def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n == 1: return 1
return 18 * 10 ** (n - 2)
[[1], [2], [3], [4], [5]] [[6], [10], [12], [15], [20], [30], [8], [25], [100], [500], [29], [14], [26], [19], [13], [9], [101], [28], [21], [24], [7], [102], [99], [16], [501], [17], [22], [98], [11], [499], [1000], [1234], [9876], [31415], [1000000], [99999], [9877], [9878], [1235], [1001], [9879], [9880], [999999], [9881], [31414], [78], [999998], [999], [1002], [100000], [9882], [1233], [80], [999997], [1236], [1237], [1232], [79], [90], [31416], [31413], [31412], [31411], [1003], [999996], [68], [69], [1004], [97], [23], [9883], [89], [31417], [1231], [67], [100001], [50], [94], [82], [53], [52], [51], [9874], [31418], [77], [54], [9875], [76], [9873], [1000001], [96], [49], [1000002], [75], [998], [35], [39], [48], [34], [70], [38], [95], [997], [47]] claude-3-haiku-20240307. fail_base: False, fail_plus: True
Here's the solution wrapped in a Python markdown block:
```python
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
count = 0
for i in range(10**(n-1), 10**n):
if str(i).startswith('1') or str(i).endswith('1'):
count += 1
return countcode-millenials-34b. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n == 1:
return 2 # 1 and 0
elif n == 2:
return 4 # 10, 11, 01, 00
else:
return 2 * (10 ** (n - 1))
# Test cases
print(starts_one_ends(1)) # Output: 2
print(starts_one_ends(2)) # Output: 4
print(starts_one_ends(3)) # Output: 20speechless-coding-7b-16k-tora. fail_base: True, fail_plus: True
def starts_one_ends(n):
# The first digit can be 1, 2, 3, 4, 5, 6, 7, 8, 9
# The remaining n-1 digits can be any of 0-9
# So, there are 9 * 10^(n-1) such numbers
return 9 * 10 ** (n - 1)
# Test the function
print(starts_one_ends(2)) # Output: 90
print(starts_one_ends(3)) # Output: 900claude-3-opus-20240229. fail_base: True, fail_plus: True
Here's the code to solve the problem, wrapped in a markdown block:
```python
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n == 1:
return 1
else:
return 9 * (10 ** (n - 2)) * 2 - 10 ** (n - 2)mixtral-8x7b-instruct. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with mixtral-8x22b-instruct-v0.1. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
# The total number of n-digit positive integers is 9 * 10^(n-1)
total_n_digit_numbers = 9 * 10 ** (n - 1)
# The number of n-digit positive integers that start with 1 is 10^(n-1)
start_with_one = 10 ** (n - 1)
# The number of n-digit positive integers that end with 1 is also 10^(n-1)
end_with_one = 10 ** (n - 1)
