so i'm creating this function that gets a user to input a number greater than 2.
The code should then print all the prime numbers starting from 2 and ending at the number the user inputted.
So here is my code and it works (yay!)
def enterNumber():
number = int(input("Enter a number greater than 2"))
lower = 2
upper = number
for number in range (lower, upper):
if number > 2:
for i in range (2, number):
if (number % i) == 0:
break
else:
if (number % i) != 0:
print(number)
enterNumber()
BUT here is my output if the user inputs 18
"Enter a number greater than 2"
user puts "18"
output: 3
5
5
5
7
7
7
7
7
9
11
11
11
11
11
11
11
11
11
13
13
13
13
13
13
13
13
13
13
13
15
17
17
17
17
17
17
17
17
17
17
17
17
17
17
17
why is my output repeating itself? Any suggestions would be greatly appreciated thanks!
Your problem is here:
for i in range (2, number):
if (number % i) == 0:
break
else:
if (number % i) != 0:
print(number)
The else being inside the loop, says that you're going to print the number every time you find a non-divisor. For instance, for number = 11, you will print 11 for i values 2-10.
You don't know whether the number is a prime until you are done with the loop. Your print has to go after the loop, on a normal exit. An often-forgotten language feature is that a loop can have an else for exactly this purpose.
for i in range (2, number):
if (number % i) == 0:
break
else:
print(number)
Output:
Enter a number greater than 218
3
5
7
11
13
17
A simple solution(there are better ways):
def enterNumber():
number = int(input("Enter a number greater than 2"))
lower = 2
upper = number
for number in range(lower, upper):
prime = True
for i in range(2, number):
if (number % i) == 0:
prime = False
break
if prime:
print(number)
enterNumber()
Related
While writing a program to find the prime numbers in a certain range, the code prints every prime number multiple times instead of just listing it once.
This is the code:
first_interval = int(input('enter the beginning of the interval'))
second_interval=int(input('enter the end of the interval'))
for digit in range((first_interval),(second_interval+1)):
if digit>1:
for i in range(2,digit):
if (digit %i)==0:
break
else: print(digit)
I was expecting every prime number in the interval (4,21) to be listed just once.
What I was expecting:
7
11
13
17
19
What I got:
5
5
5
7
7
7
7
7
9
11
11
11
11
11
11
11
11
11
13
13
13
13
13
13
13
13
13
13
13
15
17
17
17
17
17
17
17
17
17
17
17
17
17
17
17
19
19
19
19
19
19
19
19
19
19
19
19
19
19
19
19
19
21
When generating a list of prime numbers in a range you first need to ensure that the low value is at least 2. 2 is the lowest and only even prime number and therefore for greatest efficiency should be treated as a special case.
Once 2 has been dealt with (if relevant) then your loop can increment the test value by 2 because from then on you only need to work with odd numbers.
Here's an efficient function that tests for prime numbers followed by a suggestion for how you could implement your code:
from math import sqrt, floor
def isprime(n):
if n < 2:
return False
if n == 2 or n == 3:
return True
if n % 2 == 0 or n % 3 == 0:
return False
for i in range(5, floor(sqrt(n))+1, 6):
if n % i == 0 or n % (i + 2) == 0:
return False
return True
start = int(input('enter the beginning of the interval: '))
stop = int(input('enter the end of the interval: '))
# ensure that the start value is at least 2
# if it's exactly 2 just print it
if (start := min(max(2, start), stop)) == 2
print(2)
# ensure that the start value is an odd number
start |= 1
# now work in increments of 2
for n in range(start, stop+1, 2):
if isprime(n):
print(n)
Output:
enter the beginning of the interval: 4
enter the end of the interval: 21
5
7
11
13
17
19
Instead of printing out digit on every iteration, use a variable is_prime and set it to False whenever you find a divisor of digit, then print out digit after the for loop if is_prime turns out to be True.
