# Factorizing

• Mar 27th 2011, 09:50 AM
divorce4
Factorizing
I am trying to compete with my teenage daughter.

Question:

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.
• Mar 27th 2011, 09:54 AM
e^(i*pi)
Consider the amount of possible outcomes.

For example for a three digit number it is 9 * 10*10
• Mar 27th 2011, 10:04 AM
divorce4
Yes but what is the answer and what is the formula. I am just a simple father.
• Mar 27th 2011, 10:11 AM
Plato
Quote:

Originally Posted by divorce4
Yes but what is the answer and what is the formula. I am just a simple father.

There are $9\cdot 10^{N-1}$ N-digits positive integers which do not begin with zero.

Does that help?
• Mar 27th 2011, 03:27 PM
HallsofIvy
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?