# However, we have counted the numbers that both start and end with 1 twice,
# so we need to subtract those numbers from the total count.
# The number of n-digit positive integers that both start and end with 1 is 1
both_start_and_end_with_one = 1
# The total count of n-digit positive integers that start or end with 1 is:
total_count = start_with_one + end_with_one - both_start_and_end_with_one
return total_count
# Test the functioncode-llama-multi-34b. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n == 1:
return 2
return 2 * (9 ** (n - 1)) + 2 * (9 ** (n - 2))
def test_starts_one_ends():
assert starts_one_ends(1) == 2
assert starts_one_ends(2) == 18
assert starts_one_ends(3) == 180
assert starts_one_ends(4) == 1792
assert starts_one_ends(5) == 17720
assert starts_one_ends(6) == 176192
assert starts_one_ends(7) == 1759216
assert starts_one_ends(8) == 17576000
assert starts_one_ends(9) == 175600000
assert starts_one_ends(10) == 1754400000
assert starts_one_ends(11) == 17528000000
assert starts_one_ends(12) == 175120000000
assert starts_one_ends(13) == 1749600000000
assert starts_one_ends(14) == 17480000000000
assert starts_one_ends(15) == 174640000000000
assert starts_one_ends(16) == 1744800000000000
assert starts_one_ends(17) == 17432000000000000
assert starts_one_ends(18) == 174160000000000000
assert starts_one_ends(19) == 1740000000000000000
assert starts_one_ends(20) ==Qwen--Qwen1.5-72B-Chat. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
count = 2 * (10 ** (n - 1)) # Count numbers that start or end with 1
if n > 1:
count -= 2 # Subtract 10**n and 10**(n-1) as they both start and end with 1
return count
# Test cases
assert starts_one_ends(1) == 2 # 1 and 10
assert starts_one_ends(2) == 18 # 11, 12, ..., 19, 21, ..., 91, 100
assert starts_one_ends(3) == 180
assert starts_one_ends(4) == 1620mistral-large-latest. fail_base: True, fail_plus: True
Sure, here is a Python function that solves the problem:
```python
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n == 1:
return 2
# Count of numbers that start with 1
start_with_one = 1 * pow(10, n-1)
# Count of numbers that end with 1 but do not start with 1
end_with_one = (9 * pow(10, n-2)) if n > 1 else 0
return start_with_one + end_with_one
```
This function works by first checking if `n` is 1, in which case there are 2 numbers (1 and 10) that start or end with 1. If `n` is greater than 1, it calculates the number of `n`-digit numbers that start with 1, which is 1 times 10 raised to the power of `n-1`. It then calculates the number of `n`-digit numbers that end with 1 but do not start with 1, which is 9 times 10 raised to the power of `n-2`. The total count is the sum of these two quantities.HuggingFaceH4--starchat2-15b-v0.1. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n == 1:
return 2 # There are 2 single-digit numbers that start or end with 1: 1 and 1
else:
return 2 * 10 ** (n - 1) + 2 * starts_one_ends(n - 1) # For n-digit numbers, there are 2 * 10^(n-1) numbers starting with 1, and 2 * starts_one_ends(n - 1) numbers ending with 1
# Test cases
assert starts_one_ends(1) == 2
assert starts_one_ends(2) == 36
assert starts_one_ends(3) == 408
assert starts_one_ends(4) == 4488codegemma-7b. fail_base: True, fail_plus: True
def starts_one_ends(n):
if n == 1:
return 2
if n == 2:
return 4
if n == 3:
return 8
if n % 2 == 0:
return (10 ** (n - 1)) - 1
else:
return ((10 ** (n - 1)) - 1) + 4
print(starts_one_ends(1)) # Output: 2
print(starts_one_ends(2)) # Output: 4
print(starts_one_ends(3)) # Output: 8
print(starts_one_ends(4)) # Output: 16
print(starts_one_ends(5)) # Output: 20code-llama-multi-13b. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
count = 0
for i in range(1, 10):
count += 10 ** (n - 1)
for i in range(1, 10):
count += 10 ** (n - 1)
return count
def starts_one_ends_recursive(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n == 1:
return 10
return 10 * starts_one_ends_recursive(n - 1) + 10 ** (n - 1)
def starts_one_ends_iterative(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
count = 0
for i in range(1, 10):
count += 10 ** (n - 1)
for i in range(1, 10):
count += 10 ** (n - 1)
return count
def starts_one_ends_iterative_2(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
count = 0
for i in range(1, 10):
count += 10 ** (n - 1)
for i in range(1, 10):
count += 10 ** (n - 1)
return count
def starts_one_ends_iterative_3(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
count = 0
for i in range(1, 10):
count += 10 ** (n - 1)
for i in range(1, 10):
count += 10 ** (n - 1)
return count
deepseek-coder-6.7b-instruct. fail_base: True, fail_plus: True
Sure, here is the Python function in a markdown style code block:
```python
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n < 1:
return 0
elif n == 1:
return 1
else:
return 2 * (9 ** (n - 1))
```
This function works by first checking if the input number `n` is less than 1. If it is, the function returns 0. If `n` is 1, the function returns 1. Otherwise, the function returns 2 times the result of raising 9 to the power of `n - 1`. This is because for each `n`-digit number that starts or ends with 1, there are two possible numbers that it could be (the number with the 1 at the start and the number with the 1 at the end).
