L*I/3(I-H)
answer is 28
>2+8=10
>reverse of 28 is 82
>subtract 28 from 82
>82-28=54
the number is 28
The work done by A in 8 days is = 8/ 12 = 2/3
Means A alone completes 2/3 part of work.
Remaining work which is (1–2/3) = 1/3 is completed by B in 8–2 = 6 days
So the complete work done by B in 6/(1/3) = 18 days.
B alone can complete the work in 18 days.
C
Train speed 90 km /h
1st we have to change in m/s. then 90*5/18=25m/s
Then d=speed*time
=25*10=250m train length
1600 years contain 0 odd day.
300 years contain 1 odd day.
94 years = (23 leap years + 71 ordinary years)
= (46 + 71) odd days
= 117 odd days, i.e., 5 odd days
Days from 1st January 1995 to 28th February 1995
= (31 + 28) days = 59 days
= (8 weeks + 3 days) = 3 odd days
∴ Total number of odd days
= (0 + 1 + 5 + 3) = 9 odd days i.e., 2 odd days.
So, the required day is Tuesday.
3 minutes
270/(27*1000)*3600=36
I am Srilaxmi. I born and bought-up in WARANGAL. My father Agriculture cum politician and my Mom homemaker. I blessed with three brothers and one sister. My elder brother was married and working as a TA in MPDO office at Warangal. My first younger brother was also married and working with Sushee Infra Pvt Ltd as a Asset Manager at Hyderabad. Younger brother and sister are into their studies. Coming to me I completed my MBA from ICFAI university in 2008 and worked as a Audit assistant in accounting firm for two years. I had actively participated in bank audits i.e. is in CBI (Concurrent audit) and SBI(Statutory audit) during my services. Later I pursued CS Executive and attempted for five times then I give up to clear my M.Com. At present I was studying postal studies of CS Executive programme. My hobbies are reading books and listening to music.
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
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