gmatmachoman wrote:In a sequence of 13 consecutive integers, all of Which are less than 100, there are exactly three multiples of 6. How many integers in the sequence are prime?

(A) Both of the multiples of 5 Also in the sequence are multiples of either 2 or 3.
(B) Only one of the two multiples of 7 Also in the sequence is not a multiple of 2 or 3.

First number is a multiple of 6 and so is the last number

1.)10,15

15,20

40,45

70,75

75, 80

Are the possibilities
the num of Prime numbers 4, 4, 4, 3, 3 for each of the possibilities

2) 14,21
56,63
63,70

Prime numbers 4, 2, 3 for each of the possibilities

Combining

The starting num 12 end is 24 and we have 4 prime numbers between them

ajith wrote: First number is a multiple of 6 and so is the last number

how u conclude this

Hi,
From the statement we get first number is multiple of 6 so as the last,total 16 possible sequence are there satisfying this criteria.
Now from point 1 we get following pairs should be present in the sequence
10,15
15,20
40,45
45,50
70,75
75,80
For these pairs we cannot tell number of primes for sure,they are either 4 or 3

From 2nd point we get following pair should be present in the sequence
28,35
35,42
42,49
49,56
70,77
77,84
84,91
here also number of primes are 2,3 or 4
combining these two points we get only 2 sequences
42 43 44 45 46 47 48 49 50 51 52 53 54
72 73 74 75 76 77 78 79 80 81 82 83 84

here number of prime is 3 in either case.
hence combining we can give the answer.

multiple of 6 in consecutive integers will occur at an interval of 6 .Total number of integers is 13. so numbers will be 6n,6n-1 and 6n+1 providing difference between two extremes as 12.
Hopes this explains the reason for having first and last as multiple of 6

Cannot find any obvious flaw with the series

6 7 8 9 10 11 12 13 14 15 16 17 18

Since it is a DS Q, it must be imp to just estb the correct series.