Abdc Interview Questions, Process, and Tips

5c2*4c1/9c3 +5c3/9c3=25/42

these r all factors of 30
1*30=30
2*15=30
3*10=30
5*6=30
like that

500

237.6 Kmh

Ratio is 3:2:1

Total=6

Reverse of ratio is 1/3:1/2:1/1

Portion of first part is=1/3*6=>2

Portion of second part is=1/2*6=>3

Portion of third part is 1*6=>6

∴So the new ratio is 2:3:6

If division of 3 is allowed then I might do 5 divisions in one go & do in less iterations. I would assume max 2 division as you have to hands, and compare the weight.

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

1052

let (xy) be original number.
then, 10y + x +45 = 10x + y; eq 1
and x + y = 13; eq 2
solve ab equation,
then xy = 94;

d