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

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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?

(1) Both of the multiples of 5 also in the sequence are multiples of either 2 or 3.
(2) Only one of the two multiples of 7 also in the sequence is not a multiple of 2 or 3.
[Reveal] Spoiler: OA

Last edited by Bunuel on 02 Jul 2013, 14:05, edited 1 time in total.
Edited the OA.

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Re: 13 consecutive integers [#permalink]

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dimitri92 wrote:
Is there a faster way to solve this !!!


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.


Straight E. Remember only positive integers can be primes, hence if all 13 numbers in the sequence are negative there will be 0 primes but if sequence is positive, it will contain primes, for instance sequence: {6, 7, ..., 18} contains 4 primes (7, 11, 13, 17).

Not a good question.
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Re: 13 consecutive integers [#permalink]

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dimitri92 wrote:
Is there a faster way to solve this !!!


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.


Bunuel has made a good point but for those who want to tets it out here is one lengthy explanation.


First point to note, 13 consecutive integers and there are 3 multiples of 6 that means first number in the sequence and the last number in the sequnce will be multiples of 6 ( for e.g 6, 7, 8, .....18 or 30, 31, 32, ... 42)

so total permissible sequnces :
0-12, 6-18, 12-24, 18-30, 24-36, 30-42, 36-48, 42-54, 48-60, 54-66, 60-72, 66-78, 72-84, 78-90, 84-96

Also in any sequence the sequence will be of form,

6x, 6x+1 , 6x +2 ...... 6x + 12.

Out of these only 6 numbers will be even and 7 numbers will be odd.

and


Now, Lets take a look at the statement.

St 1. Both of the multiples of 5 are either a multiple of 2 or 3 which means we ignore the sequnce starting from 0-12, and any sequence in which we have multiples of 5 with prime numbers like 5*7 = 35 so sequence starting from 24-36 and 30-42 has to be ignored also we need to ignore the sequences in which we have 3 multiples of 5 like 18-30, 24-36, 30-42, 48-60, 54-66, 60-72, 78-90, 84-96
So now we are left with these sequences

6-18, 12-24, 36-48, 42-54, 66-78, 72-84,

Now the first sequnce 6-18 has 4 prime numbers and 42-54 has 3 prime numbers hence insufficient.



St 2 Only one of the two multiples of 7 Also in the sequence is not a multiple of 2 or 3

Which means multiples of 7 like 0*7, 1*7, 5*7, 7*7 are the possible values

these values lie in the sequences of 0-12, 6-18, 24-36, 30-42, 42-54, 48-60

Also 0-12 is not valid in this case because it doesnt havethe second multiple of 7

So we are left with 6-18, 24-36, 30-42, 42-54, 48-60


again , Insufficient same reasoning as St1

now St 1+ 2,

We are only left with 6-18, 42-54,

Again insufficient.

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Re: 13 consecutive integers [#permalink]

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New post 19 May 2010, 19:50
nitishmahajan wrote:
dimitri92 wrote:
Is there a faster way to solve this !!!


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.


Bunuel has made a good point but for those who want to tets it out here is one lengthy explanation.


First point to note, 13 consecutive integers and there are 3 multiples of 6 that means first number in the sequence and the last number in the sequnce will be multiples of 6 ( for e.g 6, 7, 8, .....18 or 30, 31, 32, ... 42)

so total permissible sequnces :
0-12, 6-18, 12-24, 18-30, 24-36, 30-42, 36-48, 42-54, 48-60, 54-66, 60-72, 66-78, 72-84, 78-90, 84-96

Also in any sequence the sequence will be of form,

6x, 6x+1 , 6x +2 ...... 6x + 12.

Out of these only 6 numbers will be even and 7 numbers will be odd.

and


Now, Lets take a look at the statement.

St 1. Both of the multiples of 5 are either a multiple of 2 or 3 which means we ignore the sequnce starting from 0-12, and any sequence in which we have multiples of 5 with prime numbers like 5*7 = 35 so sequence starting from 24-36 and 30-42 has to be ignored also we need to ignore the sequences in which we have 3 multiples of 5 like 18-30, 24-36, 30-42, 48-60, 54-66, 60-72, 78-90, 84-96
So now we are left with these sequences

6-18, 12-24, 36-48, 42-54, 66-78, 72-84,

Now the first sequnce 6-18 has 4 prime numbers and 42-54 has 3 prime numbers hence insufficient.