first_interval = int(input('enter the beginning of the interval: '))
second_interval = int(input('enter the end of the interval: '))
for digit in range(first_interval, second_interval+1):
if digit > 1:
is_prime = True
for i in range(2, digit):
if (digit % i) == 0:
is_prime = False
break
if is_prime:
print(digit)
Sample input/output:
enter the beginning of the interval: 1
enter the end of the interval: 100
2
3
5
7
11
13
17
19
23
29
31
37
41
43
47
53
59
61
67
71
73
79
83
89
97
You could use all to check whether digit is prime or false and print it:
first_interval = int(input('enter the beginning of the interval : '))
second_interval=int(input('enter the end of the interval : '))
for digit in range(first_interval,second_interval):
if digit > 1 and all(digit % i for i in range(2,digit)):
print(digit)
enter the beginning of the interval : 4
enter the end of the interval : 21
5
7
11
13
17
19
It is a good idea to first write a function that determines if a number is prime or not and then call this function for each number in the desired range:
def is_prime(number):
if number < 2:
return False
for i in range(2,number):
if number % i == 0:
return False
return True
first_interval = int(input('enter the beginning of the interval: '))
second_interval = int(input('enter the end of the interval: '))
for digit in range(first_interval,second_interval+1):
if is_prime(digit):
print(digit)
[1 3 6 5 7 10 9 11 14 13 15 18 17 19 22 21 23 26]
I have to display these numbers with only one print command and a while loop.
I got it with several print commands but when I was challenged to use one, I just couldn't figure it out.(beginner btw)
Your list follows a pattern, it is +2 +3 -1. So, you can just write something like the following code
index = 0
number = 1
while(index < 18):
print(number)
index += 1
if(index%3 == 1):
number += 2
elif(index%3 == 2):
number += 3
elif(index%3 == 0):
number -= 1
I am new to Python and Stackoverflow in general, so sorry if my formatting sucks and i'm not good at enlish.But i have a problem with this code.
print('Displays prime numbers from 1 to N.')
n = int(input('Please enter a value of n: '))
for n in range(1, n + 1):
if n >= 1:
for i in range(2, n):
if (n % i) == 0:
break
else:
print('They are',n,end=' ')
The result of the code when ran comes out looking like this:
Displays prime numbers from 1 to N.
Please enter a value of n:40
They are 1 They are 2 They are 3 They are 5 They are 7 They are 11 They are 13 They are 17 They are 19 They are 23 They are 29 They are 31 They are 37
but i want it like this:
Displays prime numbers from 1 to N.
Please enter a value of n:40
They are 1 2 3 5 7 11 13 17 19 23 29 31 37
If you're completely determined not to use the print function more than once inside the loop, you could set a flag to determine whether to print the first two words. Like so:
print('Displays prime numbers from 1 to N.')
n = int(input('Please enter a value of n: '))
first = 'They are '
for n in range(1, n + 1):
if n >= 1:
for i in range(2, n):
if (n % i) == 0:
break
else:
print(first + str(n), end=' ')
if len(first) > 0:
first = ''
The following solution may help you
print('Displays prime numbers from 1 to N.')
n = int(input('Please enter a value of n: '))
num = [] # Create empty list
for n in range(1, n + 1):
if n >= 1:
for i in range(2, n):
if (n % i) == 0:
break
else:
num.append(n)
# Write the print statement outside of the loop and use .join() function and for loop
#to print each element of the list look like the output you have posted
#
print('They are'," ".join(str(x) for x in num))
Output:
Displays prime numbers from 1 to N.
Please enter a value of n: 40
They are 1 2 3 5 7 11 13 17 19 23 29 31 37
Output should be as follows:
Give the lower bound of the number range: 0
Give the upper bound of the number range: 1
0 cannot be prime.
1 cannot be prime.
Could not find primes from the test range.
Give the lower bound of the number range: 0
Give the upper bound of the number range: 20
0 cannot be prime.
1 cannot be prime.
2 is a prime.
3 is a prime.
4 is not a prime, because 2 * 2 = 4
5 is a prime.
6 is not a prime, because 2 * 3 = 6
7 is a prime.
8 is not a prime, because 2 * 4 = 8
9 is not a prime, because 3 * 3 = 9
10 is not a prime, because 2 * 5 = 10
11 is a prime.
