Forest Healthcare Interview Questions, Process, and Tips

d

1/4E(E decreses by 4 times)

9

b

17and+-12=5-48=-43

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

1) time =20mins,speed=15kmph=15*(5/18)=25/6 m/sec
distance=20*(25/6)=250/3 m
then 2) given as time =15min
speed=250/3*15=(50/9)*(18/5)kmph=20kmph

A

B

125

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