UNO High School Problem of the Week Competition
Problem 3
What is the product of all the positive odd integers less than 100,000?
Solution
So we have (1x3x5x7x...x99,999)
= (1x2x3x4x5x...x100,000)/(2x4x6x...x100,000)
We can rewrite this as
= 100,000!/((1x2)x(2x2)x...x(2x50,000)
By multiplying all the 2's together in the denominator we get
= 100,000!/(250,000x(1x2x3x...x50,000))
= 100,000!/(250,000x50,000!)
