IF N=5^5*10^10*15^15*20^20*25^25*30^30*35^35*40^40*45^45*50^

alanforde800Maximus wrote:IF N=5^5*10^10*15^15*20^20*25^25*30^30*35^35*40^40*45^45*50^50 , then how many zeroes exist between decimal and first non zero digit of N ?

150
200
250
300
350

This problem is about TRAILING 0's: the number of 0's at the end of a large product.

Since 10=2*5, EVERY COMBINATION OF 2*5 contained within the prime-factorization of N will yield a 0 at the end of the integer representation of N..

N=5� * 10¹� * 15¹� * 20²� * 25²� * 30³� * 35³� * 40�� * 45�� * 50��
N=5� * 2¹�5¹� * 3¹�5¹� * 2²�2²�5²� * 5²�5²� * 2³�3³�5³� * 2��2��2��5�� * 3��3��5�� * 2��5��5��

The prime-factorization include more 2's than 5's.
Implication:
Every 2 in the prime-factorization of N can be combined with a 5 to yield a trailing zero in the integer representation of N.
Thus, the number of trailing zeros is equal to the number of 2's contained within the prime-factorization of N.
To count the number of 2's within the prime-factorization of N, we must sum the exponents for the factors in blue:
10+20+20+30+40+40+40+50 = 250.

The correct answer is C.