St 2 Only one of the two multiples of 7 Also in the sequence is not a multiple of 2 or 3

Which means multiples of 7 like 0*7, 1*7, 5*7, 7*7 are the possible values

these values lie in the sequences of 0-12, 6-18, 24-36, 30-42, 42-54, 48-60

Also 0-12 is not valid in this case because it doesnt havethe second multiple of 7

So we are left with 6-18, 24-36, 30-42, 42-54, 48-60


again , Insufficient same reasoning as St1

now St 1+ 2,

We are only left with 6-18, 42-54,

Again insufficient.



see this is what im trying to avoid ...i want an answer which is simpler and faster ....i also arrived at the answer but such a lengthy method is just going to drag me down on gmat

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Re: 13 consecutive integers [#permalink]

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New post 13 Jun 2010, 13:49
Lets assume they are +ve.

First point to note, 13 consecutive integers and there are 3 multiples of 6 that means first number in the sequence and the last number in the sequnce will be multiples of 6 ( for e.g 6, 7, 8, .....18 or 30, 31, 32, ... 42)

Now any number less than 100 will have its sqrt < 10.
Only prime numbers less than 10 are 2,3,5,7, If the number is not divisible by any of them then it is prime.

This sequence starts and ends with multiple of 6. Since it has 7 even and 6 odd numbers => 13-7 = 6 are left to be checked for prime condition.

(A) Both of the multiples of 5 Also in the sequence are multiples of either 2 or 3.[/color]
In this case 5 is either multiple of 2 or 3 => numbers such as 25,35 wont be included=> the number of 5 will not be simultaneously multiple of 2 and 3.
Now between 6k and 6k+12 exclusive there are 3 numbers divisible by 3.
as per given condition one of them will be divisible by 5 and other by 2( Between 6k and 6k+12 there will be one number divisible by 6). So we can subtract the count of the third number and that of multiple of 5 i.e. divisible by 3. So total left = 6-2 = 4. We are left with checking of 7.
for numbers >7 it will always be multiple of any other number so prime numbers = 4-1 =3
If one of the number =7 then the number of prime numbers will remain 4...so not sufficient.

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

Again starting with the checking of 6 odd numbers.
as per the given condition one of the multiple of 7 has to be 7, 49 , 63,77 , 35 etc which is not prime.
clearly if one of them is 7 and other is any one of 35,49,63 the number of primes will differ by 1.
So not sufficient.

If we combine both we have a common case of having 7 or not.
Thus E.
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Re: 13 consecutive integers [#permalink]

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New post 13 Mar 2011, 09:41
Bunuel wrote:
dimitri92 wrote:
Is there a faster way to solve this !!!


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.


Straight E. Remember only positive integers can be primes, hence if all 13 numbers in the sequence are negative there will be 0 primes but if sequence is positive, it will contain primes, for instance sequence: {6, 7, ..., 18} contains 4 primes (7, 11, 13, 17).

Not a good question.


The official answer is C with the following explination:

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. 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 3 primes in the sequence. The correct answer is C.

I got E from sequencies 6-18 4 primes, 66-78 3 primes. I guess it's a fluke then in Princeton CAT, what does everyone think?

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Re: 13 consecutive integers [#permalink]

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New post 13 Mar 2011, 10:00
abogomolov wrote:
Bunuel wrote:
dimitri92 wrote:
Is there a faster way to solve this !!!


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.


Straight E. Remember only positive integers can be primes, hence if all 13 numbers in the sequence are negative there will be 0 primes but if sequence is positive, it will contain primes, for instance sequence: {6, 7, ..., 18} contains 4 primes (7, 11, 13, 17).

Not a good question.


The official answer is C with the following explination:

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. 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 3 primes in the sequence. The correct answer is C.

I got E from sequencies 6-18 4 primes, 66-78 3 primes. I guess it's a fluke then in Princeton CAT, what does everyone think?


Answer to this question is E, not C (no matter what the OA says). It's not a good question though so I wouldn't worry about it at all.
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Re: 13 consecutive integers [#permalink]

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abogomolov wrote:

The official answer is C with the following explination:

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. 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 3 primes in the sequence. The correct answer is C.

I got E from sequencies 6-18 4 primes, 66-78 3 primes. I guess it's a fluke then in Princeton CAT, what does everyone think?


Yes, that solution is dead wrong. They seem to be assuming that a number which is a multiple of 7 cannot be prime. Well, there is one multiple of 7 which is prime: 7 itself. So those above who have chosen E are correct, and the OA is not. Regardless, it's a horrible question, since any reasonable approach is far too time-consuming.
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Re: 13 consecutive integers [#permalink]

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No Indication about negative or positive integers. Prime number could not be negative. Ans. E
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Re: 13 consecutive integers [#permalink]

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great explanation!

