RMO practice question 6

Let $n$ be the number of terms of the sequence and $x$ be the number erased. Then, we have:

$\begin{aligned} \frac{\frac{n(n+1)}{2}-x}{n-1} & = \frac{602}{17} \\ \frac{n(n+1)}{2}-x & = \frac{602(n-1)}{17} \\ \Rightarrow x & = \frac{n(n+1)}{2} - \frac{602(n-1)}{17} \end{aligned}$

For integer $x$ to be an integer, $n-1$ must be divisible by $17$ . Therefore, $n-1=17m$ $\Rightarrow n = 17m+1$ , where $m$ is a positive integer.

We also note that $0 < x < n$ , therefore,

$0 < \dfrac{n(n+1)}{2} - \dfrac{602(n-1)}{17} < n \\ 0 < \dfrac{(17m+1)(17m+2)}{2} - 602m < 17m+1 \\ 0 < 289m^2-1153m+2 < 34m+2$

$\Rightarrow \begin{cases} 289m^2-1153m+2 > 0 & & \Rightarrow m > 3.987884017 \\ 289m^2-1153m+2 < 34m + 2 & \Rightarrow 289m^2-1187m < 0 & \Rightarrow m < 4.107266436 \end{cases}$

$\begin{aligned} \Rightarrow 3.987 < m < 4.107 \quad \Rightarrow m & = 4 \\ n & = 17(4)+1 = 69 \\ x & = \frac{(69)(70)}{2} - 602(4) = \boxed{7} \end{aligned}$

Denote $n$ to be the number of numbers and $k$ to be the number erased.

Thus, this implies that $\frac{\frac{n(n+1)}{2} - k}{n-1} = \frac{602}{17}$

You can continue from this and bash using mods and number theory techniques, but I solved this using using a very elementary method.

Notice how the denominator of the average is $17$ . This implies that $n \equiv 1 \text{ (mod 17) }$ . We also know that $43 | 602 \implies n > 43$ .

Now we check $n = 52$ which does not work.

Then check $n = 69$ , and ... it works with $k = 7$ !

As you check more and more cases it becomes evident that it is not possible for the conditions to hold true, you can prove this rigorously using induction.

Therefore, the erased number is $\boxed{7}$ .

1. Since consecutive positive integers and starts from $1$ , and after removing one number the average becomes as $\frac{602}{17}=35.4118...$ , hence the total number, $n$ would be somewhere around $35.4118\times 2 = 70.8$ , say $71$ .
2. Also, since the fraction has a denominator of $17$ , so, $n-1$ is divisible by $17$ . And we can easily see that $68$ is the nearest number divisible by $17$ . Hence, $n=68+1=69$ .
3. Now, assuming $x$ be the missing number, a/q, $\frac{602}{17}=\frac{1+2+3+...+69-x}{68}=\frac{\frac{69\times 70}{2}-x}{68}$ gives the missing number to be $7$ .

First recognise that the average is $\frac{602}{17} = 35.41$ This indicates that the maximum number in the set should be around 70.

The set ${1,2,3,...71}$ leads to an average too big, even if you remove the largest number (71). The set ${1,2,3,...,68}$ leads to an average too small, even if you remove the smallest possible number number (1).

Thus the largest digit is either $n=69$ or $n=70$

Taking the same equation as line 2 of Alan's solution, make the RHS $\frac{(n-1)(602)}{17}$

For an integer solution, 17 must divide $\left(n-1\right)$ . Thus $n=69$ , giving $n-1=68$ Then solve for $k$ as Alan's equation.