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Series problem!!!

Expert replies
by akshatgupta87 » Wed Apr 27, 2011 11:29 pm
Q.)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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Source: — Data Sufficiency |

by force5 » Thu Apr 28, 2011 12:55 am
Long one. IMO -E.

check the way Ian solved this-


I assume the question intends for the 13 consecutive integers to all be positive- otherwise the answer is clearly E. Even if we assume this, as written, the answer is E anyway; using both statements, the sequence could be 6, 7, ..., 18, which contains four primes (7, 11, 13, 17), or it could be 42, 43, ..., 54, which contains three primes (43, 47, 53).

I'd like to add one further restriction to the question, that the integers are all larger than 10, which makes the question more interesting. This rules out the example of 6, 7, ..., 17, 18 used above, and guarantees that if a number in our list is divisible by 7, it isn't prime. We'll use that later. Assuming this:

First, if a number is greater than 1 and less than 100, it is either prime, or it is divisible by at least one of 2, 3, 5 or 7 (because every positive integer x larger than 1 that is not prime has a prime factor less than or equal to sqrt(x)).

We can write our sequence as follows: 6k, 6k+1, 6k+2, ..., 6k + 11, 6k + 12. Seven of these numbers must be even, and 6k+3 and 6k+9 are both divisible by 3. The only possible primes are:

6k+1
6k+5
6k+7
6k+11

From 1, we know that none of these four numbers is a multiple of 5; all the multiples of 5 in the list are divisible by 2 or 3, so we have already ruled them out. Still, one of them might be divisible by 7.

From 2, we know that exactly one of these numbers is a multiple of 7, and therefore one of the four numbers in the list above is certainly not prime (this is where I'm using the assumption that all the integers are greater than 10). We don't know whether any are multiples of 5, however.

Since from 1 and 2 together we know that exactly three of the numbers in the list above are not divisible by 2, 3, 5 or 7, the statements together must be sufficient. There are three primes. C.

One can still find examples to demonstrate that neither statement is sufficient on its own; for 1), the set could be 12, ..., 24, which contains four primes, or 42, ..., 54, which contains three primes. For 2), the set could be 48, ... 60 which contains only two primes, or 42, ..., 54 which contains three primes.
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