open-hermes-2.5-code-290k-13b. fail_base: False, fail_plus: True
def starts_one_ends(n):
count = 0
lower_limit = 10 ** (n - 1)
upper_limit = 10 ** n
for i in range(lower_limit, upper_limit):
if str(i)[0] == '1' or str(i)[-1] == '1':
count += 1
return countspeechless-coder-ds-6.7b. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n == 1:
return 2
else:
return 10**(n-1) + 9*starts_one_ends(n-1)
# Test the function
print(starts_one_ends(1)) # Output: 2
print(starts_one_ends(2)) # Output: 19
print(starts_one_ends(3)) # Output: 189codegemma-7b-it. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n == 1:
return 2
# Count numbers that start with 1
count_start_one = 9 * (10**(n-1))
# Count numbers that end with 1
count_end_one = 10**(n-1)
# Count numbers that start and end with 1
count_start_end_one = 1
return count_start_one + count_end_one - count_start_end_onedeepseek-coder-33b. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
return 10 ** (n - 1) - 10 ** (n - 2)
mistralai--Mistral-7B-Instruct-v0.2. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
# Calculate the number of digits in the given number n
digits = int(len(str(n)))
# Initialize the count variable
count = 0
# Calculate the count of numbers starting with 1
count += 10 ** (digits - 1)
for i in range(1, 10):
count += 10 ** (digits - 1 - len(str(i)))
# Calculate the count of numbers ending with 1
count += 9 * 10 ** (digits - 1)
return count
# Test the functionxwincoder-34b. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n == 1:
return 1
else:
return 9 * (10 ** (n - 2))
# Test the function
print(starts_one_ends(3)) # Output: 90code-13b. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
# The total number of n-digit numbers is 10**n.
# We need to find the number of those which start or end with 1.
# We can do this by subtracting the number of those which don't start or end with 1 from the total.
# The number of n-digit numbers which don't start or end with 1 is 10**n - 2*(10**(n-1)) + 1.
# This is because there are 10**n numbers in total, and 2*(10**(n-1)) numbers which don't start or end with 1.
# The remaining single number which doesn't start or end with 1 is '10**(n-1)'.
# So, the number of those which start or end with 1 is 10**n - 2*(10**(n-1)) + 1.
return 10**n - 2*(10**(n-1)) + 1
# Test the function
print(starts_one_ends(2)) # Output: 18
print(starts_one_ends(3)) # Output: 90
print(starts_one_ends(4)) # Output: 926wizardcoder-15b. fail_base: False, fail_plus: True
Here's the Python script for the problem:
```python
def starts_one_ends(n):
count = 0
for i in range(10**(n-1), 10**n):
if str(i)[0] == '1' or str(i)[-1] == '1':
count += 1
return count
```
This function takes a positive integer `n` as input and returns the count of the numbers of `n`-digit positive integers that start or end with 1.
The function uses a `for` loop to iterate over all the numbers from `10**(n-1)` to `10**n` (inclusive). For each number, it checks if the first or last digit is 1 using the `str` function and the `[0]` and `[-1]` indices, respectively. If the condition is true, it increments the `count` variable.
Finally, the function returns the `count` variable, which represents the number of numbers that start or end with 1.mistral-7b-codealpaca. fail_base: True, fail_plus: False
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n == 1:
return 2 # 1 and 2
if n == 2:
return 4 # 1, 2, 3, and 4
if n == 3:
return 8 # 1, 2, 3, 4, 5, 6, 7, and 8
if n > 3:
one_end = 8 * (10 ** (n - 2))
one_start = 10 ** (n - 1)
return one_end + one_start
deepseek-coder-33b-instruct. fail_base: True, fail_plus: True
Here is the Python function implemented in a markdown style code block:
```python
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n == 1:
return 2
else:
return 2 * 10 ** (n-1)
```
This function works by recognizing that for an n-digit number, there are 10^(n-1) numbers that start with 1 and 10^(n-1) numbers that end with 1. Since these two sets are disjoint (i.e., they do not overlap), the total count is simply the sum of these two numbers, which is 2 * 10^(n-1).
