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In a sequence of 13 consecutive integers, all of which are

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Director
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New post 02 Jul 2007, 03:01
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In a sequence of 13 consecutive integers, all of which are less than 100, there are exactly 3 multiples of 6. How many integers in the sequence are prime?

(1) Both of the multiples of 5 in the sequence are also multiples of either 2 or 3.

(2) Only one of the two multiples of 7 in the sequence is not also a multiple of 2 or 3.
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New post 02 Jul 2007, 06:46
C. I think only ste of #s [6,7,8.......18] satisfies given constraints (1+2).
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New post 02 Jul 2007, 06:55
how could be in a interval of 13 consecutive numbers 2 multiples of 5,3 and 2? there maybe something wrong with this question , no?
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New post 02 Jul 2007, 07:24
andrehaui wrote:
how could be in a interval of 13 consecutive numbers 2 multiples of 5,3 and 2? there maybe something wrong with this question , no?


Question is correct. Read it thoroughly.
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New post 02 Jul 2007, 07:40
Answer E is correct.

1) does not give us any new information

2) means that one of the following integers is part of the sequence: 7*5, 7*7, 7*11 or 7*13. If we take 7*5=35, there are two possible sequences in which 35 is included:
24 to 36
OR
30 to 42

In 24-36 there are two prime numbers (29, 31)
In 30-42 there are three prime numbers (31, 37, 41)

So 2) is insufficient.
  [#permalink] 02 Jul 2007, 07:40
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