1. 1g
2. 3g with 1g counter
3. 3g
4. 3g plus 1g
5. 3g plus 1 g plus 1g weighed medicine
6. 9g with 3g counter
7. 9g with 1g of counter and 1g weighed medicine
8. 9g with 1g counter
9. 9g
10. 9g plus 1g
11. 9g plus 3g with 1g on counter
12. 9g plus 3g
13. All 3 weights on one side.
s=36kmph
in meter per second is=36*5/18=>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
38 years
Let Rajan’s present age be x years. Then, his age at the time of marriage = (x – 8) years.
x = 65(x−8)
⇒5x=6x−48
⇒x=48 years
Rajan’s sister’s age at the time of his marriage = (x – 8) – 10 = (x – 18) = 30 years
∴ Rajan’s sister’s present age = (30 + 8) years = 38 years
Total=60reems
Utilized=40reems
Un utilized= 60-40=20
Percentage of remainder 20/60=.3333
0.33333*100=33.333%
665
BROTHER
A perfect square is a square number of a digit. eg 64 is a perfect number, a square of 8
Now digits AB9 is a square number of a number.
AB9 Can also be written as A multiply by B multiply by 9
Get the Square root of AB9
Assumption, A=1, B=1
1*1=1=A, 1*1=1=B Therefore,
Square root of A = A, B = B and 9=3
Therefore
An odd number is a number indivisible by 2.
for example 1,3,5,7…….
Therefore Squares A*B*9= AB9
Where a=1, b=1, 3 as digits.
Conclusion
A=1 is an odd number
198.20
twenty members
1
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