I am trying to compete with my teenage daughter.
In a 6 figure number where the first number cannot be zero but the other five
can be any number, how many different 6 figure numbers can there be ?
Also I would like the formula for calculating same.
Consider the amount of possible outcomes.
For example for a three digit number it is 9 * 10*10
Yes but what is the answer and what is the formula. I am just a simple father.
There are N-digits positive integers which do not begin with zero.
Originally Posted by divorce4
Does that help?
There are 999999 six digit numbers. The first 99999 of them have a leading 0 so 999999- 99999 of them do not have a leading 0.
What is this competition with your daughter? To see who can first get someone else to do the problem for them?
(And why was this titled "factorizing"?)