An Emirp is a prime number whose reversal is also a prime. For
example, 17 is a prime and 71 is a prime, so 17 and 71 are emirps.
Write a program that prints out the first N emirps, five on each line.
Calculate the first N emirp (prime, spelled backwards) numbers, where
N is a positive number that the user provides as input.
Implementation Details
You are required to make use of 2 functions (which you must write).
isPrime(value) # Returns true if value is a prime number. reverse
(value) # Returns the reverse of the value (i.e. if value is 35,
returns 53). You should use these functions in conjunction with logic
in your main part of the program, to perform the computation of N
emirps and print them out according to the screenshot below.
The general outline for your program would be as follows:
Step 1: Ask user for positive number (input validation) Step 2:
Initialize a variable Test to 2 Step 3: While # emirps found is less
than the input:
Call isPrime with Test, and call it again with reverse(Test).
If both are prime, print and increment number of emirps found. Test++ Hint - to reverse the number, turn it into a string and then
reverse the string. Then turn it back into an int!
MY CODE:
n = 0
count = 0
i = 1
def isPrime(value):
test = 2
count = 0
while(test < value):
if( value % test == 0):
count+=count
test+=test
if(count == 0):
return 1
else:
return 0
def reverse(value):
reverse = 0
while(value > 0):
reverse = reverse * 10 + (value % 10)
value = value / 10
return reverse
n = float(input("Please enter a positive number: "))
while(count < n):
i+=i;
if(isPrime(i)):
if(isPrime(reverse(i))):
print("i" + "\n")
count+=count
if((count % (5)) == 0 ):
print("\n")
where are the mistakes?
UPDATED CODE BUT STILL NOT RUNNING:
n = 0
count = 0
i = 1
def isPrime(value):
test = 2
while(test < value):
if( value % test != 0):
test +=1
else:
return 0
def reverse(value):
return int(str(value)[::-1])
i = int(input("Please enter a positive number: "))
count = 0
while(count < 5):
if(isPrime(i)):
if(isPrime(reverse(i))):
print(str(i) + "\n")
count += 1
i += 1
else:
i += 1
else:
i +=1
There are a lot of issues with your code. I altered the function isPrime. There is among other no need for count here and if you want to increment test by 1 every loop use test +=1:
def isPrime(value):
test = 2
while(test < value):
if( value % test != 0):
test +=1
else:
return 0
return 1
There is an easy way to reverse digits by making value a string in reversed order and to convert this into an integer:
def reverse(value):
return int(str(value)[::-1])
You among other have to assign the input to i. n in your code is a constant 5. You have to increment i by one in any condition and increment the count by one if i is an emirp only:
i = int(input("Please enter a positive number: "))
count = 0
while(count < 5):
if(isPrime(i)):
if(isPrime(reverse(i))):
print(str(i) + "\n")
count += 1
i += 1
else:
i += 1
else:
i +=1
def emrip_no(num):
i=0
j=0
for i in range(1,num+1):
a=0
for j in range(1,i+1):
if(i%j==0):
a+=1
if(a==2):
print("Number is a prime number")
k=0
l=0
rv=0
while(num!=0):
r=num%10
rv=(rv*10)+r
num=num//10
print("It's reverse is: ",rv)
for k in range(1,rv+1):
b=0
for l in range(1,k+1):
if(k%l==0):
b+=1
if(b==2):
print("It's reverse is also a prime number")
else:
print("It's reverse is not a prime number")
n=int(input("Enter a number: "))
emrip_no(n)
Related
lower = int(input("Enter lower range limit:"))
upper = int(input("Enter upper b range limit:"))
Count = 0
for i in range(lower, upper+1):
if((i%3==0) & (i%5==0)):
Count += 1
print(count)
I want to print 100 values which are divisible by 3 and 5, but in count only 7 is showing, can anyone suggest what to do, it will be appreciated!
Use this
num = int(input("How many Numbers you want :"))
i=0
Count = 0
while(Count != num):
if((i%3==0) and (i%5==0)):
Count += 1
print(i)
i = i + 1
print("Count is :",Count)
Python program to print all the numbers
divisible by 3 and 5 for a given number
Result function with N
def result(N):
# iterate from 0 to N
for num in range(N):
# Short-circuit operator is used
if num % 3 == 0 and num % 5 == 0:
print(str(num) + " ", end = "")
else:
pass
Driver code
if name == "main":
# input goes here
N = 100
# Calling function
result(N)
This is the question:
Write a Python program that reads a positive integer n and finds the
average of all odd numbers between 1 and n. Your program should not
accept a negative value for n.