12 is not a prime, because 2 * 6 = 12
13 is a prime.
14 is not a prime, because 2 * 7 = 14
15 is not a prime, because 3 * 5 = 15
16 is not a prime, because 2 * 8 = 16
17 is a prime.
18 is not a prime, because 2 * 9 = 18
19 is a prime.
20 is not a prime, because 2 * 10 = 20
Searched 21 numbers, from which 8 were primes.
The last found prime was 19.
low = input("Give the lower bound of the number range: ")
high = input("Give the upper bound of the number range: ")
low = int(low)
high = int(high)
for n in range(low,high):
if n<2:
print (n, "cannot be prime.")
for n in range(low,high):
for x in range (2, n):
if n % x == 0: # is n divisible with x? If yes, not a prime
y = n//x*x
print(n, "is not a prime, because", x, "*", n//x, "=", y)
break #this breaks the inner loop, and we continue with the outer!
else:
print(n, "is a prime.")
break
The code is not successfully running
It shows the output:
Give the lower bound of the number range: 0↩
Give the upper bound of the number range: 20↩
0 cannot be prime.↩
1 cannot be prime.↩
3 is a prime.↩
4 is not a prime, because 2 * 2 = 4↩
5 is a prime.↩
6 is not a prime, because 2 * 3 = 6↩
7 is a prime.↩
8 is not a prime, because 2 * 4 = 8↩
9 is a prime.↩
10 is not a prime, because 2 * 5 = 10↩
11 is a prime.↩
12 is not a prime, because 2 * 6 = 12↩
13 is a prime.↩
14 is not a prime, because 2 * 7 = 14↩
15 is a prime.↩
16 is not a prime, because 2 * 8 = 16↩
17 is a prime.↩
18 is not a prime, because 2 * 9 = 18↩
19 is a prime.
You are not checking if n is divisible by all numbers between 2 and n, since you exit the inner for loop after the first iteretion in both cases (when n divisible by x and when it is not).
To fix it, you need to move the else block to the outer loop (so it will run only when you don't exit the loop, meaning n is prime), and only break when n is not prime.
low = input("Give the lower bound of the number range: ")
high = input("Give the upper bound of the number range: ")
low = int(low)
high = int(high)
for n in range(low,high):
if n<2:
print (n, "cannot be prime.")
for n in range(low,high):
for x in range (2, n):
if n % x == 0: # is n divisible with x? If yes, not a prime
y = n//x*x
print(n, "is not a prime, because", x, "*", n//x, "=", y)
break #this breaks the inner loop, and we continue with the outer!
else:
print(n, "is a prime.")
try to use this one works for 0-20 range
low = input("Give the lower bound of the number range: ")
high = input("Give the upper bound of the number range: ")
low = int(low)
high = int(high)
for n in range(low, high+1):
if n <= 1:
print(n, "cannot be prime")
elif n == 2 or n == 3:
print(n, "is a prime")
elif n % 2 == 0 or n % 3 == 0:
print(n, "is not a prime")
else:
print(n, "is a prime")
This code is supposed to output fizz if number is divisible by 3, buzz if divisible by 5 and fizzbuzz if divisible by 3 and 5. Although I'm a bit unfamiliar with defining my own function and using the return appropriately. How do I remove the last 16 if the user inputs the number 16?
number = int(input("Enter a number: "))
def fizzbuzz(number):
n = 1
while n <= number:
if n % 3 != 0 and n % 5 != 0:
print(n)
elif n % 3 == 0 and n % 5 == 0:
print("fizzbuzz")
elif n % 3 == 0:
print("fizz")
elif n % 5 == 0:
print("buzz")
n = n + 1
return number
print(fizzbuzz(number))
If number = 16 it outputs
Enter a number: 16
1
2
fizz
4
buzz
fizz
7
8
fizz
buzz
11
fizz
13
14
fizzbuzz
16
16
How do I get I remove the last number 16, as it isn't supposed to be there
print("buzz")
n = n + 1
return number
print(fizzbuzz(number))
Here's your problem. Don't return the number, and don't print the return value of the function.
print("buzz")
n = n + 1
fizzbuzz(number)