How to tell if a number is a multiple of n

[Couperin]

I want to write a program that scans a list of numbers and checks if any of the numbers are multiples of 2, and then delete those numbers

Any ideas how to do this?

I use Python.

 

[D H]

Use the modulo operator (%).

 

[chroot]

newList = [ x for x in oldList if x % 2 != 0 ]

- Warren

 

[Couperin]

Thank you.
This is what I have so far:

biglist = []

for i in range (1,6001):
    biglist.append(i)

This puts the numbers 1 to 6000 in a list as you know.
The modulo operator will definitely come in handy. My problem is with syntax (I'm a newbie). How would I write something to the effect of:

If (any number in the list) % 2 > 0:
   delete (number in the list)

Thanks for your help.

 

[Couperin]

newList = [ x for x in oldList if x % 2 != 0 ]

- Warren

Thank you, I didn't see this. I'll see where I get.

 

[chroot]

Couperin:

The verbose C-derived way would be to use a for loop:

newList = []

for x in oldList:
  if x % 2 != 0:
    newList.append(x)

The list comprehension I gave above performs the same operation in a more efficient and more easily-readable way.

- Warren

 

[Couperin]

Thank you very much chroot. You've allowed me to complete my program.

I got the idea when I was trying to solve the following maths problem that was once set in an Oxbridge interview for prospective maths students:

Of the numbers 1, 2, 3, ..., 6000, how many are not multiples of 2, 3 or 5?

I gave the question a go and after a bit of work (more than was necessary) with pen and paper, I got the answer 1600. But I didn't know whether or not this was the answer, and I couldn't find the answer on the net so I decided to write this program.

I've adapted the code so that you can crack similar riddles with any range of numbers (so long as they have a difference of 1, though I'm going to see if I can program something that allows you to work with different kinds of arithmetic progressions), and as many 'multiples' as you want.

Here's the code in case anyone's interested:

# riddle cracker, by Ben H, except for def inti

biglist = []
#
def inti(prompt):
    while 1:                                # loop infinitely
        x = raw_input('> ') # request a string
        try:
            new_x = int(x)                  # convert the input to a floater
            break                           # the conversion was successfull: break out of the loop
        except ValueError:                  # this block is accessed if the integer conversion fails
            print 'Please enter an integer. Not a string.'
    return new_x    
print "Riddle cracker, by Ben H"
#
def start():
    print "List of natural numbers (difference 1) from:"
    num1 = inti('')
    print "to:"
    num2 = inti('')

    for i in range (num1,num2 + 1):
        biglist.append(i)

    print "Numbers in list:"
    print len(biglist)

    deletenums()
#
def deletenums():
    print "Delete numbers divisible by:"
    diviz = inti('')
    print diviz
    for x in biglist:
        if x % diviz ==0:
            biglist.remove(x)
    print len(biglist), " numbers remain."

    choices()
#
def choices():
    print "Options:"
    print "(1) Delete more numbers?"
    print "(2) Start over again?"
    print "(3) Exit?"

    choice = inti('')

    if choice == 1:
        deletenums()
    elif choice == 2:
        del biglist[:]
        start()
    else:
        print "See you again..."

start()

And in case anyone's interested, here's the output when the above riddle is inputed:

Riddle cracker, by Ben H
List of natural numbers (difference 1) from:
> 1
to:
> 6000
Numbers in list:
6000
Delete numbers divisible by:
> 2
2
3000  numbers remain.
Options:
(1) Delete more numbers?
(2) Start over again?
(3) Exit?
> 1
Delete numbers divisible by:
> 3
3
2000  numbers remain.
Options:
(1) Delete more numbers?
(2) Start over again?
(3) Exit?
> 1
Delete numbers divisible by:
> 5
5
1600  numbers remain.
Options:
(1) Delete more numbers?
(2) Start over again?
(3) Exit?
> 1
Delete numbers divisible by:
> 5
5
1600  numbers remain.
Options:
(1) Delete more numbers?
(2) Start over again?
(3) Exit?
> 3
See you again...

1600? Nice one