Cause and Effect Mr.11 & divisibility!

Computer Science Level 5

We call a natural number $n$ as 'awesome' if at least one of the numbers formed by permutations of digits of $n$ is divisible by 11.

There are $x$ 'awesome' numbers and $y$ 'non-awesome' numbers less than 1001.

Find the ratio $\dfrac{y}{x}$ (as a decimal value).

Details and assumptions :

All the numbers are represented in base 10.
If a number already has zeroes, they can be brought to the start during permutations. e.g $1102$ can be permuted to $0121$
You can't add zeroes at the start of the number like $01201$ being permuted to $12100$ is not allowed, because the original number is intended to be $1201$ and not $01201$ .
A number is 'non-awesome' if it is not awesome, i.e. none of its permutations is divisible by 11.

This is a part of the set 11≡ awesome (mod remainders)

The answer is 3.255319.

6 solutions

Pranjal Jain
Jun 11, 2015

Python 3.4.2

def isawesome(x):
    for i in permutations(str(x)):
        m=''.join(i)
        if int(m)%11==0:
            return 1
        else:
            continue
    return 0
n=0
for i in range(1,1001):
    if isawesome(i):
        n+=1
print((1000-n)/n) #Returns 3.255

Okay, here's something shorter, the program for determining $\text{awesome(x)}$ is one line using the any statement ...

from itertools import permutations
def awesome(x):
    return any(int(''.join(i))%11==0 for i in permutations(str(x)))
k=0
for i in range(1,1001):
         if awesome(i)==True:
                     k+=1

This returns $k=235$ , and then you calculate...

Aditya Raut - 5 years, 12 months ago

@Agnishom Chattopadhyay 'any' really made it short, didn't it?

Aditya Raut - 5 years, 12 months ago

Huh, any is a neat trick, I'll have to remember that.

Brock Brown - 5 years, 12 months ago

Yes, this is pretty interesting. I did not know the any command.

But here is a slightly simpler way:

1	`return (True in int(''.join(i))%11==0 for i in permutations(str(x)))`

Agnishom Chattopadhyay - 5 years, 12 months ago

@Agnishom Chattopadhyay – That's basically the same thing, right? You asked it to check if there's True in any of the outputs... Just that you didn't use the word any :P

Aditya Raut - 5 years, 11 months ago

@Aditya Raut – Yeah. I did not know the any function

Agnishom Chattopadhyay - 5 years, 11 months ago

Here 's a really lengthy and nooby solution in C++. Adding it here for the sake of variety. :)

EDIT: I edited the code a bit to also output the ratio directly so that we don't have to manually use a calculator to get the final required ratio.

Prasun Biswas - 5 years, 10 months ago

Bill Bell
Aug 10, 2015

from itertools import permutations
from fractions import Fraction

def isAwesome(k):
    for p in permutations(list(str(k))):
        if int(''.join(p))%11==0:
            return True
            break
    else:
        return False
x=0
y=0
for n in range(1,1001):
    if isAwesome(n):
        x+=1
    else:
        y+=1
print 1.*Fraction(y,x)

Samarth Agarwal
Jul 26, 2015

//java: int i=0,a,b,c,d,e,f,g,h,j,k,x=0;

for(i=100;i<999;i++)

{

a=i%10;//ones

d=i/10;

b=d%10;//tens

c=i/100;//hundreds

e=((100*a)+(10*b)+c)%11;

f=((100*a)+(10*c)+b)%11;

g=((100*b)+(10*a)+c)%11;

h=((100*b)+(10*c)+a)%11;

k=((100*c)+(10*a)+b)%11;

j=((100*c)+(10*b)+a)%11;

if(e==0||f==0||g==0||h==0||k==0||j==0)

  x=x+1;

}

System.out.println(x);//this gives awesomes in 100 to 999

1000 is not awsome in 1-99 there are 9 awesomes so y/x can be found

Roel Baars
Jul 1, 2015

Mathematica

1 2	`awesome = Sum[If[Or @@ Divisible[FromDigits /@ Permutations[IntegerDigits[n]], 11], 1, 0], {n, 1, 1001}] N[(1001 - awesome)/awesome]`

Arulx Z
Jun 30, 2015

>>> from itertools import permutations
>>> awe = 0.0
>>> for x in xrange(1, 1001):
        for y in list(map("".join, permutations(str(x)))):
            if int(y) % 11 == 0:
                awe += 1
                break

>>> print (1000 - awe) / awe
3.25531914894

I wonder why python doesn't have ++ operator

Brock Brown
Jun 16, 2015

I did basically the same thing that Pranjal did.

Python 3.4:

from itertools import permutations as combos
def awesome(n):
    for combo in combos(str(n)):
        test = ''.join(combo)
        if test[0] != '0' and int(test) % 11 == 0:
            return True
    return False
x = 0
y = 0
for n in range(1001):
    if awesome(n):
        x += 1
    else:
        y += 1
print("Answer:", y/x)

So upgraded to Python 3.4?

Pranjal Jain - 5 years, 12 months ago

Yup. Apparently, Python 3 is the present and future of Python or something.

I've got to say that the thing I'll miss the most is being able to use...

1	`print "Hello world!"`

Rather than...

1	`print("Hello world!")`

Brock Brown - 5 years, 12 months ago

Cause and Effect Mr.11 & divisibility!

This is a part of the set 11≡ awesome (mod remainders)

The answer is 3.255319.

6 solutions

0 pending reports