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Since first two terms of \(32t^3 - 16t^2 + 8t - 4\) are divisible by \(t^2\) then the question becomes whether \(8t-4\) is divisible by \(t^2\).
(1) \(t^2 \lt 25\). Since \(t\) is a prime number then \(t=2\) or \(t=3\). If \(t=2\) then \(8t-4\) is divisible by \(t^2=4\) but if \(t=3\) then \(8t-4\) is NOT divisible by \(t^2=9\). Not sufficient.
(2) \(t^2-8t+12=0\). Solving gives \(t=2\) (\(t=6\) is not a valid solution since 6 is not a prime number). Sufficient.
I got there a different way. Can anyone clarify if my method was sound or just lucky?
First I factored out a 4 from the entire equation to yield 4*(8t^3 -4t^2 + 2t -1). Thus, the equation is a multiple of 4. If t^2 divides 4 evenly, the question can be definitively answered.
1. The stem translates to -5<t<5. Since t = prime, t = 2 or 3.
2^2 divides 4 evenly, 3^2 does not. Insuff
2. t= 2 or 6. Since 6 isn't prime t must 2 which is suff.
I believe your reasoning to be sound. Both ideas hinge on recognizing that a term that can be factored out renders the rest of the equation divisible by the term factored out. I arrived at my solution similarly.
1) -4<t<4... t must be prime so t=2 or 3. 3^2=9 can not be factored out of each term in given information. 2^2=4 can be. Insuff.
2) t= 6 or 2. T can't = 6 due to prime # restriction. 2^2=4 can be factored out of each term in original information. 4(8t^3-4t^2+2t-1). Sufficient.
I actually plugged the potential numbers in for T, this took a little longer, but it assured me of my answer. I'm posting this in case it helps someone see through this problem clearly.
From the stem: T must be positive and is a prime= 2,3,5,7,11....
Statement (1): (T^2)< 25 therefore T must be either 2 or 3
If T=2: [32(2^3)]-[16(2^2)]+[8(2)]-4 [32(8)]-[16(4)]+16-4 256-64+16-4=204 204/4 = 51
If T=3: [32(3^3)]-[16(3^2)]+[8(3)]-4 [32(27)]-[16(9)]+24-4 864-144+24-4= 740 740/9≠Integer
We can see here that statement (1) does not give us a clear outcome.
Statement (2): (T^2)-8T+12=0 (T-6)(T-2)=0 T=6 or T=2
Since we know T is a prime number, T must be 2
Statement (2) is sufficient. The correct answer is B.
Since first two terms of \(32t^3 - 16t^2 + 8t - 4\) are divisible by \(t^2\) then the question becomes whether \(8t-4\) is divisible by \(t^2\).
(1) \(t^2 \lt 25\). Since \(t\) is a prime number then \(t=2\) or \(t=3\). If \(t=2\) then \(8t-4\) is divisible by \(t^2=4\) but if \(t=3\) then \(8t-4\) is NOT divisible by \(t^2=9\). Not sufficient.
(2) \(t^2-8t+12=0\). Solving gives \(t=2\) (\(t=6\) is not a valid solution since 6 is not a prime number). Sufficient.
Answer: B
Can't we just instantly test that t equals 2 from looking at the equation? All of its factors are divisible by 2. If t=2 fails, we know it's NOT divisible.
Is my logic crazy or did I just get lucky answering B?