- 5^5 = 5^5
- 10^10 = 2^10 * 5^10
- 15^15 = 3^15 * 5^15
- 20^20 = 2^40 * 5^20
- 25^25 = 5^50
- 30^30 = 2^30 * 3^30 * 5^30
- 35^35 = 5^35 * 7^35
- 40^40 = 2^120 * 5^40
- 45^45 = 3^90 * 5^45
- 50^50 = 2^50 * 5^100
Trailing zeros are obtained from powers of 10, or for every pair of 2 and 5 we can take from the factorizations above. Our product contains
[tex]2^{10+40+30+120+50}5^{5+10+15+20+50+30+35+40+45+100}=2^{250}5^{350}=10^{250}5^{100}[/tex]
which means there are 250 trailing zeros. Probably C is the correct answer, though [tex]10^{250}\neq250[/tex]; it's probably supposed to say 250.
