Question

Find the number of trailing zeros in the product of

(1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)………. *(50^50).

A) 10^150
B) 10^200
C) 10^250
D) 10^245
E) 10^225

Answer 1

• 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

`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)`

which means there are 250 trailing zeros. Probably C is the correct answer, though
`10^(250)\\eq250`; it's probably supposed to say 250.