How to find number of digits in a Natural Number
You might have solved last year CAT papers(when it was available) or any AIMCAT/SIMCAT/..CAT.
1 question that troubles you is to find the number of digits in a number.
Well there are various types of questions that can be framed in which you need to find the number of digit, last digit or last 3 digit and there are various methods to solve this problem.
Today we will discuss the question type in which you need to find the number of digit in a natural number which is written in form of power.
For e.g. Number of digits in 3^86 ?
Don’t you feel to skip this question as soon as you encounter it. But if you actually know the method to solve such question. You can do it in seconds and get extra points.
Step 1: Remember the logarithmic value of following numbers log 2, log 3 , log 5, log 7. It is very important to know these values and remember it so that you can make fast calculations and solve the problem cited above.
To make it simple, I am providing the values. Just don’t forget to remember the values.
Log 2 (base 10) = 0.3010
Log 3 (base 10) = 0.4771
Log 5 (base 10) = 0.6990
Log 7 (base 10) = 0.8451
Step 2. Use the formula : No. of digits in a natural number N = (Characteristic of log N (base 10)) +1
Coming back to the question.
log (3^86) (base 10) = 86 x log 3 (base 10)
= 86 x 0.4771
characteristic of log (3^86) (base 10) = 41
So, Number of digits in 3^86 = 41 +1 = 42.
Questions for practice :
- Find the number of digits in 5^63.
- Find the number of digits in 10^32.
- Find the number of digits in 12^456.