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# b, c, and d are consecutive even integers such that 2 < b <

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b, c, and d are consecutive even integers such that 2 < b < [#permalink]  12 Aug 2017, 10:05
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b, c, and d are consecutive even integers such that 2 < b < c < d. What is the largest positive integer that must be a divisor of bcd?

[Reveal] Spoiler: OA
48

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Re: b, c, and d are consecutive even integers such that 2 < b < [#permalink]  16 Oct 2017, 00:54
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b, c and d are consecutive even integers such that 2<b<c<d.
Let's assume b = 2n, then c = 2n+2 , d = 2n+4 and n>1 since 2<b<c<d.

Then the expression bcd = 2n(2n+2)(2n+4)= 8n(n+1)(n+2). Now n(n+1)(n+2) is the product of 3 consecutive integers which is always divisible by 6. Therefore 8n(n+1)(n+2) is always divisible by 6*8 or 48.
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Re: b, c, and d are consecutive even integers such that 2 < b < [#permalink]  26 Sep 2018, 11:30
i can't understand this one.. can someone else offer a different explanation?
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Re: b, c, and d are consecutive even integers such that 2 < b < [#permalink]  27 Sep 2018, 09:28
Expert's post
Picking numbers

Consecutive even integers

2 < 4 < 6 < 8

4,6, and 8 actually are = $$2^6 * 3$$

The largest divisor of b*c*d is 48.

Hope this helps.
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Re: b, c, and d are consecutive even integers such that 2 < b < [#permalink]  15 Dec 2018, 17:08
IlCreatore wrote:
b, c and d are consecutive even integers such that 2<b<c<d.
Let's assume b = 2n, then c = 2n+2 , d = 2n+4 and n>1 since 2<b<c<d.

Then the expression bcd = 2n(2n+2)(2n+4)= 8n(n+1)(n+2). Now n(n+1)(n+2) is the product of 3 consecutive integers which is always divisible by 6. Therefore 8n(n+1)(n+2) is always divisible by 6*8 or 48.

Why is it always divisible by 6? Please explain with those not great with number property problems such as myself.
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Re: b, c, and d are consecutive even integers such that 2 < b < [#permalink]  16 Dec 2018, 01:42
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n(n+1)(n+2) and n=1

1*2*3=6 divisible by 6

n=2

2*3*4=24 divisible by 6

And so forth
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Re: b, c, and d are consecutive even integers such that 2 < b < [#permalink]  16 Dec 2018, 09:04
Carcass wrote:
n(n+1)(n+2) and n=1

1*2*3=6 divisible by 6

n=2

2*3*4=24 divisible by 6

And so forth

Thank you, that is reminiscent of another problem I have done, I should definitely remember it now for the GRE.
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Re: b, c, and d are consecutive even integers such that 2 < b < [#permalink]  16 Dec 2018, 09:14
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What you did say right now could be a huge problem.

In your preparation for the GRE, GMAT , LSAT or whatever it would be if you do not gain experience from a previous problem, then you will not climb the learning curve.

In other words, you will stay always on the ground.

Regards
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Re: b, c, and d are consecutive even integers such that 2 < b < [#permalink]  16 Dec 2018, 10:46
Carcass wrote:
What you did say right now could be a huge problem.

In your preparation for the GRE, GMAT , LSAT or whatever it would be if you do not gain experience from a previous problem, then you will not climb the learning curve.

In other words, you will stay always on the ground.

Regards

True, I need to stay on my toes and redo problems I have gotten wrong in the past. I usually wait a couple days to make sure I have understood what I missed or did wrong rather than just having it memorized.
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Re: b, c, and d are consecutive even integers such that 2 < b < [#permalink]  01 Jan 2019, 05:52
Carcass wrote:
Picking numbers

Consecutive even integers

2 < 4 < 6 < 8

4,6, and 8 actually are = $$2^6 * 3$$

The largest divisor of b*c*d is 48.

Hope this helps.

I thought like this
consecutive integers are 2,4,6,8
hence The largest divisor of b*c*d is 4*6*8= 192
I am really confuse why 192 is not a divisor of 192.
Please let me know why I am wrong.

