Interesting PS

ani781 wrote: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

Can someone help me how to solve this question?
OA is C

Hmmm, are you sure those are the answer choices?
I believe the answer should be [spoiler]250 not 10^250[/spoiler]

Cheers,
Brent

ani781 wrote: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

Wow, there has been a disproportionate number of "trailing zero" questions on the forums lately.

To determine the number of trailing zeros of an integer, N, we must find the number of 10's "hiding" in the prime factorization of N.

Some examples
800 = (2)(5)(2)(5)(2)(2)(2) = (10)(10)(2)(2)(2). There are two 10's hiding in the prime factorization of 800, and 800 has two trailing zeros.

15000 = (2)(5)(2)(5)(2)(5)(3)(5) = (10)(10)(10)(3)(5). There are three 10's hiding in the prime factorization of 15000, and 15000 has three trailing zeros.

So, to find trailing zeros, we need only focus on the number of 2's and 5's in the prime factorization.

What about (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)………. *(50^50)?
Well, we might automatically recognize that this product will have more 5's than 2's, so we need only focus on the 2's since we have more 5's than can possibly pair with the 2's. That said, let's find the number of twos AND fives in the prime factorization.

1^1 has 0 twos and 0 fives
5^5 has 0 twos and 5 fives
10^10 = (2^1 x 5^1)^10 = 2^10 x 5^10: 10 twos and 10 fives
15^15 = (3^1 x 5^1)^15 = 3^15 x 5^15: 0 twos and 15 fives
20^20 = (2^2 x 5^1)^20 = 2^40 x 5^20: 40 twos and 20 fives
25^25 = (5^2)^25 = 5^50: 0 twos and 50 fives
30^30 = (2^1 x 3^1 x 5^1)^30 = 2^30 x 3^30 x 5^30: 30 twos and 30 fives
35^35 = (5^1 x 7^1)^35 = 5^35 x 7^35: 0 twos and 35 fives
40^40 = (2^3 x 5^1)^40 = 2^120 x 5^40: 120 twos and 40 fives
45^45 = (3^2 x 5^1)^45 = 3^90 x 5^45: 0 twos and 45 fives
50^50 = (2^1 x 5^2)^50 = 2^50 x 5^100: 50 twos and 100 fives

TOTAL: 250 twos and 350 fives
So, we can create a total of 250 pairs of twos and fives (with 100 excess fives).
In other words, there are 250 tens hiding in the prime factorization of our original number.
So, the product has 250 trailing zeros.

Cheers,
Brent