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# If a + 1 = 20/a and b is the average of a set of c consecutive integer

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If a + 1 = 20/a and b is the average of a set of c consecutive integer  [#permalink]

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22 Mar 2017, 00:13
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Difficulty:

65% (hard)

Question Stats:

63% (02:46) correct 37% (02:53) wrong based on 168 sessions

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If $$a + 1 = \frac{20}{a}$$ and b is the average of a set of c consecutive integers, where c is odd, which of the following must be true?

I. $$(a^2)(b^2)(c^2)$$ is even.

II. $$a + b + c$$ is odd.

III. $$ab(c^2 + c)$$ is even.

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

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Re: If a + 1 = 20/a and b is the average of a set of c consecutive integer  [#permalink]

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22 Mar 2017, 01:05
1
ziyuen wrote:
If $$a + 1 = \frac{20}{a}$$ and b is the average of a set of c consecutive integers, where c is odd, which of the following must be true?

I. $$(a^2)(b^2)(c^2)$$ is even.
II. a + b + c is odd.
III. $$ab(c^2 + c)$$ is even.

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

$$a + 1 = \frac{20}{a}$$

$$a^2 + a - 20 = 0$$

a = -5 or 4

a could be even or odd.

c is odd.

b is the average of c consecutive integers. Say 3 consecutive integers (1, 2, 3, then b = 2). Say 5 consecutive integers (1, 2, 3, 4, 5, then b = 3)
b could be even or odd.

I. $$(a^2)(b^2)(c^2)$$ is even.
a, b and c all could be odd in which case this will be odd. So not necessarily true.

II. a + b + c is odd.
If a is 4 and b and c are odd, this will be even. So not necessarily true.

III. $$ab(c^2 + c)$$ is even.
Since c is odd, $$c^2$$ is also odd.
Odd + Odd = Even

Since $$(c^2 + c)$$ is even, and any integer multiplied by an even number gives an even result, $$ab(c^2 + c)$$ is certainly even.

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Re: If a + 1 = 20/a and b is the average of a set of c consecutive integer  [#permalink]

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22 Mar 2017, 01:17
1
2
ziyuen wrote:
If $$a + 1 = \frac{20}{a}$$ and b is the average of a set of c consecutive integers, where c is odd, which of the following must be true?

I. $$(a^2)(b^2)(c^2)$$ is even.
II. a + b + c is odd.
III. $$ab(c^2 + c)$$ is even.

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

Hi

$$a + 1 =\frac{20}{a}$$

$$a^2 + a - 20 = 0$$

$$(a + 5)(a - 4) = 0$$

$$a= -5$$ or $$a=4$$

$$b$$ is the meadian of set S, which consists of $$c$$ consecutive integers, where $$c$$ is odd: S = {1,2,3} --> $$b=2$$ or S={1,2,3,4,5} ----> $$b = 3$$

We have: $$c = odd$$, $$a = even/odd$$ and $$b = even/odd$$.

I. $$(a^2)(b^2)(c^2)$$ is even. We can have $$odd^2*odd^2*odd^2 = odd$$ No.

II. $$a + b + c$$ is odd. We can have $$even + odd + odd = even$$. No.

III. $$ab(c^2 + c)$$ is even.

$$a*b*c*(c + 1) = even$$. $$c$$ is odd hence $$c + 1$$ will always be even. Yes.

Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1321
Location: Malaysia
Re: If a + 1 = 20/a and b is the average of a set of c consecutive integer  [#permalink]

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23 Mar 2017, 17:53
ziyuen wrote:
If $$a + 1 = \frac{20}{a}$$ and b is the average of a set of c consecutive integers, where c is odd, which of the following must be true?

I. $$(a^2)(b^2)(c^2)$$ is even.
II. a + b + c is odd.
III. $$ab(c^2 + c)$$ is even.

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

Source : Manhattan Advanced Quant Question No. 2

OFFICIAL SOLUTION

The three statements concern even/odd status, so we need to consider the odd/even status of a, b, and c. Start by solving the quadratic equation for a:

$$a + 1 = \frac{20}{a}$$
$$a^2 + a = 20$$
$$a^2 + a – 20 = 0$$
$$( a + 5)( a – 4) = 0$$
a = –5 or 4

So a can be either odd or even.

We were told that c is odd.

The average of an odd number of consecutive integers is simply equal to the middle term. That middle term must be an integer, but could be odd or even. Thus, b can be either odd or even.

We might have been able to glance ahead and see that $$c^2 + c$$ is always even for any integer c. There are at least two rationales: first, since $$c^2$$ will have the same odd/even property as c, we are adding either an even and an even or an odd and an odd. The result will always be even. Alternatively, we could factor $$c^2 + c$$ into $$c( c + 1)$$, which will always give us the even product of an odd and an even.

This problem involves a number of different topics: even/odds, averages, and quadratics. None of these topics is tested at an extremely difficult level, but we have to switch among them, making this hybrid problem more difficult than otherwise.

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Re: If a + 1 = 20/a and b is the average of a set of c consecutive integer  [#permalink]

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12 Apr 2018, 05:25
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Re: If a + 1 = 20/a and b is the average of a set of c consecutive integer   [#permalink] 12 Apr 2018, 05:25
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