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16 Sep 2014, 00:26
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If \(x^2 = y + 5\), \(y = z - 2\) and \(z = 2x\), is \(x^3 + y^2 + z\) an integer divisible by 7?
(1) \(x \gt 0\)
(2) \(y = 4\)
(1) \(x \gt 0\)
(2) \(y = 4\)
21 Sep 2014, 04:16
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Official Solution:
If \(x^2 = y + 5\), \(y = z - 2\) and \(z = 2x\), is \(x^3 + y^2 + z\) an integer divisible by 7?
We have a system of three distinct equations with three unknowns. Let's solve:
(i) \(x^2 = y + 5\);
(ii) \(z = 2x\);
(iii) \(y = z - 2\).
Substituting (ii) into (iii), we get \(y = 2x - 2\). Now, substituting this into (i), we have \(x^2 = (2x - 2) + 5\). Simplifying this equation gives us \(x^2 - 2x - 3 = 0\), which has solutions \(x = 3\) or \(x = -1\).
If \(x=3\), then \(y=2x-2=4\), and \(z=2x=6\). In this case, \(x^3 + y^2 + z = 27 + 16 + 6 = 49\), which is divisible by 7.
If \(x=-1\), then \(y=2x-2=-4\), and \(z=2x=-2\). In this case, \(x^3 + y^2 + z = -1 + 16 - 2 = 13\), which is NOT divisible by 7.
Hence, to answer the question, we need to determine which of the above cases we have.
(1) \(x \gt 0\). This means that \(x = 3\), and \(x^3 + y^2 + z=49\), which is divisible by 7. Sufficient.
(2) \(y = 4\). This means that \(x = 3\), and \(x^3 + y^2 + z=49\), which is divisible by 7. Sufficient.
Answer: D
If \(x^2 = y + 5\), \(y = z - 2\) and \(z = 2x\), is \(x^3 + y^2 + z\) an integer divisible by 7?
We have a system of three distinct equations with three unknowns. Let's solve:
(i) \(x^2 = y + 5\);
(ii) \(z = 2x\);
(iii) \(y = z - 2\).
Substituting (ii) into (iii), we get \(y = 2x - 2\). Now, substituting this into (i), we have \(x^2 = (2x - 2) + 5\). Simplifying this equation gives us \(x^2 - 2x - 3 = 0\), which has solutions \(x = 3\) or \(x = -1\).
If \(x=3\), then \(y=2x-2=4\), and \(z=2x=6\). In this case, \(x^3 + y^2 + z = 27 + 16 + 6 = 49\), which is divisible by 7.
If \(x=-1\), then \(y=2x-2=-4\), and \(z=2x=-2\). In this case, \(x^3 + y^2 + z = -1 + 16 - 2 = 13\), which is NOT divisible by 7.
Hence, to answer the question, we need to determine which of the above cases we have.
(1) \(x \gt 0\). This means that \(x = 3\), and \(x^3 + y^2 + z=49\), which is divisible by 7. Sufficient.
(2) \(y = 4\). This means that \(x = 3\), and \(x^3 + y^2 + z=49\), which is divisible by 7. Sufficient.
Answer: D
14 Sep 2019, 17:07
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Bunuel
Another approach - It is a good idea in general to assess what you may be able to piece together from the given information before you jump in further by taking on either statement. Here, the three given equations each contain just two unknowns, so substitution is a straightforward process. As soon as I saw a quadratic (\(x^2\)), I thought that x would likely have two solutions. Thus, I looked to get the first equation in terms of x only. This led me from the second to the third equation and back again pretty quickly: if \(z = 2x\), then, backtracking a bit, \(y = (2x) - 2\), and, reaching back a little further, \(x^2 = (2x - 2) + 5\), or \(x^2 - 2x -3 = 0\). This is a factorable quadratic, with binomial roots of (x - 3) and (x + 1). The solutions will be 3 or -1. With all the hard work out of the way, I took on Statement (1) first, rather than what I consider to be the easier statement in (2).
Why would it be important that x would be greater than 0? Because that would lead only to the conclusion that x = 3. It is thus not even necessary to test what could happen if x were to equal -1. In fact, I did not even bother to see what would happen when I substituted 3 for x in the modified quadratic, since I knew that regardless of the outcome, it would be possible to tell whether the eventual answer would be divisible by 7. Statement (1) had to be SUFFICIENT, narrowing the answer choices down to (A) or (D).
Statement (2) requires little effort to justify, since all three given equations must be considered. The middle equation would tell us that z must equal 6, and then the third equation that x must equal 3. At that point, the original quadratic would no longer have two justifiable solutions for x, and we would also be able to answer the question. Thus, (D) had to be the correct response, and both the effort and time it took to arrive at such a conclusion were minimal.
In Yes/No questions, all you need to be able to determine is whether there is enough evidence to reach a certain conclusion. It may not be necessary to actually work the numbers out.
I sometimes call myself "the lazy mathematician" in my tutoring sessions. I love math, but I would rather work less to get a correct solution than put in more effort and get... the same thing. I have adopted this approach not only because of some innate tendency, but also no doubt due to knowing a brilliant mathematician in college, a guy who tied for 24th in the Putnam collegiate math competition as a sophomore, and who later finished first in the nation in another year. I discovered (via a friendship that extended into chess club) that his extraordinary abilities stemmed not from seeing something in the aether that no one else could see, but from finding ways to break down complex matters into simpler pieces that he could readily put together to lead to a solution.
Good luck with your studies.
- Andrew
