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If k is a multiple of 24 but not a multiple of 16, which of the follow

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If k is a multiple of 24 but not a multiple of 16, which of the follow  [#permalink]

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New post 13 Aug 2018, 03:54
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If k is a multiple of 24 but not a multiple of 16, which of the following cannot be an integer?


(A) \(\frac{k}{8}\)

(B) \(\frac{k}{9}\)

(C) \(\frac{k}{32}\)

(D) \(\frac{k}{36}\)

(E) \(\frac{k}{81}\)

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Re: If k is a multiple of 24 but not a multiple of 16, which of the follow  [#permalink]

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New post 13 Aug 2018, 07:07
Bunuel wrote:
If k is a multiple of 24 but not a multiple of 16, which of the following cannot be an integer?


(A) \(\frac{k}{8}\)

(B) \(\frac{k}{9}\)

(C) \(\frac{k}{32}\)

(D) \(\frac{k}{36}\)

(E) \(\frac{k}{81}\)


k is a multiple of 24 but not a multiple of 16.
k=2x2x2x3xp
k is not equal to 2x2x2x2xq
A) Integer when k=24
B) Integer when k=72
C) Not an integer since each multiple of 32 will definitely be a multiple of 16
D) Integer when k=72
E) Integer when \(k=2X2X2X3^4\)

Answer C.
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If k is a multiple of 24 but not a multiple of 16, which of the follow  [#permalink]

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New post 13 Aug 2018, 19:43
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Bunuel wrote:
If k is a multiple of 24 but not a multiple of 16, which of the following cannot be an integer?


(A) \(\frac{k}{8}\)

(B) \(\frac{k}{9}\)

(C) \(\frac{k}{32}\)

(D) \(\frac{k}{36}\)

(E) \(\frac{k}{81}\)


We have \(k=2^3*3^1*\)(any odd integer)Let's check the options:-
A. k/8 is an integer when k=24*any odd integer
B. k/9 is an integer when k=24*3*any odd integer
C. \(\frac{k}{32}\) can never be an integer . (\(32=2^5\), we can't make the odd integer=\(2^2\) since k is not a multiple of 16)
D. \(36=2^2*3^2=2^3*3^1*\frac{3}{2}\) , so k/36 is an integer.
E. 81=3^4=2^3*3^1*\frac{3^3}{2^3}[/m], so k/36 is an integer.

Ans. (C)
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Re: If k is a multiple of 24 but not a multiple of 16, which of the follow  [#permalink]

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New post 18 Aug 2018, 18:29
Bunuel wrote:
If k is a multiple of 24 but not a multiple of 16, which of the following cannot be an integer?


(A) \(\frac{k}{8}\)

(B) \(\frac{k}{9}\)

(C) \(\frac{k}{32}\)

(D) \(\frac{k}{36}\)

(E) \(\frac{k}{81}\)


If k is a multiple of 24, then k could be:

24, 48, 72, 96, 120, 144, …

However, since k is not a multiple of 16, then k can’t be:

48, 96, 144, …

So we see that k could only be:

24, 72, 120, …

That is, k is an odd multiple of 24.

We see that k/8 is an integer since 24 is divisible by 8. Furthermore, k/9 and k/81 could each be an integer if k is 9 x 24 and 81 x 24, respectively. Lastly, k/36 could be an integer if k is 9 x 24 (notice that 9 x 24 = 9 x 4 x 6 = 36 x 6). Therefore, k/32 can’t be integer (notice that 32 = 16 x 2, so if k is not a multiple of 16, then k can’t be a multiple of 32).

Answer: C
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Re: If k is a multiple of 24 but not a multiple of 16, which of the follow  [#permalink]

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New post 19 Aug 2018, 05:46

Solution



Given:
    • k is a multiple of 24 but not a multiple of 16

To find:
    • The option which cannot be an integer.

Approach and Working:
    • K is multiple of 24 then K can be written as \(2^{(3+a)}*3^{(1+b)}* some other prime factors\).
    o However, k is not a multiple of 16.
    o So, a must be 0.
    • Thus, K= \(2^3+*3^{(1+b)}* some other prime factors\).

Now, among the given options, only \(\frac{K}{32}\) will not be an integer.

Hence, the correct answer is option C.

Answer: C
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Re: If k is a multiple of 24 but not a multiple of 16, which of the follow  [#permalink]

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New post 19 Aug 2018, 06:19
Bunuel wrote:
If k is a multiple of 24 but not a multiple of 16, which of the following cannot be an integer?


(A) \(\frac{k}{8}\)

(B) \(\frac{k}{9}\)

(C) \(\frac{k}{32}\)

(D) \(\frac{k}{36}\)

(E) \(\frac{k}{81}\)



We are given: k is a multiple of 24, k can be any of the following: 24, 48, 72, 96, 120, 144...

We are also told: k is not a multiple of 16. Thus, we can eliminate: 48, 96, 144...

We are left with values: 24, 72, 120...

On substituting in the values above in the given options, 24/32 , 72/32, 120/32 are not integers.

Thus, the answer is C.
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Re: If k is a multiple of 24 but not a multiple of 16, which of the follow &nbs [#permalink] 19 Aug 2018, 06:19
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