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 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 also not a multiple of 2 or 3?
D
agree with manisha
I felt “E” is teh answer.
Given :- there are three multiples of 6 which can be anyone out of
{6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96}
Considering A, multiples of 5 which is also a multiple of 2 or 3 can be one from (10,15,30,40,45,50,60,65,70,75,80,90}
For example there is no coincide between the two sets we can have 3 + 2 = 5 non primes, that make our answer as 13-5 = 8 primes, but if there is a coincide, we can have 3 primes and 13-3 = 10 primes, so A cannot answer. Same analysis can be done with B. Please prove me wrong.
Can manisha or Patel elucidate us on how would the answer be “D”.
According to me it should be “E”.I do agree with Rajib
Ans D
Answer is either C or E.
I saw this question in a princeton exam i gave and according to them you need to take 13 numbers in which one of the multiples of 7 is not also a multiple of 2 and 3.
For example
6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Since you need 3 multiples of 6 therefore the options are sequences starting with a multiple of 6.
Since there are only 2 multiples of 5 and both the multiples of 5 need to be multiples of 2 and 3 as well you cannot have a sequence containing 25,35
Also you need to choose a sequence which has only 2 multiples of 7 and in which one of the multiples is NOT also a multiple of 2 and 3.
Therefore need to choose a sequence which contains 7,35 or 49.
( You cannot choose the equation with 35 because it is a multiple of 5 and is not also a multiple of 2 and 3 according to statement 2 above)
In the sequence above if you ignore the multiples of 2 and 3 and 5.
You can prime numbers = 7,11,13,17 = 4
Now consider the sequence :
42,43,44,45,46,47,48,49,50,51,52,53,54
sequence above satisfies condition 1 and 2
prime nos = 43,47,53 = 3
Therefore there are 2 sequences which satisfy condition 1 and 2 but have different prime nos therefore the answer is E?
I agree……”D” was wrong…….thr could be 4 or 3 Prime Nos. as per the statements given…Answer should be “E” only!!
But as per Rajib and Vijay, i could not find any sequence of 13 nos. which can give me 8 or 10 primes….Can you please explain with an example of the sequence which can give us 8-10 primes…..
Hi Guys, this is pavan, I’ve joined the community a few hours ago.
This is my first comment and I would like go with pranaysad. There will be two sequences satisfying the conditions stated above…
I was able to view the question but not able to find the options for the answers. Some one pls let me knw whr to find the options…
Since I’m not able to view teh options – I cant give my option but I’ll agree with pranaysad. we can get 3 or 4 primes…
My sequences are {6-18}, {42-54} and {48-60}
where in {6-18} -> 4 primes,
{42-54} -> 3 primes and {48-60} -> 3 primes
answer is B
the only consecutive sequence of 13 consecutive numbers must definitely start with a multiple of 6 and end with a multiple of 6.
Stmt 1:
many possiblities are possible – and no unique sequence – Hence not sufficient.
Stmt 2 :
It states tat 7 is in the sequence. thus making the sequence – 6 7 8 9 10 ….. 18.
And the number of prime nos can be found out from this unique sequence. Hence sufficient to answer the prob.
Hence B is the answer
E
I am sorry my earlier answer is wrong…
answr is E – no unique sequence available even for stmt 2 and also combining them…
what i understand is
statement A: both multiples of 5 are either multiples of 2 or 3 i.e there are two multiples of 5 and they are either multiple of 2 or 3 (or both)
options are 10–15, or 15,20(not considering 30 because neither 35 nor 25 is a multiple of 2 or 3), 40–45 or 45–50, 70–75 or 75–80
Clearly there are multiple possibilities of sequences..so A is not sufficient
Statement B: Only one of the two multiples of 7 in the sequence is also not a multiple of 2 or 3?
options are 7–14 , 28–35, 35–42, 42–49,49–56,70–77 and so forth(just need to keep the 3,6,9 X 7 out)
combine the two
you can have sequences such as 7 to 19 or 6 to 18 and in both cases we get a different no. of prime numbers.
so answer is E
Thirteen consecutive integers
3 are multiples of 6,let us start from 6,7,8,9,10,11,12,13,14,15,16,17,18
In this sequence 6,12,18 are multiples of 6.
