3000(4x/100) + 5000(2x/100)=10
To determine how many consecutive zeros the product of S will end with, we need to find the highest power of 10 that divides the product. This is equivalent to finding the highest power of 5 that divides the product, since the number of factors of 2 will always be greater than the number of factors of 5.
The primes in S are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}.
There are 24 primes in S, so the product of S is:
2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 x 43 x 47 x 53 x 59 x 61 x 67 x 71 x 73 x 79 x 83 x 89 x 97
We need to find the highest power of 5 that divides this product. To do this, we count the number of factors of 5 in the prime factorization of each number in S.
5 appears once: 5
5 appears once: 25
5 appears once: 35
5 appears once: 55
5 appears once: 65
5 appears once: 85
So, there are six factors of 5 in the product of S. However, we also need to consider the powers of 5 that arise from the factors 25, 35, 55, and 65.
25 = 5 x 5 appears once: 25
35 = 5 x 7 appears once: 35
55 = 5 x 11 appears once: 55
65 = 5 x 13 appears once: 65
Each of these numbers contributes an additional factor of 5 to the product of S. Therefore, there are 6 + 4 = 10 factors of 5 in the product of S.
Since each factor of 5 corresponds to a factor of 10, we know that the product of S will end with 10 zeros. Therefore, the product of S will end with 10 consecutive zeros
4/52 * 3/51
=12.05×5.4/0.6
=12.05×9.0
=108.45
No loss no gain.
I belive it is 3:45 hrmins
75%
ans: 40
6/3*12/3*15/3=40
R
T,P,R,Q,S
NTFS — New Technology File System
Fat — file allocation table
NTFS having a quota, commpress system with securites base as
administrator giving, multipal user, groups, to set permission
Statements :
All mangoes are golden in colour. No golden coloured things are cheap.
Conclusions :
I. All mangoes are cheap.
II. Golden coloured mangoes are not cheap.
As fast as possible. For example, I will suitable with the working environment within one month.
Sum of all numbers in series= 870 (I.e 312+162+132+142+122)
Average=sum of numbers/count of numbers
Average= 870/5
= 174