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The average of $3$ consecutive odd numbers is $14$ more than one third of the first of these numbers, what is the last of these numbers?

$17/19/15/$data inadequate/none of these

Let three consecutive odd numbers be $a-2,a,a+2$. Their average is $a.$ So, $a=\frac13(a-2)+14\implies a=20.$ But that's not possible as $a$ is odd.

What am I missing?

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You're missing nothing at all, from the looks of it. – Zach L. Nov 14 '13 at 5:53
The wording, as is too often the case, is rather poor. There are no such numbers. None of these is probably the intended choice, although for sure "data" that describe an impossible situation are pretty inadequate. – André Nicolas Nov 14 '13 at 6:01
@AndréNicolas- Answer is given as 'data inadequate.' Are you suggesting that? I think that if we change the statement to 'three consecutive even numbers,' then we may say for sure that the answer is $22.$ – Ramit Nov 14 '13 at 9:56
The data is adequate to determine there is no solution. Usually "insufficient" means not enough information. Here you have enough data. But here the data is bad. In the sense of "bad" one could quibble the data is "inadequate". Andre is pointing out that the wording is bad. I'd prefer if there was a "no solution/inpossible" option. – fleablood Jun 5 at 20:52

The average of 3 consecutive odd numbers is 14 more than one third of the first of these numbers, what is the last of these numbers?

Let $a, a + 2, a + 4$ be three consecutive odd numbers

$\frac{3a + 6}{3} = 14 + \frac{a}{3}$

$3a + 6 = 42 + a$

$2a = 36$

$a = 18$, which is impossible because the numbers are odd

So, the answer would be "none of these."

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We take $3$ odd numbers $a,b,c$ such that $a= (2n+1), b=(2n+3), c=(2n+5)$ for some integer $n$. From information given, we know $$\frac{(2n+1)+(2n+3)+(2n+5)}{3} = 14 + \frac{1}{3}(2n + 1) \\ \frac{6n+9}{3}=\frac{43+2n}{3} \\ \frac{4n-34}{3} = 0 \\ n = \frac{34}{4} \\ n=8.5$$ Since n is not an integer, the solution cannot exist.

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