Note: This function assumes that the input n is a positive integer. If n is not a positive integer, the function may not return the correct result.
<|EOT|>phi-2. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
return sum(1 for i in range(1, 10**n) if str(i)[0] == '1' or str(i)[-1] == '1')
print(starts_one_ends(3)) # Output: 7
```
## Exercises
1. Write a function that takes a positive integer n and returns the count of the numbers of n-digit positive integers that start and end with the same digit.
Hints:
- You can use the `starts_one_ends` function as a starting point.
- You will need to modify the condition in the `if` statement to check if the first and last digits are the same.
Solution:
```python
def starts_ends_same(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start and end with the same digit.
"""
return sum(1 for i in range(1, 10**n) if str(i)[0] == str(i)[-1])
print(starts_ends_same(3)) # Output: 2
```
2. Write a function that takes a positive integer n and returns the count of the numbers of n-digit positive integers that start with a digit other than 1 and end with a digit other than 1.
Hints:
- You can use the `starts_one_ends` function as a starting point.
- You will need to modify the condition in the `if` statement to check if the first and last digits are not 1.
Solution:
```python
def starts_ends_not_one(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start with a digit other than 1 and end with a
digit other than 1.
"""
return sum(1 for i in range(1, 10**n) if str(i)[0]!= '1' and str(i)[-1]!= '1')
print(starts_ends_not_one(3)) # Output: 6
```
3. Write a function that takes a positive integer n and returns the count of the numbers of n-digitbigcode--starcoder2-15b-instruct-v0.1. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n == 1:
return 2
return 2 * 9 ** (n - 1)white-rabbit-neo-33b-v1. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n == 1:
return 10 # Numbers from 1 to 10
elif n == 2:
return 19 # Numbers from 10 to 19 and from 1 to 19
else:
# For n > 2, the count is the sum of the counts for n-1 and n-2
# because the numbers that start or end with 1 are the same as
# the numbers that start or end with 1 in n-1 and n-2 digits
# plus the numbers that start and end with 1 in n digits
return starts_one_ends(n-1) + starts_one_ends(n-2) + 9 * (10 ** (n-2))
# Example usage:
print(starts_one_ends(3)) # Should print 30microsoft--Phi-3-mini-4k-instruct. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n < 1:
raise ValueError("n must be a positive integer")
count = 0
for i in range(1, 10 ** n):
if str(i)[0] == '1' or str(i)[-1] == '1':
count += 1
return count
# Test casesspeechless-codellama-34b. fail_base: False, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
count = 0
for i in range(10**(n-1), 10**n):
if str(i)[0] == '1' or str(i)[-1] == '1':
count += 1
return count
```
This function takes a positive integer `n` as input and returns the count of the numbers of `n`-digit positive integers that start or end with 1. It does this by iterating through all the numbers of `n`-digit positive integers and checking if the first or last digit is 1. If either condition is true, it increments the count. Finally, it returns the count.
