My code and output

This is the question, Write a Python program to find numbers in between 100 and 400 (both inclusive) where each digit of the numbers is an even number. The numbers obtained should be printed in a comma-separated sequence. I'm only allowed to use while loops so I'm not sure how.

CodePudding user response：

You need to collect the values in a list, and at the end print them comma-separated

x = 100
keep = []
while x <= 400:
    strx = str(x)
    if all(int(digit) % 2 == 0 for digit in strx):
        keep.append(strx)
    x  = 2
print(",".join(keep))
# 200,202,204,206,208,220,222,224,226,228,240,242,244,246,248,260,262,264,266,268,280,282,284,286,288,400

CodePudding user response：

Here is a solution with only while loops and without converting to string:

i = 100
while i <= 400:
    d = i
    flag = True
    while d > 0:
        d,r = divmod(d,10)
        if r%2:
            flag = False
    if flag:
        print(i)
    i  = 2 # optimization as only even numbers are valid

output (as single line):

200 202 204 206 208 220 222 224 226 228 240 242 244 246 248 260 262 264 266 268 280 282 284 286 288 400 

CodePudding user response：

This is just some correction to have a list with all the values at the end. With only using a while loop.

x = 100
l = []
while x < 400:
    x  = 2
    str_x = str(x)
    if int(str_x[0]) % 2 == 0 and int(str_x[1]) % 2 == 0 and int(str_x[2]) % 2 == 0:
        l.append(x)
print(l)

A more pythonic way in doing this:

l = lambda a, b : [
    x for x in range(a, b 1) 
    if all(int(digit) % 2 == 0 for digit in str(x)) 
]
print(l(200, 401))