Question

N=999,999,999,999,999,999. How many 9s are there in the decimal expansion of N ^2?

Answer 1

There will be zero 9's in the decimal expansion of $N^2$.

Given:

N=999,999,999,999,999,999

To find:

The number of 9's in the decimal expansion of $N^2$

Solution:

• A decimal expansion is a combination of a natural number N followed by an infinite sequence d of digits.
• Where d is the sequence whose values can be any whole number less than 10.
• A decimal expansion is denoted by:

         $N.d_{1} d_{2} d_{3} d_{4} d_{5} .....$

We can write 999,999,999,999,999,999 in decimal form as

999,999,999,999,999,999.000000                                               ...(I)

Here the zeros after decimal point can be any infinitely many.

Now, $N^2=(999,999,999,999,999,999.00000)^2$

• Without calculating actual square, If we square this number then irrespective of the 9's before decimal point there will not be any 9's after the decimal point.
• We can write decimal form of the respective answer like as in the  statement (I)
• We will have Zero number of 9's in the decimal expansion of the given number as there will be again zeros after the decimal point in the calculated answer.

Hence, there will be zero 9's in the decimal expansion of $N^2$.

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