Rosauers supermarkets inc Interview Questions, Process, and Tips

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Second

Let and A and B are playing and A has 3$ and B won 3 times.

Total money earn by A = no of games A won – no of games A loss
if A loss game means B won.
Therefore
Total money earn by A = no of games A won – no of games B won
3 = no of games A won -3
no of games A won=6
So
A won = 6 games
B won = 3 games

Hence total no of games is 9

Bopri is farthest to the west followed by Kakran, Akram, Tokhada, and Paranda to the east.

C)1440
2*2=4
4*3=12
12*4=48
48*5=240
240*6=1440

Assuming the youngest has X years, so we have
X+(X+3)+(X+6)+(X+9)+(X+12)=50
5X+30=50
5X=20
X=4
The youngest has 4 years

4

e. 56 is 7*8 which is the product of 7 with even no.

FLRIHO=cooler

(999)2-(998)2
=998001-996004
=1997

insufficient information distance between LA and NY is not
given it should be given to find the no solution

one cat kill one rat six minutes
so 1 cat kill 100 rat willbe needed 6*100=600minutes
then 100 rates kill 50minutes means then 600/50=12
so the answer is 12

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