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Re: integer k ... DS [#permalink]
22 Sep 2009, 07:31
This post received KUDOS
Start off by factoring the original equation in the stem where (k^2+4k+3)=(k+3)(k+1) So, (k+2(k^2+4k+3)=(k+1)(k+2)(k+3) where (k+1)(k+2)(k+3) are essentially consecutive integers.
As consecutive integers, (k+1)(k+2)(k+3) can be (odd)(even)(odd) in which case k has to be even. It can also be (even)(odd)(even) in which case k has to be odd. If k is odd,(k+1)(k+2)(k+3) will definitely be divisible by 4 as the product terms carry at least two 2s. This is because there are two even numbers in the product terms and they will each carry at least one 2 in their prime factorization and 4=(2)(2).
If k is even however,(k+1)(k+2)(k+3) may or may not be divisible by 4. It depends if the middle term carries two 2s, (i.e. is a multiple of 4) or not.(k+1) and (k+3) will definitely not carry a 2 in this case as they are odd numbers. For example, if k=2,(k+1)(k+2)(k+3) can be (3)(4)(5) which is divisible by 4 as the middle term, 4, is a multiple of 4. The same logic applies for k=6. If k=8 however,(k+1)(k+2)(k+3) = (9)(10)(11) which is NOT divisible by 4 as the middle term is 10=5x2 and there is only one 2 in its prime factorization.
1) All multiples of 8 are even. Hence, k is even. Possible values of (k+1)(k+2)(k+3) are: If k=8, (k+1)(k+2)(k+3) = (9)(10)(11) If k =16, (k+1)(k+2)(k+3) =(17)(18)(19) If k=24,(k+1)(k+2)(k+3) = (25)(26)(27) If k=32, (k+1)(k+2)(k+3) =(33)(34)(25)
Looking at the middle, or the only even term, of each series of consecutive integers, 10, 18 or 26, we see that in their prime factorization, they only contain one 2. Hence, the entire product of 3 consecutive integers will NOT be divisible by 4 if k is a multiple of 8.
Logically, this is also because multiples of 4 which are 8 or greater include - 8,12,16,20, 24 Multiples of 8 include 8, 16, 24. The equation given in the stem asks for the product of 3 consecutive integers. The difference of multiples of 4 and 8 are either 0 or a multiple of 8 apart. Never 3 apart. Hence, the middle number of (k+1)(k+2)(k+3) when k is a multiple of 8 will never contain a multiple of 4. Statement A is sufficient.
2) I assume the statement means (k+1)/3 is odd. So, k+1=3(odd number) k+1=odd k=odd-1 k = even
For reasons explained in the beginning, if k is even,(k+1)(k+2)(k+3) may or may not be divisible by 4.
Re: integer k ... DS [#permalink]
22 Sep 2009, 16:03
I agree that the correct answer is A.
Is (k + 2)(k2 + 4k + 3) divisible by 4?
(k + 2)(k^2 + 4k + 3) = (k + 1)(k + 2)(k + 3)
Now to simplify matters, one of the terms NEEDS to be divisible by 4 in order for the whole expression to be divisible by 4. Why?
In order for an integer to be divisible by 4, its product must contain two factors of 2. This would appear to be satisfied by two even numbers. In the instance that we had two even numbers (only possible in this case if k+1 and k+3 are even, then we would already have a number divisible by 4 since those numbers are two consecutive even numbers.
Statement 1: k is divisible by 8:
If k is divisible by 8, then k = 8n, where n is a positive integer. This immediately makes k+1 and k+3 odd numbers, which are not divisible by 4. Now we test the (k+2) term:
k+2 = 8n + 2 = 4(2n) + 2
As a result, if k is divisible by 8, the expression when divided by 4 will have a remainder of 2. Therefore we can can conclude it IS NOT divisible by 4, and Statement 1 is sufficient.
Statement 2: (k + 1)/3 is an odd integer:
We can represent an odd integer as 2n + 1, where n is an integer. Therefore, (k+1)/3 = 2n + 1 k = 6n + 2
Subbing this into the original equation:
(k+1)(k+2)(k+3) = (6n + 3)(6n + 4)(6n + 5)
Since 6n + 3 and 6n + 5 both represent odd numbers, if 6n + 4 is divisible by 4, the expression is divisible by 4. Since we cannot determine this (if n = 1, it is not divisible by 4, if n = 2, it is divisible by 4, etc.), Statement B is insufficient.