250

5 |125

5 |25

5 |5

1

So, 4500 can be represented as 2⁵ × 5³

where

• m = 5
• n = 3

KNOW MORE:

• Since the fraction 241 ÷ 4000 can be represented as 241 ÷ (2⁵ × 5³) it has a terminating decimal expansion.
• Terminating decimal expansion means that when the fraction is divided by long division, it would end at a point and does not have a recurring decimal expansion.