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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  [#permalink]

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Updated on: 25 Mar 2017, 05:37
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Question Stats:

44% (02:34) correct 56% (02:50) wrong based on 155 sessions

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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  [#permalink]

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24 Mar 2017, 06:15
4
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.

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Re: For positive integers n and k, if f (n) represents the remainder when  [#permalink]

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25 Mar 2017, 05:42
1
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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27 Mar 2017, 16:32
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1
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.

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Re: For positive integers n and k, if f (n) represents the remainder when  [#permalink]

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17 Jun 2018, 17:03
Hi Vyshak, hazelnut, JeffTargetTestPrep

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)

$$(k+12)^2-kQ=16$$

We have 2 cases:

Case 1
$$(odd)^2-odd=even$$ --> This works if k is odd and Q is odd.

Case 2
$$(even)^2-even=even$$ --> 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  [#permalink]

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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 $$(odd)^2-odd=16$$ - 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)

$$(k+12)^2-kQ=16$$

We have 2 cases:

Case 1
$$(odd)^2-odd=even$$ --> This works if k is odd and Q is odd.

Case 2
$$(even)^2-even=even$$ --> 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  [#permalink]

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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

Each statement by itself is sufficient.
Re: For positive integers n and k, if f (n) represents the remainder when &nbs [#permalink] 18 Jun 2018, 18:51
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