Last edited by AE on 01 Jan 2019, 10:30, edited 1 time in total.
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Re: b, c, and d are consecutive even integers such that 2 < b < [#permalink]  01 Jan 2019, 07:30
Expert's post
AE wrote:
Carcass wrote:
Picking numbers

Consecutive even integers

2 < 4 < 6 < 8

4,6, and 8 actually are = $$2^6 * 3$$

The largest divisor of b*c*d is 48.

Hope this helps.

I though like this
consecutive integers are 2,4,6,8
hence The largest divisor of b*c*d is 4*6*8= 192
I am really confuse why 192 is not a divisor of 192.
Please let me know why am I wrong.

Hi..

The numbers chosen 4,6,8 are not the correct choice for answering this question, as you say why not 192 then...
Ok let us see..
We have three consecutive even integers, so atleast one will be multiple of 3, but what about number of 2s in it.
Since we are looking for MUST, we will look for least number of 2s that will always be there.

Ok for that I have to ensure that there is no multiple of 8, because a multiple of 4 is surely there..
So let the numbers be 8n+2, 8n+4, 8n+6...
So the product becomes (8n+2)(8n+4)(8n+6)=$$2(4n+1)*2(4n+2)*2(4n+3)=8(4n+1)(4n+2)(4n+3)$$..
Now (4n+1)(4n+2)(4n+3) is a product of three consecutive numbers. So the product of three consecutive numbers will be multiple of 2 and 3..

So overall 8*2*3=48..

For example ..
26*28*30=48*13*7*5
And 4*6*8=48*4
So you see both of the above are necessarily multiple of 48..
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Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

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Re: b, c, and d are consecutive even integers such that 2 < b < [#permalink]  11 Jan 2019, 23:37
Expert's post
chetan2u wrote:
AE wrote:
Carcass wrote:
Picking numbers

Consecutive even integers

2 < 4 < 6 < 8

4,6, and 8 actually are = $$2^6 * 3$$

The largest divisor of b*c*d is 48.

Hope this helps.

I though like this
consecutive integers are 2,4,6,8
hence The largest divisor of b*c*d is 4*6*8= 192
I am really confuse why 192 is not a divisor of 192.
Please let me know why am I wrong.

Hi..

The numbers chosen 4,6,8 are not the correct choice for answering this question, as you say why not 192 then...
Ok let us see..
We have three consecutive even integers, so atleast one will be multiple of 3, but what about number of 2s in it.
Since we are looking for MUST, we will look for least number of 2s that will always be there.

Ok for that I have to ensure that there is no multiple of 8, because a multiple of 4 is surely there..
So let the numbers be 8n+2, 8n+4, 8n+6...
So the product becomes (8n+2)(8n+4)(8n+6)=$$2(4n+1)*2(4n+2)*2(4n+3)=8(4n+1)(4n+2)(4n+3)$$..
Now (4n+1)(4n+2)(4n+3) is a product of three consecutive numbers. So the product of three consecutive numbers will be multiple of 2 and 3..

So overall 8*2*3=48..

For example ..
26*28*30=48*13*7*5
And 4*6*8=48*4
So you see both of the above are necessarily multiple of 48..

Son of a gun this is a brilliant answer
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Re: b, c, and d are consecutive even integers such that 2 < b < [#permalink]  11 Jan 2019, 23:56
Expert's post
since 2<b<c<d
why not try 2 < 4< 6< 8??

this is the smallest possible value right?

lets rewrite it this way

2<(2*2)<(3*2)<(4*2) replace 2 with N since these numbers are multiples of 2

dont think to rewrite the expression like this

2< 2n <3n <4n because once n is greater than 2 the numbers wont be consecutive.

rewrite as

2 < 2n < 2(n+1)< 2(n+2)

then multiply 2n*2(n+1)*2(n+2)=

8(n^3+3n^2+2n) plug in 1 for n and you get 8(1+3+2)= 8*6=48
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Re: b, c, and d are consecutive even integers such that 2 < b < [#permalink]  21 Jan 2019, 22:24
Expert's post
I was looking at this problem again, and just thought,

what if instead of all the fancy number property stuff, why not try this shortcut?
If you have a string of consecutive even numbers, the greatest common factor of the last three numbers will just be the two lesser numbers times the number immediately preceding the last string of 3 numbers.
Re: b, c, and d are consecutive even integers such that 2 < b <   [#permalink] 21 Jan 2019, 22:24
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