To determine the number of digits in a number whose square results in a 24-digit number, we can use properties of logarithms and exponents. Here is the step-by-step process:
1. Understand the Given Condition:
We are given that the square of a certain number results in a 24-digit number. Let this number be [tex]\( n \)[/tex].
2. Range of the 24-Digit Number:
A 24-digit number lies in the range:
[tex]\[
10^{23} \leq n^2 < 10^{24}
\][/tex]
This is because any number with 24 digits is at least [tex]\( 10^{23} \)[/tex] and less than [tex]\( 10^{24} \)[/tex].
3. Taking the Square Root:
To find the range for [tex]\( n \)[/tex], we take the square root of both sides of the inequality:
[tex]\[
\sqrt{10^{23}} \leq n < \sqrt{10^{24}}
\][/tex]
Using the properties of exponents, the square roots can be simplified:
[tex]\[
10^{11.5} \leq n < 10^{12}
\][/tex]
4. Determine the Number of Digits:
We need to find the number of digits in [tex]\( n \)[/tex]. The number of digits in a number [tex]\( x \)[/tex] is given by [tex]\( \lfloor \log_{10}x \rfloor + 1 \)[/tex].
- For the lower bound [tex]\( 10^{11.5} \)[/tex]:
[tex]\[
\lfloor \log_{10}(10^{11.5}) \rfloor + 1 = \lfloor 11.5 \rfloor + 1 = 11 + 1 = 12
\][/tex]
- For the upper bound [tex]\( 10^{12} \)[/tex]:
[tex]\[
\lfloor \log_{10}(10^{12}) \rfloor + 1 = \lfloor 12 \rfloor + 1 = 12 + 1 = 13
\][/tex]
5. Conclusion:
Since [tex]\( n \)[/tex] lies between [tex]\( 10^{11.5} \)[/tex] and [tex]\( 10^{12} \)[/tex], the only digit count that fits within this range is 12.
Therefore, the number of digits in the certain number [tex]\( n \)[/tex] is 12.
