Question

without actual division find after how many digits decimal will terminate 147/2^5×5^2×7^2​

Answer 1

Question:

Without actual division find after how many digits does the decimal expansion of 147/(2^5)(5^2)(7^2) terminates.

Answer:

5

Note:

• Rational no : The number which can be written in the form of p/q where p and q are integers but q≠0 are called rational number .

Eg : 1/2 , 4/3 , 3.44 , 2.3333333....... etc

• Rational number can be characterized as;

1) Terminating

Eg : 1/2 , 3.44 , 5/4 etc

2) Non-terminating but repeating

( or Non-terminating but recurring )

Eg : 5/3 , 11/7 , 4.3333...... etc

• The rational number p/q (in its simplest form) will terminate if its denominator q can be written as 2^m × 5^n and its decimal expansion would terminate after m digits (ie, the power of 2) .

• Irrational no. : All those numbers which are not rational are irrational numbers . Such type of number which cannot be written in the form of p/q where p and q are integers but q≠0 are called irrational number .

Eg : √3 , π , 3√2 , 1.207086499...... etc

• Irrational numbers are characterized as non-terminating non-recurring or non-terminating non-repeating .

Solution:

The given rational number is :

47/(2^5)(5^2)(7^2).

The simplest form of the given rational number is;

1/(2^5)×(5^2).

Here,

The denominator is 2^5 × 5^2.

Clearly,

The denominator is of the form 2^m × 5^n ,

Where m = 5 and n = 2 .

Thus,

The given rational number is terminating and its decimal expansion would terminates after 5 digits.

Hence,

The decimal expansion of 47/(2^5)(5^2)(7^2) would terminates after 5 digits .