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Re: 13 consecutive integers [#permalink]

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New post 14 Mar 2011, 09:47
to be honest this question is so weird to me, or i didnt understand the whole question.
the question is that there are 13 consecutive numbers : a , a+1,a+2...a+12. all these number is <100
1. Both of the multiples of 5 Also in the sequence are multiples of either 2 or 3.
it means there are 2 numbers x and y which can divide by 5, 2, and 3 so those number is only 30 and 60 and the distance between 30 and 60 are so far, this is impossible.
2 numbers x and y can divide 5 and 2 so the number could be 10 and 20 and so on. start a =10 so we can find 5 prime numbers
2 numbers x and y can divide 3 and 5 so it could be 15 and 30 and so on
start a= 15 so we find only 4 prime number
( this is crazy) insufficient.

statements 2. Only one of the two multiples of 7 Also in the sequence is not a multiple of 2 or 3.
numbers can divide 7 is 7,14, and so on,,,
the same like statement 1
hmhm
it took me 5 mins to think of this solution and yuck! this question is tough

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Re: In a sequence of 13 consecutive integers, all of which are [#permalink]

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In a sequence of 13 consecutive integers, all of which are [#permalink]

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New post 22 Sep 2017, 01:44
Yes as Bunuel pointed out, straight E, if no mention of negative or positive integers.

However, let us assume that question mentions that all are positive integers and try to solve.

It is given there are 3 exact multiples of 6 => so 1st , 7th and 13th number is multiple of 6.

Let the starting number be 'a', so sequence is a, a+1, a+2, ....... a+12


Statement 1: Both of the multiples of 5 also in the sequence are multiples of either 2 or 3.


it implies there are only two multiples of 5. so let us examine the positions of multiples of 5.
if a is multiple of 5 => (a+5), (a+10) also multiple of 5 => then 3 multiples of 3, only 2 multiples are mentioned, so a not multiple of 5.
if a + 1 is multiple of 5 => (a+6), (a+11) also multiple of 5 => then 3 multiples of 3, only 2 multiples are mentioned, so a+1 not multiple of 5.
if a + 2 is multiple of 5 => (a+7), (a+12) also multiple of 5 => then 3 multiples of 3, only 2 multiples are mentioned, so a+2 not multiple of 5.

so (a+3) or (a+4) is multiple of 5

Let a + 3 be multiple of 5,
then given a is multiple of 6, and a+3 is multiple of 5,
one such seq is : (12,13,14,15,...18....20......24), also 20 multiple of 5 and 2, 15 multiple of 5 and 3, as satisfying the statement.

Now next such sequence will occur at
12 + LCM(5,6) => (42......45...48.....50......54) 4 prime numbers
12 + 2 * LCM(5,6) => (72.....75...78....80.....84), 4 prime numbers
and stops here as, next seq (12 + 3 * LCM(5,6)) > 100

Now Let a+4 be multiple of 5,
so one such sequence is (6......10,.12..15......18) - 4 prime numbers
Next sequences
6 + LCM(5,6) => (36 ..... 40..42...45....48) - 4 prime numbers
6 + 2 * LCM(5,6) => (66 ..... 70..72...75....78) - 3 prime numbers
and stops here.

So from above, we can see that 3 or 4 prime numbers in the seq satisfying this statement - Not Suff

Statement 2: Only one of the two multiples of 7 Also in the sequence is not a multiple of 2 or 3.

One such sequence is (6,7....12....14....18) - only one multiple of 7 and 2 - 4 prime numbers
Next sequence:
6 + LCM(6,7) -> (48, 49...54....56...60) - only one multiple of 7 and 2 - 2 prime numbers
and this pattern stops here.

Another sequence: (24...28..30...35...36) - only one multiple of 7 and 2 - 2 prime numbers
Next sequence
24 + LCM(6,7) -> (66,...70,..72,....77..78) - only one multiple of 7 and 2 - 3 prime numbers

From above, statement 2 cleary insuff as there 2,3 or 4 prime numbers.

Now combining both the statements,
common sequences
(6,....12....18) - 4 prime numbers
(66,...72...78) - 3 prime numbers.

Still Insuff, So answer E

Kudos [?]: 2 [0], given: 9

In a sequence of 13 consecutive integers, all of which are   [#permalink] 22 Sep 2017, 01:44
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In a sequence of 13 consecutive integers, all of which are

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