A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came along and erased one number. The average of the remaining numbers is <span class="math-tex">\(35 \dfrac{7}{17}\)</span>. What was the number erased?
Step-by-step Solution:
To solve this problem, we can set up an equation for the average of the remaining numbers and use it to find the original number of integers (n) and the erased number (x).
The equation for the average is: $\frac{S_{new}}{n-1} = \frac{\frac{n(n+1)}{2} - x}{n-1} = \frac{602}{17}$.
The average of the remaining numbers ($\approx 35.4$) must be very close to the average of the original set of numbers, which is approximately $\frac{n}{2}$.
So, $\frac{n}{2} \approx 35.4$, which means $n \approx 70.8$. Thus, 'n' is likely to be 69, 70 or 71.
From our average equation, the sum of the remaining numbers is $S_{new} = (n-1) \times \frac{602}{17}$.
Since the sum of integers must be an integer, (n-1) must be a multiple of 17.
Let's look for a multiple of 17 near our estimate for n ($ \approx 70 $):
So, the original number of integers, n, must be 69.
Now that we know n=69, we can find the original sum and the new sum.
The erased number 'x' is the difference between the original sum and the new sum:
x = Original Sum - New Sum = 2415 - 2408 = 7.
The number that was erased is 7.