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Four consecutive even numbers are removed from the set $\left\{1, 2, 3, ..., n\right\}$. If the arithmetic mean of the remaining numbers is 51.5625, determine the value of n and which numbers are removed:

Attemp: $ 515625= (1+1+(n-1) \cdot r) \cdot \frac{n}{2}$. But it doesn't work

Answer: N = 100, and the numbers taken are 22, 24, 25 and 28

Lambert macuse
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1 Answers1

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Suppose that the four evens are $$ r-3, r-1, r+1, r+3 $$ where $r$ is some odd number. Then the sum of the set-with-items-removed is $$ Q = \left( \frac{n(n+1)}{2} \right) - 4r $$ and the number of items is $n-4$, so the average is $$ \frac{\left( \frac{n(n+1)}{2} \right) - 4r}{n-4} = 51 \frac{9}{16} $$ Cross-multiply to get $$ 16 \left(\left( \frac{n(n+1)}{2} \right) - 4r\right) = 459(n-4) $$ so \begin{align} 8n(n+1) - 64r &= 459(n-4)\\ 8n^2 + 8n - 64r &= 459n-1836 \\ 8n^2 - 451n - 64r + 1836 &= 0 \\ \end{align} Rewriting $r = 2k+1$ (because it's odd), we get \begin{align} 0 &= 8n^2 - 451n - 64r + 1836 \\ 0 &= 8n^2 - 451n - 64(2k+1) + 1836 \\ 0 &= 8n^2 - 451n - 128k - 64 + 1836 \\ 0 &= 8n^2 - 451 n - 128k + 1772 \\ 451n - 1772 &= 8n^2 - 128k \\ 451n - 1772 &= 8(n^2 - 16k) \\ \end{align} so $8$ must divide $451n - 1772$. Taken mod 8, this says that $0 = 3n - 4 (\bmod 8)$.

Then I'd observe that for $n > 120$, the average of the numbers is at least $58.5$, and that for $n < 90$, the average is at most $7$, so you've only got a few values of $n$ to try, namely $n = 92, 100, 108, 116$ (because of the mod-8 restriction).

Plug in each of these $n$ values and see whether there's a nonnegative integer $k$ that gets you the result you want.

John Hughes
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  • Note that you already had a $n\equiv 4 \bmod 16$ restriction from the observation that $51.5625=\frac{825}{16}=51 \frac{9}{16}$ – hardmath Apr 27 '21 at 15:09
  • Indeed...but I was just writing down what I thought as I thought of it, and I didn't notice that. :( – John Hughes Apr 27 '21 at 15:32