For any natural number \( n \), let \( a_n \) be the largest integer not exceeding \( \sqrt{n} \). Then the value of \( a_1 + a_2 + \cdots + a_{50} \) is
Solution
Solution
We are given:
For any natural number \( n \), let \( a_n = \left\lfloor \sqrt{n} \right\rfloor \)
We need to find:
\[ \sum_{n=1}^{50} \left\lfloor \sqrt{n} \right\rfloor \]
Group the terms based on constant value of \( \left\lfloor \sqrt{n} \right\rfloor \):
| \( \left\lfloor \sqrt{n} \right\rfloor = k \) | Range of \( n \) | Count of Terms | Total Sum Contributed |
|---|---|---|---|
| 1 | 1 to 3 | 3 | \( 1 \times 3 = 3 \) |
| 2 | 4 to 8 | 5 | \( 2 \times 5 = 10 \) |
| 3 | 9 to 15 | 7 | \( 3 \times 7 = 21 \) |
| 4 | 16 to 24 | 9 | \( 4 \times 9 = 36 \) |
| 5 | 25 to 35 | 11 | \( 5 \times 11 = 55 \) |
| 6 | 36 to 48 | 13 | \( 6 \times 13 = 78 \) |
| 7 | 49 to 50 | 2 | \( 7 \times 2 = 14 \) |
Now, sum all contributions:
\[ 3 + 10 + 21 + 36 + 55 + 78 + 14 = \boxed{217} \]
Final Answer:
\[ \boxed{217} \]
When 10^100 is divided by 7, find the remainder.
a. 1
b. 6
c. 3
d. 4
Solution
Finding the Remainder of \( 10^{100} \div 7 \)
Let’s analyze the cyclic pattern of \( 10^n \mod 7 \):
| Power \( n \) | \( 10^n \mod 7 \) |
|---|---|
| 1 | \( 3 \) |
| 2 | \( 2 \) |
| 3 | \( 6 \) |
| 4 | \( 4 \) |
| 5 | \( 5 \) |
| 6 | \( 1 \) |
Since the cycle length is 6, we reduce:
\( 100 \mod 6 = 4 \), so we calculate \( 10^4 \mod 7 \)
Therefore, the remainder when \( 10^{100} \) is divided by 7 is \( \boxed{4} \).
