Cambridge Logic Interview Questions, Process, and Tips

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

237.6 Kmh

answer 10

6

RABBIT

( b ) P

Given:
Speed of man downstream= 30/3=10km/h
Speed of man in upstream=18/3=6km/h
Speed of man in still water=1/2 (speed of upstream+ speed of down stream)
= 1/2 (10+6)
=1/2*16
=8km/h

3606

Had is correct ! Anil went in right direction but ended up
with wrong answers
Soln: B lied ( see anil’s comment ) which implies B has not
stolen mule. So B could steal camel or sheep.
1) Lets assume B stole camel.
In that case, A lies ( refer – A says “B had stolen sheep”
), which implies A had stolen sheep ( becoz a lier cant
steal mule ). So A->sheep B->camel C->mule. But this cant
happen because C cant lie ( refer – C says ” B had stolen
mule” ).
2) Lets assume B stole sheep.
In that case, A is true ( refer – A says “B had stolen
sheep” ), which implies A had stolen mule. So A->Mule
B->Sheep C->camel. Here C lies ( refer – C says ” B had
stolen mule” )
SO ANS: answer A- mule, B-sheep, C camel