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For positive integers n and k, if f (n) represents the remainder when
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Updated on: 25 Mar 2017, 05:37
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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) \(f(k+11)=9\)
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Originally posted by hazelnut on 24 Mar 2017, 04:56.
Last edited by hazelnut on 25 Mar 2017, 05:37, edited 2 times in total.



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Re: For positive integers n and k, if f (n) represents the remainder when
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24 Mar 2017, 06:15
f(n) = Remainder(n^2/k)
St1: f(k + 12) = 16 Remainder((k+12)^2/k) = 16 (k^2 + 24k + 144)/k = 16 Remainder of 16 is contributed by 144/k k > 16 and final divisible value before 144 = 144  16 = 128 Factors of 128 that are greater than 16 = 128, 64, 32 > k is even Sufficient
St2: f(k + 11) = 9 Remainder of 9 is contributed by 121/k k > 9 and final divisible value before 121 = 121  9 = 112 Factors of 112 greater than 9 = 112, 56, 28, 16, 14 > k is even Sufficient.
Answer: D



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Re: For positive integers n and k, if f (n) represents the remainder when
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25 Mar 2017, 05:42
ziyuen wrote: 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) \(f(k+11)=9\) 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 \(14416=kP\), \(128=kP, 2^7=kP\), 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 5051level question related to “CMT 4 (B: if you get A or B too easy, consider D)”.
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Re: For positive integers n and k, if f (n) represents the remainder when
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27 Mar 2017, 16:32
ziyuen wrote: 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) \(f(k+11)=9\) We are given that f(n) represents the remainder when n^2 is divided by k. We need to determine whether k is even. Statement One Alone:f(k + 12) = 16 This means when (k + 12)^2 is divided by k, the remainder is 16. Let’s expand (k + 12)^2: (k + 12)^2 = k^2 + 24k + 144 Since the first two terms contain the factor k, the remainder when (k + 12)^2 is divided by k is 144/k. Since the remainder is 16, we can say: 144/k = q + 16/k for some integer q Let’s simplify this expression by multiplying both sides by k: 144 = qk + 16 128 = qk We see that k is a factor of 128. Notice that since 16 is the remainder, k, as a divisor, must be greater than 16. Since 128 = 2^7 and k > 16 = 2^4, k must be 2^5 = 32, 2^6 = 64, or 2^7 = 128. Therefore, regardless which value k is, k is even. Statement one alone is sufficient. Statement Two Alone:f(k + 11) = 9 This means when (k + 11)^2 is divided by k, the remainder is 9. Let’s expand (k + 11)^2: (k + 11)^2 = k^2 + 22k + 121 Since the first two terms contain the factor k, the remainder when (k + 11)^2 is divided by k is 121/k. Since the remainder is 9, we can say: 121/k = q + 9/k for some integer q Let’s simplify this expression by multiplying both sides by k: 121 = qk + 9 112 = qk We see that k is a factor of 112. Notice that since 9 is the remainder, k, as a divisor, must be greater than 9. Since 112 = 2^4 x 7 and k > 9, k has to be greater than 7 also. Since k > 7 and it has to be a factor of 112, it must contain a factor of 2 (e.g., k can be 2 x 7 = 14 or 2^4 = 16). Therefore, regardless what value k is, k is even. Statement two alone is sufficient. Answer: D
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Re: For positive integers n and k, if f (n) represents the remainder when
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17 Jun 2018, 17:03
Hi Vyshak, hazelnut, JeffTargetTestPrepHow come this approach is wrong  I got to this part, but instead of expanding, I tried to solve by using even/odd hazelnut wrote: In the case of con 1), \((k+12)^2=kQ+16\) (Q: any positive integer) \((k+12)^2kQ=16\) We have 2 cases: Case 1\((odd)^2odd=even\) > This works if k is odd and Q is odd. Case 2\((even)^2even=even\) > This works if k is even. So k can be even or odd.



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Re: For positive integers n and k, if f (n) represents the remainder when
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18 Jun 2018, 17:25
Your approach holds true if you are checking for odd or even. Here, there is a limit set in the form of remainder. The remainder we need to arrive is 16. In other words, you should try \((odd)^2odd=16\)  Identify the odd values of k and Q that satisfy this equation. You won't find any because k can be only 128, 64 or 32 for St1. aserghe1 wrote: How come this approach is wrong  I got to this part, but instead of expanding, I tried to solve by using even/odd hazelnut wrote: In the case of con 1), \((k+12)^2=kQ+16\) (Q: any positive integer) \((k+12)^2kQ=16\) We have 2 cases: Case 1\((odd)^2odd=even\) > This works if k is odd and Q is odd. Case 2\((even)^2even=even\) > This works if k is even. So k can be even or odd.



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Re: For positive integers n and k, if f (n) represents the remainder when
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18 Jun 2018, 18:51
k > 16 (k+12)^2 = ky + 16 y = (k+12)^2 / k  16 /k y = (k^2 + 24 k + 144  16) / k y = (k^2 + 24 k + 128) / k y = k + 24 + 128 / k Thus k is even because 128 has no odd factors except 1 Sufficient
k > 9 (k+11)^2 = ky + 9 y = (k+11)^2 / k  9 / k y = (k^2 + 22 k + 121  9) / k y = (k^2 + 22 k + 112) / k y = k + 22 + 112 / k The only odd integer 7 divides 112; but k > 9 k is even Sufficient
Answer D Each statement by itself is sufficient.




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