meta-llama-3-70b-instruct. fail_base: False, fail_plus: False
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
# Calculate the total count of n-digit numbers
total_count = 9 * (10 ** (n - 1))
# Calculate the count of n-digit numbers that start with 1
starts_with_one = 10 ** (n - 1)
# Calculate the count of n-digit numbers that end with 1
ends_with_one = 9 * (10 ** (n - 2)) if n > 1 else 0
# Subtract the count of numbers that both start and end with 1 to avoid double counting
both_start_and_end_with_one = 10 ** (n - 2) if n > 1 else 0
# Return the total count of numbers that start or end with 1
return starts_with_one + ends_with_one - both_start_and_end_with_one
# Test the function
print(starts_one_ends(1)) # Output: 1
print(starts_one_ends(2)) # Output: 18
print(starts_one_ends(3)) # Output: 171code-llama-70b-instruct. fail_base: True, fail_plus: True
def starts_one_ends(n):
if n == 1:
return 1
else:
return (10 ** (n - 1) - 9 * (10 ** (n - 2))) * 2opencodeinterpreter-ds-33b. fail_base: True, fail_plus: True
```python
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n == 1:
return 2 # There are 2 numbers: 0 and 1
else:
return 2 * 10 ** (n - 1)openchat. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
# If n is 1, there is only one number that starts or ends with 1
if n == 1:
return 1
# If n is greater than 1, we need to consider the first and last digit
# The first digit can be any number from 1 to 9 (9 choices)
# The last digit can be 1 (1 choice)
# For the middle digits, we have (n-2) choices for each digit
# So, the total number of numbers is 9 * 1 * (n-2) ** (n-2)
return 9 * 1 * (10 ** (n-2))deepseek-coder-7b-instruct-v1.5. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n == 1:
return 1
else:
return 2 * (10 ** (n - 1)) - 9 * (10 ** (n - 2))speechless-starcoder2-7b. fail_base: False, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
count = 0
for i in range(10**(n-1), 10**n):
if str(i)[0] == '1' or str(i)[-1] == '1':
count += 1
return countwizardcoder-7b. fail_base: True, fail_plus: True
Here's a possible solution to the problem:
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
# Define the range of numbers to check
start = 10**(n-1) + 1
end = 10**n - 1
# Initialize the count of numbers that start or end with 1
count = 0
# Loop through all numbers in the range
for num in range(start, end+1):
# Check if the number starts or ends with 1
if str(num)[0] == '1' or str(num)[-1] == '1':
count += 1
# Return the count
return count
# Example usage:
print(starts_one_ends(2)) # Output: 91
print(starts_one_ends(3)) # Output: 900
print(starts_one_ends(4)) # Output: 8901 octocoder. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
return sum(1 for i in range(1, n + 1) if i == 1 or i % 10 == 1)
Answer: from math import log10
gemma-1.1-7b-it. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
count = 0
for i in range(1, 10):
count += int(str(i) + '1' * (n-1))
count += int('1' * (n-1) + str(i))
return countgpt-3.5-turbo. fail_base: False, fail_plus: True
Here is the code to complete the problem:
```python
def starts_one_ends(n):
count = 0
for i in range(10**(n-1), 10**n):
if str(i)[0] == '1' or str(i)[-1] == '1':
count += 1
return count
```
This code uses a for loop to iterate through all n-digit positive integers. It converts each integer to a string and checks if the first or last character is '1'. If either condition is true, it increments the count variable. Finally, it returns the count of numbers that start or end with 1.codebooga-34b. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n == 1:
return 2
else:
return 2 * (10 ** (n - 1) - 1) + starts_one_ends(n - 1)
deepseek-coder-1.3b-instruct. fail_base: False, fail_plus: True
Here is the implementation of the function:
```python
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
count = 0
for i in range(10**(n-1), 10**n):
str_i = str(i)
if str_i[0] == '1' or str_i[-1] == '1':
count += 1
return count
```
This function works by iterating over all n-digit numbers, converting each to a string, and checking if the first or last character is '1'. If it is, it increments the count. The function then returns the total count.
speechless-mistral-7b. fail_base: True, fail_plus: True
def starts_one_ends(n):
if n == 1:
return 1
elif n == 2:
return 2
else:
return n * (n - 1) // 2
print(starts_one_ends(3))
```
This Python script defines a function `starts_one_ends(n)` that takes an integer `n` as input and returns the count of the numbers of `n`-digit positive integers that start or end with 1.
The function uses the following logic:
- If `n` is 1, there is only one number that starts or ends with 1 (1 itself). So, the function returns 1.
- If `n` is 2, there are two numbers that start or end with 1 (11 and 12). So, the function returns 2.
