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

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Senior Manager
Joined: 18 Jul 2018
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Location: India
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If m is set of integers from 500<m<900  [#permalink]

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16 Sep 2018, 19:29
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If m is set of integers from 500<m<900. How many elements will m have, If m is multiple of 4 but not a multiple of 6.

1) 33
2) 34
3) 48
4) 66
5) 67

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Joined: 14 Jun 2018
Posts: 120
Re: If m is set of integers from 500<m<900  [#permalink]

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20 Sep 2018, 07:14
First Mutiple of 4 between 500 and 900 = 504 and last one is 896. Total multiple of 4 = (896-504)/4 + 1 = 99
To remove the multiple of 6 , we take the lcm of 4,6 which is 12.
First multiple of 12 is 504 and last is 888. Total multiple of 12's (888-504)/12 + 1 = 33
Total multiple of 4 but not 6 = 99-33 = 66
##### General Discussion
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Joined: 02 Aug 2009
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If m is set of integers from 500<m<900  [#permalink]

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16 Sep 2018, 20:58
3
1
Afc0892 wrote:
If m is set of integers from 500<m<900. How many elements will m have, If m is multiple of 4 but not a multiple of 6.

1) 33
2) 34
3) 48
4) 66
5) 67

Total integers from 501 to 900 =400, but exclude 900, so 400-1=399
Number of multiple of 4 in 400=$$\frac{400}{4}=100$$, but this is including 900, so $$100-1=99$$

Now lowest common multiple of 4 and 6 = LCM of (4,6)=12
So in every $$\frac{12}{4}=3$$ multiple of 4, one is a multiple of 6.. so 2 out of 3 is what we are looking for..
Therefore in 99, $$99*\frac{2}{3}=66$$

D
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If m is set of integers from 500<m<900  [#permalink]

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16 Sep 2018, 22:43
2
1
Afc0892 wrote:
If m is set of integers from 500<m<900. How many elements will m have, If m is multiple of 4 but not a multiple of 6.

1) 33
2) 34
3) 48
4) 66
5) 67

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

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. $$(99*\frac{2}{3})=66$$

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If m is set of integers from 500<m<900  [#permalink]

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17 Sep 2018, 11:53
1
generis wrote:
Afc0892 wrote:
If m is set of integers from 500<m<900. How many elements will m have, If m is multiple of 4 but not a multiple of 6.

1) 33
2) 34
3) 48
4) 66
5) 67

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

First Term must be 504 not divisible by 6)
Last term is 896 (we know 900 is divisible by 4; subtract 4) (also not divisible by 6)

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

To exclude one of two integers as a factor

1) use LCM. LCM of 4 and 6 is 12
(4 * 3)=12
Every third term must be omitted. $$\frac{2}{3}$$ of terms will remain.
$$(\frac{2}{3}*99)=66$$
Total number of terms in m: (99 - 33) = 66

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

Which terms are divisible by 6?

Divisibility rule for 6: must be divisible
by 2 (ends in even)
and 3 (digits sum to 3 or multiple of 3)

Six terms are listed.
-- Not divisible by 6: 508, 512, 520, and 524
-- Divisible by 6: 504, 516

Six terms. Two are divisible by 6.
$$\frac{2}{6}=\frac{1}{3}$$ of the terms must be excluded. $$\frac{2}{3}$$ are kept.
That pattern will hold.
Number of terms: $$(\frac{2}{3}* 99)=66$$ terms

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
Senior SC Moderator
Joined: 22 May 2016
Posts: 2036
If m is set of integers from 500<m<900  [#permalink]

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18 Sep 2018, 11:40
1
generis wrote:
Afc0892 wrote:
If m is set of integers from 500<m<900. How many elements will m have, If m is multiple of 4 but not a multiple of 6.

1) 33
2) 34
3) 48
4) 66
5) 67

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

First Term must be 504 not divisible by 6)
Last term is 896

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

To exclude one of two integers as a factor

1) use LCM. LCM of 4 and 6 is 12
(4 * 3)=12
Every third term must be omitted. $$\frac{2}{3}$$ of terms will remain.
$$(\frac{2}{3}*99)=66$$
Total number of terms in m: (99 - 33) = 66
OR
2) Find the pattern: list the first few terms
504, 508, 512, 516, 520, 524

Which terms are divisible by 6?
Six terms are listed.
-- Not divisible by 6: 504 and 508
-- Divisible by 6: 512
-- Not divisible by 6: 516 and 520
-- Divisible by 6: 524

Six terms. Two are divisible by 6.
$$\frac{2}{6}=\frac{1}{3}$$ of the terms must be excluded. $$\frac{2}{3}$$ are kept.
That pattern will hold.
Number of terms: $$(\frac{2}{3}* 99)=66$$ terms

dave13 wrote:
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.]
Quote:
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 ?

Whoops. You forgot to "count" 4 itself.
Quote:
$$\frac{2}{3}$$ 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
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
$$99*\frac{2}{3}=66$$

$$m = 66$$
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For what are we born if not to aid one another?
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If m is set of integers from 500<m<900 &nbs [#permalink] 18 Sep 2018, 11:40
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# If m is set of integers from 500<m<900

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