And here is my code, which curiously doesn't work:
k = int(input('Enter a positive integer: '))
while k <= 0:
print('Please enter a positive integer!! \n')
k = int(input('Enter a positive integer: '))
else:
b = 1
sum1 = 0
while b <= k:
if b % 2 == 1:
sum1 = sum1+b
b += 1
avg = sum/k
print(avg)
Example: input: 8 and output: 2.5, while it should be 4. Any tips?
if we use while True, the program will run until it receives a positive number. and when we get a positive number, then we execute the instructions and turn off the loop using a break
1 version with list:
n = int(input())
while True:
if n <= 0:
n = int(input('Enter a positive number: '))
else:
numbers = [i for i in range(1, n + 1) if i % 2 == 1]
print(sum(numbers) / len(numbers))
break
2 version with list:
n = int(input())
while True:
if n <= 0:
n = int(input('Enter a positive number: '))
else:
numbers = []
for i in range(1, n+1):
if i % 2 == 1:
numbers.append(i)
break
print(sum(numbers)/len(numbers))
3 version with counter
n = int(input())
while True:
if n <= 0:
n = int(input('Enter a positive number: '))
else:
summ = 0
c = 0
for i in range(1, n+1):
if i % 2 == 1:
summ += i
c += 1
print(summ/c)
break
You have used sum (builtin function name) instead of sum1 (your variable name) in the 2nd last line. Additionally, you need to count the total number of odd numbers and divide using that number instead of the input.
Okay I reviewed the question and here is the answer:
k = int(input('Enter a positive integer: '))
while k <= 0:
print('Please enter a positive integer!! \n')
k = int(input('Enter a positive integer: '))
else:
b = 1
sum1 = 0
c = 0
while b <= k:
if b % 2 == 1: #determines if odd
sum1 = sum1+b
c += 1 #variable which counts the odd elements
b += 1 #counter
avg = sum1/c
print(avg)
sum = int(input("Enter a positive integer: "))
Oddtotal = 0
for number in range(1, sum+1):
if(number % 2 != 0):
print("{0}".format(number))
Oddtotal = Oddtotal + number
I'm a beginner in Python and I'm practicing to code this problem I saw. Next prime is needed, but there are limitations on the input. I have searched for similar questions, but my code is still not working. Hope you can help. Thank you!
The problem I get is when I enter 32, the results show 33 when the next prime is 37...
Here's my code so far.
num = int(input("Enter a positive number:"))
import math
def nextprime(n):
if n < 0:
raise ValueError
for next in range(n + 1, n +200):
if next > 1:
for i in range(2, next):
if (next % i) == 0:
break
else:
return next
In your code when you arrive to a number that reminder is not zero you return that number. You need a flag for every number this flag is True if can be divide flag convert to False for the first number that flag not convert to false return that number like below.
Don't use next because this is a builtin function.
Try this: (I don't improve your code)
def nextprime(n):
if n < 0:
raise ValueError
for i in range(n + 1, n +200):
if i > 1:
pr = True
for j in range(2, i):
if (i % j) == 0:
pr = False
break
if pr:
return i
return 'not found'
You can also try this code, write function to check that a number is prime or not like def is_prime then for number of larger that you input num find min number next. (this answer from this thread.)
def is_prime(x):
return all(x % i for i in range(2, x))
def next_prime(x):
return min([a for a in range(x+1, 2*x) if is_prime(a)])
print(next_prime(32))
You can also use sympy like below: (this answer from this thread.)
from sympy import *
nextprime(32)
def next_prime(n):
while True:
n=n+1
for i in range (2,int(n/2)):
if n%i==0:
break
else:
return n
print(next_prime(67))
Few off-topic tips:
as user1740577 mentioned, don't use next as a variable name
refrain from using eval when possible, it's okay here, but in real project this will lead to big no-no.
Place imports at the very top of your script
Consider using variable names i and j only for iterations.
For duplicate except blocks use (Error, Error)
As for solution to your problem, with some adjustments, if you don't mind
def next_prime(n: int) -> int:
if n < 0:
raise ValueError('Negative numbers can not be primes')
# Base case
if n <= 1:
return 2
# For i as every odd number between n + 1 and n + 200
for i in range(n + 1 + (n % 2), n + 200, 2):
# For every odd number from 3 to i (3 because we covered base case)
for j in range(3, i, 2):
# If remained is equals to 0
if not i % j:
# break current loop
break
# If loop j didn't break [nobreak: ]
else:
return i
raise RuntimeError('Failed to compute next prime number :c')
def main():
while True:
try:
num = int(input('Enter positive number: '))
print(f'Next prime is: {next_prime(num)}')
break
except ValueError:
print('Please enter a positive integer!')
if __name__ == '__main__':
main()
Made some speed improvements to the code from #rajendra-kumbar:
#!/usr/bin/env python
import sys
import time
import math
def next_prime(number):
if number < 0:
raise ValueError('Negative numbers can not be primes')
# Base case
if number <= 1:
return 2
# if even go back 1
if number % 2 == 0:
number -= 1
while True:
# only odds
number += 2
#only need to check up to and including the sqrt
max_check = int(math.sqrt(number))+2
# don't need to check even numbers
for divider in range(3, max_check, 2):
# if 'divider' divides 'number', then 'number' is not prime
if number % divider == 0:
break
# if the for loop didn't break, then 'number' is prime
else:
return number
if __name__ == '__main__':
number = int(sys.argv[1].strip())
t0 = time.time()
print('{0:d} is the next prime from {1:d}'.format(next_prime(number), number))
run_time = time.time() - t0
print('run_time = {0:.8f}'.format(run_time))
it is about twice as fast
You can try something like simple:
def is_prime(number:int):
check = 0
for i in range(2,number):
if number % i == 0:
check += 1
if check == 0:
return True
else:
return False
def next_prime(value):
check = value + 1
while is_prime(check) is False:
check += 1
return check
value = int(input("Insert the number: "))
print(next_prime(value))
The sum of digit is running well, but Why is the factorial section not working? Is there any mistakes that I made with the if else, or the break ?