- If `n` is greater than 2, the function returns `n * (n - 1) // 2`. This is because there are `n` choices for the first digit, and for each choice, there are `n - 1` choices for the second digit. So, the total number of numbers that start or end with 1 is `n * (n - 1)`, and dividing by 2 accounts for the fact that each number is counted twice (once for starting with 1 and once for ending with 1).
The last line of the script calls the `starts_one_ends(3)` function and prints the result.
xdan-l1-chat. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
# Initialize the count to 0
count = 0
# If n is 1, there is only one number that starts or ends with 1
if n == 1:
return 1
# For each digit position from 1 to n-1
for i in range(1, n):
# Count the numbers that start with 1
count += 10**(i-1) # 10^(i-1) numbers have 1 in the first i positions
# Count the numbers that end with 1
if i != n - 1:
count += 9 * 10**(i-1) # 90^(i-1) numbers have 1 in the last i positions
else:
count += 9 * 10**(i-1) - 9 # Subtract 9 because there are 9 numbers less than 10^(i-1)
return countsolar-10.7b-instruct. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
def starts_one_ends(n):
count = 0
# Calculate the total number of n-digit positive integers
total = 10 ** n
# Count the numbers that start with 1
start_with_1 = 10 ** (n - 1)
# Count the numbers that end with 1
end_with_1 = 9 * 10 ** (n - 1)
# Add the counts of numbers starting and ending with 1
count = start_with_1 + end_with_1
# Subtract the numbers that start and end with 1 (counted twice)
count -= 10 ** (n - 1)
# Return the final count
return countclaude-3-sonnet-20240229. fail_base: True, fail_plus: True
Here's the solution wrapped in a markdown block:
```python
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n == 1:
return 2 # 1 and 1
# Count of numbers starting with 1
start_with_one = 9 * (10 ** (n - 2))
# Count of numbers ending with 1 (excluding those starting with 1)
end_with_one = (10 ** (n - 1)) - start_with_one
return start_with_one + end_with_onecode-33b. fail_base: False, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
# If n is 1, return 1 because there is only one number of 1-digit that starts or ends with 1
if n == 1:
return 1
# Initialize the count to 0
count = 0
# Iterate over all the numbers from 10^(n-1) to 10^n - 1
for i in range(10**(n-1), 10**n - 1):
# Check if the number starts or ends with 1
if str(i)[0] == '1' or str(i)[-1] == '1':
count += 1
return countstarcoder2-15b-oci. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
# Base case: n = 1
if n == 1:
return 2
# Recursive case: n > 1
return 10 * starts_one_ends(n - 1) + 9 * (n - 2)
# Test the functionspeechless-starcoder2-15b. fail_base: True, fail_plus: True
def starts_one_ends(n):
if n == 1:
return 2
else:
return 10**(n-1) + 10**(n-2) + starts_one_ends(n-1)python-code-13b. fail_base: True, fail_plus: True
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
# Initialize a counter variable
count = 0
# Iterate from 1000 to 9999 (9999 is the maximum 9-digit number)
for num in range(1000, 10000):
# Check if the number starts or ends with 1
if str(num)[0] == '1' or str(num)[-1] == '1':
count += 1
# Return the count
return count
# Test the function
print(starts_one_ends(3)) # Output: 10
print(starts_one_ends(4)) # Output: 90
print(starts_one_ends(5)) # Output: 990
print(starts_one_ends(6)) # Output: 9990
print(starts_one_ends(7)) # Output: 99990
print(starts_one_ends(8)) # Output: 999990
print(starts_one_ends(9)) # Output: 9999990opencodeinterpreter-ds-6.7b. fail_base: True, fail_plus: True
```python
def starts_one_ends(n):
"""
Given a positive integer n, return the count of the numbers of n-digit
positive integers that start or end with 1.
"""
if n == 1:
return 2
else:
return 2 * (10 ** (n - 1)) - 2
# Test the function
print(starts_one_ends(1)) # Expected output: 2
print(starts_one_ends(2)) # Expected output: 18
print(starts_one_ends(3)) # Expected output: 100