DS - multiple

Question

sensible way?

hobbit · Answer

answer seem to be E

consider thw two sequences:
6...18 (has 7,11,13,17 inside)
42...54 (has 43,47,53 inside)
both sequences have 13 consecutive integers ith 3 multiples of 6.

both statements satisfy 1 and 2:
6..18 has 10,15 both are divisible by either 2 or 3. and has only 14 (but not 7) that is divisible by 2 or 3.
42...54 has 45 and 50 both are divisible by 2 or 3. and only 42, but not 49, is divisible by 2 or 3.

these are examples that show the insufficiency of both statements together.

to find these examples i realized the following:
- from the 3 multiples of 6 i inferred that every sequence must begin with a multiple of 6.
- 6..18 was my first trial, and i found it satisfies both 1 and 2 and has 4 primes.
- i searched for another example: my candidate was something that contains 49 as it is an obvoius multiple of 7 that is not divisible by 2 or 3.
- i checked 42...54 and it was what i needed (note that 48..60 clearly doesn't satisfy st1 as it has 3 multiple of 5 inside)

mbagal1 · Answer

I plugged in numbers and I seem to get E also. There are multiple possibilities even when 1 and 2 are combined.

anindyat · Answer

Toughy...
Going for E too ....agree with hobbit

Tried to approach it using n, n +1 ....n +12
Getting lost somewhere

n_sathya · Answer

Problem statement is: 13 consecutive integers, less than 100, exactly 3 multiples of 6.
that means, sequence starts with multiple of 6 for exmple
6, 7, 8, 9 ,10,11,12,13,14,15,16,17,18 you can form ALL the series in this fashion such as
12,13.....24
18,19....30 etc..

therefore, we need piece of information to determine which series is actually under discussion.

1. says both multiples of 5 are multiples of either 2 or 3 - not enough 
2. actually gives TWO piece of information:
 a. there are ONLY TWO multiples of 7 in the sequence (though its implied in problem description itself)
 b. ONLY ONE of the two multiples arent multiple of 2 or 3

Taking 2 alone results in two possible series:
6,7... 18 (7 is multiple of 1) or 30,31,32...36 (35 is multple of 5)

Using (1) on this piece of information, the series should also have multples of 5 with 2 or 3, only one series remains:
6,7,8,9,10,11,12,13..18 - using which we can determine number of primes

SO THE ANSWER IS C

gk3.14 · Answer

OA is C..


Here is the OE

 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. Weâ€™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 donâ€™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 primes in the sequence. The correct answer is C.

BLISSFUL · Answer

when we have have 2 samples 6-18( 4 prime nos: 7,11,13,17)
and 42-54 (3 prime nos: 43,47,53)

that satisfy both 1) and 2) and the stem , how can the answer be C?

It must be E. and  offcial answer must be wrong  !

hobbit · Answer

OE is a crap (and wrong)... who wrote it anyway...
it says that the number of primes must be 2.... but we have an example that shows 4 and example that show 3....

well... before throwing OE lets check ourselves again:
6...18 has two multiple of 5 inside (10,15). 10 is divisible by 2 and 15 is divisible by 3. so clearly statement 1 holds.... because both multiple of 5 in the sequence divisible by either 2 or 3.
we have 2 multiples of 7 (7 & 14). exactly one of them is divisible by either 2 or 3 (thats 14). so statement 2 holds....
and we certainly have 4 primes inside (7,11,13,17)

ok... OE is wrong, but what is wrong with it??????

we have indeed 7 multiples of 2 (6,8,10,12,14,16,18) and 5 multiple of 3 (6.9.12.15.18) of which are 3 multiple of 6 (6,12,18) we have 2 multiple of 5 (10,15) which already counted. and two multiples of 7 (7,14) which we already counted 1.
7+5-3 + (2-2) +(2-1) = 10 ... so according to OE there should all the other 3 (out of initial 13) should be prime (note the typo in the message above). but we have 4.... what did we miss?
oh yes....
the multiple of 7 which we counted is itelf a prime !!!!!!
so OE is right always EXCEPT when the "counted" multiple of 7 is indeed a prime. this happens exactly twice: for sequence 0...12 and for sequence 6..18

so OE is wrong. OA is wrong. and now we can understand why.

hobbit · Answer

also...
OE also fails to mention that some of the sequences may contain 3 multiple of 5 (like 18..30)


however, even if OE was correct... it does you bad service in listening to it. doing "counting tricks" like these, excluding things and including others... is complicated. they are so error prone (look what it did to the "officials"). 

!!!never use a technique that can make you do mistakes !!!!
there is always another way around. i promise.

this time it was simple checking. there are total 15 sequences to check (it is given that the integers are less than 100). go check them all. 
simple, clean, no mistakes (unless of course you don't know what are prime numbers or how to check if a number is divisible by 2,3,5,6,7... which is fairly easy.

in no time (in fact it may take some time... but time spent on simple tasks) you can find a counter example - or no that there is no such example.

don't think about going the OE way... 
my assumption is always: if the explanation is complicated -either it is not a real gmat question or the explanation is not good (not necessarily wrong... but not good. there should be a simpler way to do the question).

HongHu · Answer

Very impressive hobbit. Thanks for the explanations!

jainan24 · Answer

This question must be from McGrawhill GMAT (at least the font looks liek their software), the book is fraught with horrendous errors in maths and verbal is easier than typical GMAT

prude_sb · Answer

The set of 13 numbers has to start with a multiple of 6 

6 ....12....18....24.....30....36..............96

we can keep taking sets of 3 that satisfy conditions specified in 1 & or 2

just 1 or just 2 doesnt seem sufficient 

with 2 we see that 49 (c1) or 35 (c2) or 7 (c3) should be included

now with 1 we see that c2 is ruled out since both multiples of 5 are multiples of either 2 or 3

with 49 we can have 

42...45.. 48...50... 54 (other case ruled out since 55 will be included) 
primes: 43, 47,53 

with 7 we can have 
6 7 ...10...12....15...18....
primes 7,11,13,17


so E ?

DS - multiple

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