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Bunuel
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What is the smallest possible integer n such that n^3 - 11n^2 + 32n - 06 Apr 2020, 11:14
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C
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65% (hard)

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61% (02:20) correct 39% (02:18) wrong based on 263 sessions

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What is the smallest possible integer n such that \(n^3 - 11n^2 + 32n - 28 > 0\) ?

A. 6
B. 7
C. 8
D. 9
E. 10

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C
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What is the smallest possible integer n such that n^3 - 11n^2 + 32n - 06 Apr 2020, 11:49
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Bunuel
What is the smallest possible integer n such that \(n^3 - 11n^2 + 32n - 28 > 0\) ?

A. 6
B. 7
C. 8
D. 9
E. 10

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\(n^3 - 11n^2 + 32n - 28 > 0\)
n*(n^2-11n+32)>28

use plug in ; at n=8 we get 8*(64-88+32) ; 64
sufficient
OPTION C
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What is the smallest possible integer n such that n^3 - 11n^2 + 32n - 06 Apr 2020, 14:12
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n^3−11n^2+32n−28>0
n*(n^2-11n+32)>28

n=6
12<28

n=7
28=28

n=8
64>28
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What is the smallest possible integer n such that n^3 - 11n^2 + 32n - 06 Apr 2020, 19:14
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[(11= 2+2+7) and (28=2*2*7)]

\(n^3 - 11n^2 + 32n - 28 \) = \((n-2)^2 (n-7) \)

(n-2)^2 is always non-negative; Hence if \(n^3 - 11n^2 + 32n - 28 >0 \) , then (n-7)>0.

Smallest value n can take is 8.




Bunuel
What is the smallest possible integer n such that \(n^3 - 11n^2 + 32n - 28 > 0\) ?

A. 6
B. 7
C. 8
D. 9
E. 10

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What is the smallest possible integer n such that n^3 - 11n^2 + 32n - 06 Apr 2020, 20:09
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Bunuel
What is the smallest possible integer n such that \(n^3 - 11n^2 + 32n - 28 > 0\) ?

A. 6
B. 7
C. 8
D. 9
E. 10

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n^3-11n^2+32n-28>0\(\)

n(n^2-11n+32)-28>0\(\)

n(n^2-11n+ 28+4)-28>0\(\)

n(n-4)(n-7)+4n-28>0
n(n-4)(n-7)+4 (n-7)>0
(n-7)(n(n-4)+4)>0
(n-7) (n^2-4n+4)>0\(\)
(n-7)(n-2)^2>0\(\)

least value of n should be more than 7

from the answers the least value more than 7 is 8. Hence 8 is the answer
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What is the smallest possible integer n such that n^3 - 11n^2 + 32n - 09 Apr 2020, 04:17
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Bunuel
What is the smallest possible integer n such that \(n^3 - 11n^2 + 32n - 28 > 0\) ?

A. 6
B. 7
C. 8
D. 9
E. 10

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Solution:

There is no easy factoring technique for this 3rd degree polynomial, so we are left with direct substitution.

We can directly substitute the numbers from our given choices into the left hand side of the inequality and see if the value is greater than 0. Since we are looking for the smallest possible integer, let’s start with 6.

A. n = 6: 6^3 - 11(6)^2 + 32(6) - 28 = 216 - 396 + 192 - 28 = 408 - 424 = -16

Since -16 < 0, choice A is not the correct answer.

B. n = 7: 7^3 - 11(7)^2 + 32(7) - 28 = 343 - 539 + 224 - 28 = 567 - 567 = 0

Since 0 = 0, choice B is not the correct answer.

C. n = 8: 8^3 - 11(8)^2 + 32(8) - 28 = 512 - 704 + 256 - 28 = 768 - 732 = 36

Since 36 > 0, choice C is the correct answer.

Answer: C
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What is the smallest possible integer n such that n^3 - 11n^2 + 32n - 16 Jul 2024, 09:58
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n*(n^2-11n+32)>28

Generally a good way to start while plugging values is to try with the middle value (8 in this case). It will give a sense of direction.

Co-incidentally 8 is the answer in this case. Might not happen so in other problems.
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