7 months
9 sec=4 times
x sec=12 times
x=(9*12)/4=27 sec
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
63:45:55
$57.30
Camel is the answer
Camel can store water for many days or months in there intestines
Take one screw out from the rest of the three tyres and put the screws in the fourth tyre. Now all tyres will be held by three screws each.
A certain sum amounts to Rs. 1725 in 3 years
and amounts to Rs.1875 in 5 years
so interest of 2 years = 1875 -1725
= 150
so interest of 1 year = 75
so interest of 3 years = 75 × 3 =225 rs
so , Principal = Amount – SI
= 1725 – 225
= 1500 rs
now ,
S.I. = P × N × R /100
75 = 1500 × 1 × R /100
R = 75 / 15
R = 5%
divisible by 8 – 124
” 12 – 83
” both – 41
124 + 83 – 41 = 166
999 – 166 = 833
ans is 833…..
total sum = 205 Rs.
23
it is based on the right shift operation on the perticular digit
30.5