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Updated on: 22 Apr 2014, 03:03
Originally posted by kishankolli on 22 Sep 2009, 03:51.
Last edited by Bunuel on 22 Apr 2014, 03:03, edited 1 time in total.
Last edited by Bunuel on 22 Apr 2014, 03:03, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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A
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For positive integer k, is the expression (k + 2)(k^2 + 4k + 3) divisible by 4?
(1) k is divisible by 8.
(2) (k + 1)/3 is an odd integer
(1) k is divisible by 8.
(2) (k + 1)/3 is an odd integer
22 Apr 2014, 03:10
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pretzel
No, the correct answer is A.
For positive integer k, is the expression (k + 2)(k^2 + 4k + 3) divisible by 4?
\((k + 2)(k^2 + 4k + 3)=(k+1)(k+2)(k+3)\), so the expression is the product of three consecutive integers.
(1) k is divisible by 8 --> \(k=8n=even\) --> \((k+1)(k+2)(k+3)=odd*even*odd\). Now, \(k+2=8n+2\), though even, is not a multiple of 4 (it's 2 greater than a multiple of 8), therefore the expression is not divisible by 4. Sufficient.
(2) (k + 1)/3 is an odd integer --> \(k+1=3*odd=odd\) --> \(k=even\) --> \((k+1)(k+2)(k+3)=odd*even*odd\). Now, \(k+2=even\) may or may not be divisible by 4. For example, consider \(k=2\) and \(k=8\). Not sufficient.
Answer: A.
Hope it's clear.
22 Sep 2009, 08:31
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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.
So, Statement B is insufficient.
Is the answer A?
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.
So, Statement B is insufficient.
Is the answer A?
