If m is set of integers from 500<m<900

Number of terms = \(\frac{Last Term-First Term}{Increment}+1\)

First Term must be 504
Last term is 896 (we know 900 is divisible by 4; subtract 4 from 900)

Number of terms
\((\frac{896-504}{4}+1)=(\frac{392}{4}+1)=(98+1)=99\)

Some of those 99 multiples of 4 are also multiples of 6. Not allowed. To exclude terms that have 6 as a factor in the set of multiples of 4,

1) use LCM. LCM of 4 and 6 is 12. \(\frac{12}{4}=3\)
In a string of multiples of 4, every third term must be omitted because it is also a multiple of 6.
If \(\frac{1}{3}\) of terms are removed, then --
\(\frac{2}{3}\) of terms will remain.
\((\frac{2}{3}*99)=66\)

\(m\) has \(66\) terms.

OR
2) Find the pattern: list the first few terms
504, 508, 512, 516, 520, 524

Which terms are divisible by 6?
504 and 516 (divisible by 2 and 3)

Six multiples of 4 are listed. Two of the six terms ARE divisible by 6:
\(\frac{2}{6}=\frac{1}{3}\) of the terms that are divisible by 4 are also divisible by 6 and must be excluded.

If \(\frac{1}{3}\) of terms are omitted, then \(\frac{2}{3}\) are kept.
That pattern will hold.
Number of terms: \((99* \frac{2}{3}=66\) terms

Answer D

EDIT: Smaller numbers are easier to test. The pattern that emerges holds whether we use 4, 8, 12, etc or 504, 508, 512. List some multiples of 4 until you see a pattern:
4, 8, 12,
16, 20, 24,
28, 32, 36
In a string of multiples of 4, 1 out of 3 terms is impermissibly also divisible by 6 and must be excluded.

If \(\frac{1}{3}\) of terms are excluded, then \(\frac{2}{3}\) of terms remain. \((99*\frac{2}{3})=66\)

hello generis

i will be rephrasing and/or commenting your explanation , that way i may understand the solution

To exclude one of two integers as a factor. Got it. Very logical, to solve a major problem, one should get rid of second problem

1) use LCM. LCM of 4 and 6 is 12 got it in other words we need to find a common language between two numbers

(4 * 3)=12 (ok, here you multiply 4 by 3 to get 12 ) why 4*3 and not 6*2, or 12*1 4*3 is like main spoken lamguage, and 6*2, or 12*1 are dialects of 4*3 ?

Every third term must be omitted. (what do you mean every third term must be omitted? Why every third ? how did you see that ? if 4*3 = 12 and if every third term must be ommited to be divisible only by 4 then following this pattern i will have 4+4 =8, 4+4+4 = 12 = OMITTED (but its not third term) why ? , 4+4+4+4 = 16 , 4+4+4+4+4= 20, 4+4+4+4+4+4 = 24 (OMITTED), 4*7 = 28, 4*8= 32 4*9 = 36(OMITTED) 4*10 = 40, 4*11 =44, 4*12 = 48 (OMITTED) hmm looks like now i see a pattern

\(\frac{2}{3}\) of terms will remain. whats is the logic behind fraction without reading your second part of your explanation after '"OR" what does 2 as numerator mean and 3 as denominator how do you deduce that 2/3 is a number of terms divisible only by 4


and last question why cant i understand the solution ? perhaps because its evening here, and i am like flashlight on solar battery ...

thank you

UPDATE: I GOT IT generis YAY

dave13 Excellent! +1

And your timing could not be worse. MOMENTS ago I finished writing the reply.

In case others are confused about issues that you raised, I am posting the reply.

[The sidebar self-talk is too funny: "its evening here, and i am like a flashlight on solar battery"
-- "hmm looks like now i see a pattern."
It's like watching a Sherlock Holmes movie while Robert Downey Jr talks to himself.]

Whoops. You forgot to "count" 4 itself.I'll back up. The prompt says that \(m\)
-- is a set of numbers between 500 and 900
-- m's terms MUST be multiples of 4 but CANNOT be multiples of 6
-- problem: OVERLAP. SOME multiples of 4 will also be multiples of 6

Solution?
1) find all the multiples of 4, then
2) remove, from that set, terms (multiples of 4) that are also multiples of 6

Restated: If a term is divisible by 4 AND 6, it gets removed from my set.

HOW? Below is an approach that is more intuitive than LCM

1) list a certain number of multiples of 4

2) check each multiple of 4: is it divisible by 6? Omit it

BIG TIP (a method that I use often and one that you employed most of the way):

Find the pattern with smaller, easier numbers.

Let's say we want all the multiples of 4 that are between 1 and 50, but they cannot be multiples of 6

=> we will OMIT multiples of 4 that are also multiples of 6

Start by listing 12 of those multiples of 4:

4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48

Which ones are also multiples of 6?

4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48

What is the pattern?
{4, 8, 12}
{16, 20, 24}
{28, 32, 36}
{40, 44, 48}

In a string of three multiples of 4, every third multiple (every third "term") is also a multiple of 6

1 term omitted out of 3 terms listed = \(\frac{1}{3}\) of the 3 terms are also multiples of 6 [to omit]

So - however many multiples of 4 we find in this problem, \(\frac{1}{3}\) of them must be omitted.

"I omit one out of the three terms" = "every third term gets omitted."

OR (more efficient): KEEP \(\frac{2}{3}\) of however many multiples of 4 that we find

\(\frac{2}{3}\) or "2 out of 3," we will keep numbers NOT in bold type

{4, 8, 12}
-- keep 4 and 8, which aren't multiples of 6
-- we kept 2 of 3 terms
-- Omit 12. It's a multiple of both 4 and 6
-- we omitted 1 of the 3 terms

{16, 20, 24}
-- keep 16 and 20. (We kept 2 of 3 terms)
-- Omit 24. It's a multiple of both 4 and 6. (We omitted 1 of the 3 terms)

I therefore keep 2 out of 3 of the terms in the set that I found.

We found 99 multiples of 4.
Keep \(\frac{2}{3}\) of them
\(99*\frac{2}{3}=66\)

\(m = 66\)

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.