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

E

stem:In a sequence of 13 consecutive integers, all of which are less than 100, there are exactly 3 multiples of 6.


sequence has 13 numbers and have 3 multiples of 6...means it has to start with multiple of 6 and end with it too.... hence possibilities
6,7,8,9.....12,13,14.....17,18
12,13,14,15...........24
18,19,20................30


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

Multiples of 5 which are also multiples of 2 or 3.....are 10,15,20,30,40,45,50,60,70,75,80,90

so series may include..10 and 15...series: 6-18
or 15 and 20 series: 12-24
40 and 45 series: 36-48
45 and 50 series:42-54
70 and 75 series: 66-78
75 and 80 series: 72-84

Insufficient as Stem and Stat1 will give multiple possibilities

Stat (2) Only one of the two multiples of 7 in the sequence is not also a multiple of 2 or 3.

Values that fit : 7 : series: 6-18
35 series: 24-36 or 30-42
49 series: 42-54 or 48-60
77 series: 72-84 or 66-78
91 series: 84-96

Insufficient

Combining 1 and 2, we have series 6-18 or 42-54 or 66-78 or 72-84

There are different number of primes in each group.

Hence insuff

I'll post the OA and OE once we gather more input from the posters.
Be patient. This is one of the hard math questions.

i guess D

multiples of 6 - 6, 12..96 so pairs of (6, 18), (12, 24) etc..
multiples of 5 and 2 or 3 - 10, 15, 20, 40, 45, 50, 70 and 80

so now multiples of 6 can be only 6, 18; 12,24; 36,48; 72,84 remaining pairs are all out of focus and we have more than one solution
a alone will not work

second only one multiple of 7
so it can be pairs of 7, 14; 28,35; 49,56; 70,77, and 91,98


So only one multiple of 6 stands(6,18) answer choice D

Any altenative methods..??

OA: C.

OE:

Yes. This one is a group problem in disguise.

First, there’s a lot of information buried in the question. If there are exactly three multiples of 6 in the sequence, then the sequence must start with a multiple of 6. So, there will always be 7 multiples of 2 and 5 multiples of 3 in the sequence. Those are two of the groups in the group formula. The total is, of course, 13.

We’ll also need to subtract the numbers that are both multiples of 2 and 3 (aka the multiples of 6) from the total. That means that the statements need to tell us about the multiples of 5 and 7. Everything else in the sequence (neither in the group formula) will be prime. <- Is this true? isn't 7 a prime and also multiples of 7?

Statement (1) tells us that we don’t need to count either of the multiples of 5 since they have already been counted but it tells us nothing about the multiples of 7 in the sequence. Hence, BCE.

Statement (2) tells us to count one of the multiples of 7 but tells us nothing about the multiples of 5. Cross off B.

Finally, when we put the statements together we know that 13 = 7(the multiples of 2) + 5 (the multiples of 3) + 2 (the multiples of 5) + 2 (the multiples of 7) − 3 (the multiples of 6) − 2 (the multiples of 5 and 2 or 3) − 1 (the multiples of 7 and 2 or 3) + Primes. Hence, there are 2 ( it should be 3, i think 2 is a typo in the OE) primes in the sequence.

The correct answer is C.


Free to discuss. 6-18 certainly have 4 primes, because 7 is a multiple of 7 and a prime. From this, I think E should be the correct one despite the OA.

I posted a solution to this on Beat the GMAT, and I'll just paste it here. It's a badly phrased question - I'm sure the question writer only intends positive numbers to be considered, for example, but the question doesn't rule out negatives. The answer is E, regardless, no matter what the OA.

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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.