Condition 1.
Multiples of 5 in this sequence are 10,15…10 is multiple of 2 and 15 is multiple of 3.
Condition 2.
one of the multiples of 7 in the sequence is also multiple o0f 2 or 3
7 and 14 are multiples of 7, in that 14 is multiple of 2.
From the above sequence 7,11,13,17 are the four prime numbers.
answer is 4.
CORRECT ME IF I AM WRONG.
E..
2 sets are possible, even on combining the two stmts:
6-18, 72-84.
approach:
series starts and ends with M(6). stmt 2 suggests series has 1 M(7) from 14, 21, 28, 42, 56, 63, 70, 84, 98. Other M(7) is from 7, 35, 49, 77, 91. this gives 4 possible series: 6-18, 30-42, 48-60, 72-84.
By stmt 1, 30-42 and 48-84 are out.. Hence the two sets, as already stated, are possible. One has 3 and other has 4 prime nos.
3 Set are possible, after combining the two stmts.
{6-18}, {42-54}, {66-78}.
{6-18}
6, 12, 18 are 3 mult of 6
10, 15 stmt #1
7, 14 stmt #2
{42-54}
42, 48, 54 are 3 mult of 6
45, 50 stmt #1
42, 49 stmt #2
{66-78}
66, 72, 78 are 3 mult of 6
70, 75 stmt #1
70, 77 stmt #2
So 3 or 4 prime numbers
E
No unique sequence even after combining the stmts
We get the sequences -: 6-18, 42-54 and 72-84 : all three satisfying both the stmts…
Thus E
As per me the answer should be B.
Take GMAT team, your inputs are very much required.
this was tough… m sure, i would have guessed it during GMAT!
The Answer is “C”
if we take sequence of 13 consecutive integers, all of which are less than 100 which are multiples of 6 then we have the following options
Numbers Stmt1 Multiple of 5 Stmt2 (multiple of 7)
a) 06 12 18 10 15 7 14
b) 12 18 24 15 20 14 21
c) 18 24 30 20 25 30
d) 24 30 36 25 30 35
e) 30 36 42 30 35 40
f) 36 42 48 40 45 42
g) 42 48 54 45 50 42 49
h) 48 54 60 50 55 60
i) 54 60 66 55 60 65
j) 60 66 72 60 65 70
k) 66 72 78 65 70 75
l) 72 78 84 75 80 77 84
m) 78 84 90 80 85 90
n) 84 90 96 85 90 95
Based on Statement 1 we can eliminate the following
c,d,e,h,i,j,k,m,n
This is beacuse it says Both the multiples of 5, in the choices eliminated there are 3 multiples of 5
Based on Statement 2 we can eliminate the following
f – because there is only 1 multiple of 7
We are now left with the following choices
Numbers Stmt1 Multiple of 5 Stmt2 (multiple of 7)
a) 06 12 18 10 15 7 14
b) 12 18 24 15 20 14 21
g) 42 48 54 45 50 42 49
l) 72 78 84 75 80 77 84
B can be eliminated as
G can be eliminated as 42 is a multiple of Both 2 and 3
L for the same reason as 84 is a multiple of Both 2 and 3
as both 14 and 21 are a multiple of 2 or 3
We are Left with only choice A
i.e. 6 7 8 9 10 11 12 13 14 15 16 17 18
There are 4 Prime numbers
7, 11, 13, 17
Hence C is the Answer.
the spacing got a little mixed up.. above
Let me know if there is any confusion.
Hi Vivek Pant ,
Option 2 says that one 7′s multiples doesn’t have 2 or 3 as a factor ..
It doesn’t say about the other 7′s multiple .ie, It doesn’t matter whether 2/3 or both are factors of 2nd 7′s muliple :
2) Only one of the two multiples of 7 in the sequence is also not a multiple of 2 or 3
Therefore , You can’t rule out l
l) 72 78 84 75 80 77 84 which as 3 prime #s 73 ,79 83
hence you have 2 options (even after using both 1 & 2 ) :
- 3 primes
- 4 primes ..
Hence E should be the ans ..