На доске было написано 10 последовательных натуральных чисел. Когда стерли одно из них, то сумма девяти оставшихся оказалась равна 2002. Какие числа остались на доске?
На доске было написано 10 последовательных натуральных чисел. Когда стерли одно из них, то сумма девяти оставшихся оказалась равна 2002. Какие числа остались на доске?
На доске было написано 10 последовательных натуральных чисел. Когда стерли одно из них, то сумма девяти оставшихся оказалась равна 2002. Какие числа остались на доске? ...
GPT-4.1
To solve this problem, we start by considering the 10 consecutive natural numbers as \( n, n+1, n+2, \ldots, n+9 \). The total sum of these numbers is:
\[
\text{Total Sum} = n + (n+1) + (n+2) + \ldots + (n+9) = 10n + 45
\]
One of these numbers was erased, and the sum of the remaining nine numbers is 2002. Let's denote the erased number as \( n + k \), where \( k \) ranges from 0 to 9 (since there are 10 numbers). The sum of the remaining numbers is:
\[
\text{Sum of Remaining Numbers} = (10n + 45) - (n + k) = 9n + 45 - k = 2002
\]
Solving for \( n \):
\[
9n = 2002 + k - 45 = 1957 + k
\]
\[
n = \frac{1957 + k}{9}
\]
We need \( n \) to be a natural number, so \( 1957 + k \) must be divisible by 9. Testing values of \( k \) from 0 to 9:
\[
\begin{align*}
k = 5: & \quad n = \frac{1957 + 5}{9} = \frac{1962}{9} = 218 \\
\end{align*}
\]
This is the only natural number solution for \( n \) in this range. Therefore, the numbers are from 218 to 227, and the erased number is \( n + 5 = 223 \). The remaining numbers are:
\[
218,\ 219,\ 220,\ 221,\ 222,\ 224,\ 225,\ 226,\ 227
\]
**Answer:** 218, 219, 220, 221, 222, 224, 225, 226, and 227—the sequence missing the number 223.
To solve this problem, we start by considering the 10 consecutive natural numbers as \( n, n+1, n+2, \ldots, n+9 \). The total sum of these numbers is:
\[
\text{Total Sum} = n + (n+1) + (n+2) + \ldots + (n+9) = 10n + 45
\]
One of these numbers was erased, and the sum of the remaining nine numbers is 2002. Let's denote the erased number as \( n + k \), where \( k \) ranges from 0 to 9 (since there are 10 numbers). The sum of the remaining numbers is:
\[
\text{Sum of Remaining Numbers} = (10n + 45) - (n + k) = 9n + 45 - k = 2002
\]
Solving for \( n \):
\[
9n = 2002 + k - 45 = 1957 + k
\]
\[
n = \frac{1957 + k}{9}
\]
We need \( n \) to be a natural number, so \( 1957 + k \) must be divisible by 9. Testing values of \( k \) from 0 to 9:
\[
\begin{align*}
k = 5: & \quad n = \frac{1957 + 5}{9} = \frac{1962}{9} = 218 \\
\end{align*}
\]
This is the only natural number solution for \( n \) in this range. Therefore, the numbers are from 218 to 227, and the erased number is \( n + 5 = 223 \). The remaining numbers are:
\[
218,\ 219,\ 220,\ 221,\ 222,\ 224,\ 225,\ 226,\ 227
\]
**Answer:** 218, 219, 220, 221, 222, 224, 225, 226, and 227—the sequence missing the number 223.
To solve this problem, we start by considering the 10 consecutive natural numbers as $ n, n+1, n+2, \ldots, n+9 $. The total sum of these numbers is:
One of these numbers was erased, and the sum of the remaining nine numbers is 2002. Let's denote the erased number as $ n + k $, where $ k $ ranges from 0 to 9 (since there are 10 numbers). The sum of the remaining numbers is:
[
9n = 2002 + k - 45 = 1957 + k
]
[
n = \frac{1957 + k}{9}
]
We need $ n $ to be a natural number, so $ 1957 + k $ must be divisible by 9. Testing values of $ k $ from 0 to 9:
[
\begin{align*}
k = 5: & \quad n = \frac{1957 + 5}{9} = \frac{1962}{9} = 218 \
\end{align*}
]
This is the only natural number solution for $ n $ in this range. Therefore, the numbers are from 218 to 227, and the erased number is $ n + 5 = 223 $. The remaining numbers are:
