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.
Please make it clear-
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..