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.

Why is it always divisible by 6? Please explain with those not great with number property problems such as myself.

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.

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.

Please make it clear-

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.

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

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

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.

I tried a lot of combinations and all of them are divisible by 96
could it be the correct answer (96)!!!

(10)(12)(14) = 1680, and 1680 it's not the visible by 96

THANKS

b>2 and it is even. So it must have a multiple of 2. Same goes for c and d.
Now since there are 3 consecutive integers. So they must be a multiple of 3.
Since we are having 3 integers. So it must be a multiple of another 2.
So total 2*2*2*2*3 = 48.
I hope this helps

Given that b, c, and d are consecutive even integers such that 2 < b < c < d and we need to find What is the largest positive integer that must be a divisor of bcd

As b, c and d are consecutive even numbers so one of them will be a multiple of 2, one will be a multiple of 4 and one will be a multiple of 6

Ex: 4, 6, 8. 4 -> multiple of 2 and 4, 6 -> multiple of 6, 8 -> multiple of 2 and 4

=> b*c*d will be a multiple of 2*4*6 = 48

So, Answer will be 48
Hope it helps!

Watch the following video to learn How to Sequence problems

So can we just then learn this as a rule that whenever their are three consecutive even or odd intergers, there will always be a multiple of 3,4 and 6?

Since these explanations are tricky.