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For positive integers n and k, if f (n) represents the remainder when

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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) \(f(k+11)=9\)
[Reveal] Spoiler: OA

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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 [#permalink]

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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 [#permalink]

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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 \(144-16=kP\), \(128=kP, 2^7=kP\), 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)”.
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Re: For positive integers n and k, if f (n) represents the remainder when [#permalink]

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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   [#permalink] 27 Mar 2017, 16:32
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