How many three-digit odd numbers are possible if the leftmost digit cannot be zero?
Two ways:
First way:
There are 9 choices for the first digit: 1,2,3,4,5,6,7,8,9
There are 10 choices for the second digit: 0,1,2,3,4,5,6,7,8,9
There are 5 choices for the third digit: 1,3,5,7,9
That's 9×10×5 = 450
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Second way:
That's the arithmetic sequence
101,103,...,997,999
with a1 = 101, an = 999, and d=2
an = a1 + (n-1)d
999 = 101 + (n-1)2
Solve for n, get n = 450.
Edwin