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

\(a + 1 = \frac{20}{a}\)

\(a^2 + a - 20 = 0\)

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.

Answer (C)
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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.

Answer C.

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.

The correct answer is C.

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