NUM = int (input ("Enter a number "))
if NUM > 1:
for x in range (2, NUM):
if (NUM % x) == 0:
temp = NUM
sod = 0
while temp > 0:
remainder = temp % 10
sod += remainder
temp = temp//10
print (sod)
break
else:
factorial = 1
for I in range (1, NUM + 1,1):
factorial *= I
print (factorial)
else:
temp = NUM
sod = 0
while temp > 0:
remainder = temp % 10
sod += remainder
temp = temp//10
print (sod)
Your break statement on line 12 is outside the if statement, so the for loop will break after the first pass regardless of the value of NUM. Are you sure you didn't mean to indent the break another four spaces?
# example of factorial with prime numbers
num = int(input("Enter a number"))
factorial = 1
if num%2 == 0:
for i in range(1, num + 1):
factorial = factorial*i
print("The factorial of ",num," is",factorial)
Below code checks if the number is prime or odd number. if its prime it finds the factorial of that number and if its odd number then it sums the range of the number. Hope its what you were looking for.
NUM = int(input("Enter number here"))
if NUM > 0:
if NUM % 2 == 0:
factorial = 1
for i in range(1, NUM +1):
factorial = factorial*i
print(factorial)
else:
sum_of_NUM = 0
for x in range(NUM+1):
sum_of_NUM += x
print(sum_of_NUM)
else:
print("something you want to put")
I'm working on a credit project of CS50 and I have some problem with the verification of the credit card.
Here the function I create:
def main():
while True :
cardnumber = input("Please enter a credit card number: ")
if cardnumber.isdecimal() and int(cardnumber) > 0 :
break
count = len(cardnumber)
if count != 13 and count != 15 and count != 16:
print("INVALID")
else:
check(count, cardnumber)
def check(length, number):
lenght_max = 15
if length == 15 and int(number[0]) == 3 and (int(number[1]) == 4 or int(number[1]) == 7):
if validator(number):
print("AMEX")
elif length == 16 and int(number[0]) == 5 and int(number[1]) <= 5:
if validator(number):
print("MASTERCARD")
elif length == 16 or length == 13 and int(number[0]) == 4:
if validator(number):
print("VISA")
else:
print("INVALID")
return number
def validator(num):
sum = 0
while num > 0:
sum += num % 10
num = num // 10
return sum
odd = [int(num[i]) * 2 for i in range(1, len(num), 2)]
even = [int(num[i]) for i in range(0, len(num), 2)]
new_sum = sum(validator(x) for x in odd) + sum(even)
if(new_sum % 10 == 0):
return True
else:
print("INVALID")
main()
I found the way to print the evens and the odds(multiply also time 2) but now I have to sum booth and check if the remainder is 0
Here the complete instruction:
http://docs.cs50.net/problems/credit/credit.html
Write a helper function to sum your digits. You'll need to use it extensively.
def dig_sum(num):
sum = 0
while num > 0:
sum += num % 10
num = num // 10
return sum
num = '378282246310005' # your credit card number
odd = [int(num[i]) * 2 for i in range(1, len(num), 2)] # these two remain the same
even = [int(num[i]) for i in range(0, len(num), 2)]
new_sum = sum(dig_sum(x) for x in odd) + sum(even)
if(new_sum % 10 == 0):
print('Valid') #valid!
sum(dig_sum(x) for x in odd) will get the digit sum for each number in your odd list and sum(...) that finds the resultant sum.
Input:
'378282246310005'
Output:
Valid
A first problem with your function is that you do not store the even/odd digits somewhere: you construct a list with one element each time, and print that element.
Now since two times a digit, can only result in a two digit number, we can use:
def sum2(x):
return (2*x)//10 + (2*x)%10
You can construct a list of all the digits at odd index with:
odd = [int(number[i]) for i in range(1,length,2)]
The same for digits at even index:
even = [int(number[i]) for i in range(0,length,2)]
Now we can simply use the sum(..) builtin function to sum up the digits:
total = sum(sum2(oddi) for oddi in odd) + sum(even)
and check if it is a multiple of 10:
return total%10 == 0
Or putting it all together:
def validator(number, length):
odd = [int(number[i]) for i in range(1,length,2)]
even = [int(number[i]) for i in range(0,length,2)]
total = sum(sum2(oddi) for oddi in odd) + sum(even)
return total%10 == 0
Or we can use a the following one liner for experts:
from itertools import zip_longest
def validator(number,length):
numbi = iter(numbi)
return sum(x+sum2(y) for x,y in zip_longest(numbi,numbi,fillvalue=0))%10 == 0