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26 May 2017, 04:42
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For positive integers \(n\) and \(k\), if \(f (n)\) represents the remainder when \(n^{2}\) is divided by \(k\), is \(k\) an even number?
(1) \(f(k+12) = 16\)
(2) \(f(k+11) = 9\)
(1) \(f(k+12) = 16\)
(2) \(f(k+11) = 9\)
26 May 2017, 04:42
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Official Solution:
In the original condition, there are 2 variables \((n, k)\). By solving con 1) and con 2), you get \(n = k + 12\) and \(n = k + 11\), and since you only need to know \(K\) in order to find \(n\), you get 1 variable. In order to match the number of variables and the number of equations, D is most likely to be the answer.
In the case of con 1), \((k+12)^{2} = kQ + 16\) (\(Q\): any positive integer) can be expanded into \(k^{2} + 24k + 144 = kQ + 16\). Since \(k^{2}\) and \(24k\) can be divided by \(k\), you get \(144 = kP + 16\) (\(P\): any positive integer). If so, from \(144 - 16 = kP, 128 = kP, 2^{7}=kP\), you get \(k = 2^{5}, 2^{6}, 2^{7}\), which are all even numbers, hence yes, it is sufficient.
In the case of con 2), \((k+11)^{2} = kR + 9\) (R: any positive integer) can be expanded into \(k^{2} + 22k + 121 = kR + 9\). Since \(k^{2}\) and \(22k\) can be divided by \(k\), you get \(121 = kT + 9\) (T: any positive integer). If so, from \(121 - 9 = kT, 112 = kT, 112 = (2^{4})(7) = kT, k = 7 < 9\) is impossible, and you get \(k = 14, 16, 28, 56, 112\), which are all even numbers, hence yes, it is sufficient. Therefore, the answer is D. This question is also a 5051-level question related to "CMT 4 (B: if you get A or B too easy, consider D)".
Answer: D
In the original condition, there are 2 variables \((n, k)\). By solving con 1) and con 2), you get \(n = k + 12\) and \(n = k + 11\), and since you only need to know \(K\) in order to find \(n\), you get 1 variable. In order to match the number of variables and the number of equations, D is most likely to be the answer.
In the case of con 1), \((k+12)^{2} = kQ + 16\) (\(Q\): any positive integer) can be expanded into \(k^{2} + 24k + 144 = kQ + 16\). Since \(k^{2}\) and \(24k\) can be divided by \(k\), you get \(144 = kP + 16\) (\(P\): any positive integer). If so, from \(144 - 16 = kP, 128 = kP, 2^{7}=kP\), you get \(k = 2^{5}, 2^{6}, 2^{7}\), which are all even numbers, hence yes, it is sufficient.
In the case of con 2), \((k+11)^{2} = kR + 9\) (R: any positive integer) can be expanded into \(k^{2} + 22k + 121 = kR + 9\). Since \(k^{2}\) and \(22k\) can be divided by \(k\), you get \(121 = kT + 9\) (T: any positive integer). If so, from \(121 - 9 = kT, 112 = kT, 112 = (2^{4})(7) = kT, k = 7 < 9\) is impossible, and you get \(k = 14, 16, 28, 56, 112\), which are all even numbers, hence yes, it is sufficient. Therefore, the answer is D. This question is also a 5051-level question related to "CMT 4 (B: if you get A or B too easy, consider D)".
Answer: D
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