Count and count...

Every time we write a $1$ , it is accompanied by a $2$

But we have an extra number $100000$ and thus $A-B=1$

It seems that till $99999$ , all the digits are degenerate (Chemistry term for equal energy orbitals, here used for equal probability) . While writing $100000$ , we will write digit $1$ extra time, so $A-B=1$

For the units digit, a 1 will appear once every 10 numbers --> 0.1 of the time.

For the tens digit, a 1 will appear 10 times every 100 numbers --> 0.1 of the time.

For the hundreds digit, a 1 will appear 100 times every 1000 numbers --> 0.1 of the time...

Number of 1s = 10^5, plus the 1 in 100000, so 100001 ones appear.

The same goes for 2s, but there is no 2 in the number 100000, so 2 appears 100000 times.

A - B = 100001 - 100000 = 1

Python 2.7:

1
2
3
4
5
6

A = 0
B = 0
for n in xrange(100001):
    A += str(n).count('1')
    B += str(n).count('2')
print "Answer:", A-B

This one is easily solvable with the following observation: for numbers 1 through 99999, the number of 1s and 2s is equal. To verify that this is true, it suffices to check that the function that takes a number between 1 and 99999, swaps every 1 in that number with a 2, and vice versa, is bijective.

We are then left with 100000, which gives us a-b=1.

Let f(n) be the number obtained by replacing the digit 1 in n with the digit 2, and vice versa. Then f is a bijection on {1,...,99999}, so A-B=1.

1 will be written the same times as 2 from 1 to 99999. $A - B = 1$ because of 100000.