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Divisibility revisited!!

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by amit2k9 » Fri Jun 17, 2011 3:05 am
I believe the question needs rephrasing - instead of must be it should be could be.

reason being -

(n+6)(n+8)(n+10) where n = 1,2,10,14 or so.

for n=2, 8*10*12 it is.

24= 2^2 * 3 ; 32 = 2^5 ; 96 = 2^5 * 3

all are present as factors in 8*10*12 = 2^3 * 2*5 * 2^2*3.

hence E suits the picture here.
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by thephoenix » Fri Jun 17, 2011 5:36 am
the best way to solve these type is to use some value satisfying the conditions
here n can be 2,10,14,22....
now insert few value and u will get answer
like if n is 2 the the number will 8*10*12 (smallest possible number) and this number is div by 24,32 and 96
so st away go for E
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by Frankenstein » Fri Jun 17, 2011 5:45 am
amit2k9 wrote:I believe the question needs rephrasing - instead of must be it should be could be.

reason being -

(n+6)(n+8)(n+10) where n = 1,2,10,14 or so.
Hi,
Question is correct. There is no need to use 'could be'. For all valid values of n, the conditions holds.
n = 2,10,14,22,...
Any positive even number not divisible by 3 or 4 will be of the following 2 forms
1) n = 12p+2 where p is a non-negative integer
So, (n + 6)(n + 8)(n + 10) = (12p+8)(12p+10)(12p+12) = 4*2*12(2p+1)(6p+5)(p+1) = 96*integer
2) n = 12p-2 where is any positive integer
So, (n + 6)(n + 8)(n + 10) = (12p+4)(12p+6)(12p+8) = 4*6*4(3p+1)(2p+1)(3p+2) = 96*integer
So, So, (n + 6)(n + 8)(n + 10) is always divisible by 96, which in turn is divisible by 24 and 32.

(or)

Let n=2p where p is odd, as n is not divisible by 4
So, (n + 6)(n + 8)(n + 10) = (2p + 6)(2p + 8)(2p + 10)= 2*2*2*(p+3)(p+4)(p+5) = 8(p+3)(p+4)(p+5)
p+3,p+4,p+5 are consecutive numbers. Product of 3 consecutive numbers should be divisible by 3! =6
As p is odd, p+3,p+5 are consecutive evens, product should be divisible by 8.
So, (p+3)(p+4)(p+5) should be divisible by 8 as well as 6.
So, (p+3)(p+4)(p+5) is divisible by LCM(8,6)=24
So, 8(p+3)(p+4)(p+5) is divisible by 24*8 = 192
So, (n + 6)(n + 8)(n + 10) be divisible by 24,32,96.

The above methods are intended to use only to show that it holds for all valid values of 'n'. It is preferable to plug-in values in the test.
Cheers!

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by Brian@VeritasPrep » Fri Jun 17, 2011 8:36 am
Hey all,

Great question - I love these divisibility problems and I've found that the more of these you do the more they just "click" and they become less about testing numbers and more about following the logic that allows you to predict exactly what the product here will look like. A few thoughts here:

Amit2k9 - your only real mistake there is that you used 1; remember, the question says positive EVEN number, so 1 doesn't apply.

Overall:

If you look at what we'll get, we're taking an even number n, adding three different even numbers to it, and multiplying them all together. So we know that we'll have:

Even * Even * Even --> So at a minimum we'll have at least 2*2*2 represented in the prime factorization.

But wait...there's more! We're multiplying together three CONSECUTIVE even numbers. So, actually, we can guarantee that one of them will be a multiple of 4 and bring an extra multiple of 2 to the table. Try it - any set of three consecutive integers will provide at least one that's divisible by 4, because every SECOND even number is divisible by 4: (6, 8*, 10); (8*, 10, 12*); (10, 12*, 14). So now we know that we have at least 2*2*2*2 present in the factorization of this product.

But wait...there's more! Note that if we start with n as a number NOT divisible by 4, then that's an even integer that's off the divisible-by-4 cycle:

Even, Even AND divisible by 4, Even, Even AND divisible by 4, Even...
2, 4, 6, 8, 10...

So we know, then, that n + 6 and n + 10 will get us back ON that cycle, so the bookends of this (n+6)(n+8)(n+10) are both divisible by 4. So our product is going to look like:

(Even AND divisible by 4)(Even)(Even AND divisible by 4)

So...we know that, as prime factors, we'll have at minimum:

(2*2)(2)(2*2) ---> our product will be divisible by 2^5.


By similar logic, and knowing that we're testing for divisibility by 24 (which is 3*2^3) and 96 (3*2^5), we can use the consecutive-even-integers setup to prove that we'll definitely have a 3 in there, too. If you take any three integers in a row, one of them is divisible by 3 because every THIRD integer is divisible by 3: 1, 2, 3*, 4, 5, 6*, 7, 8, 9*.....

And for consecutive even integers, we're just taking a set of three straight integers and multiplying each by 2. So we know for sure that one - and exactly one - of the series (n+6), (n+8), (n+10) will be divisible by 3, so we'll have exactly one factor of 3 to go with our factors of 2^5. So we can prove that the product will be divisible by at least 2*2*2*2*2*3, and since those factors satisfy 24, 32, and 96, the answer is E.



Now, as a teaching tool - if you don't 100% follow this logic in the abstract form without plugging in some numbers to make it more concrete, that's perfectly fine. It takes some time to think this way. But you know that the GMAT tests divisibility and number properties, so when you're plugging in numbers in practice, try to think behind them to see if you can use that trial-and-error to become more aware of properties and rules like we just discussed. The more you learn to think like the testmaker, the easier this test becomes...
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by Sergio. » Fri Jun 17, 2011 9:44 am
if n is not divisible by 3 or 4, n could be at least 2
it cant be -2 (statement says it's positive); it can't be 6, 8 (it's not divisible by 3, 4), though it can be also 10 (not divisible by 3, 4; pos., even) and 10 also got 2 in it's prime factorization, so I'd use 2 instead of 10 (cause there's no other fact provided, I'd use the least possible value and we don't actually need another 5 as prime, so 10 is pointless to use in this case)

then just replace n with 2 and and divide it into primes (you don't need to multiply the numbers, just make an addition and divide the results into primes)

so (n+6)(n+8)(n+10) has at least next primes:
(8): 2, 2, 2
(10): 2, 5
(12): 2, 2, 3


24 has: 2, 2, 2, 3 which is enough to be divisor of the equation
32: 2, 2 ,2 ,2 , 2 enough too
96: 2, 2, 2, 2, 2, 3 and enough again

all three have required primes to be divisors of (n+6)(n+8)(n+10), so the answer is "E"
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