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

Answer D

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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