Divisible By A Part Of Me?

Computer Science Level 3

Count the number of 3-digit positive integers such that it is divisible by all of its digits.

For example, 132 is such a number because it is divisible by its digits 1, 3 and 2.

By default, it is impossible to divisible by 0, so the 3-digit integer can't have a digit 0.


The answer is 56.

Ankit Nigam by Ankit Nigam , 23, India

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5 solutions

Rushikesh Jogdand
May 15, 2016

The catch is that divisibility by 0 0 is undefined. So, define it! 

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def div(num,dig):
    if dig==0:
        return False
    elif num%dig==0:
        return True
    return False

count=0

for i in range(100,1000):
    if div(i,i%10) and div(i,(i//10)%10) and div(i//100):
        print(i)
        count+=1

2 Helpful 0 Interesting 0 Brilliant 0 Confused

There's actually a one-liner python solution to this problem. 

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print(len([i for i in range(100,1000) if ('0' not in str(i) and all(i%int(x)==0 for x in str(i)))]))

This will give you the answer 56 56 to this problem.

In case we want to print out those numbers too, we can modify it as, 

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count=0
for item in (i for i in range(100,1000) if ('0' not in str(i) and all(i%int(x)==0 for x in str(i)))):
    count+=1
    print(item)
print('\nAnswer =',count)

Prasun Biswas - 5 years, 1 month ago

you may use also ZeroDivisionError to catch that exception

Vincent Miller Moral - 4 years, 12 months ago
Yuriy Kazakov
Jun 3, 2018 
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import math
a=[1,2,3,4,5,6,7,8,9]
w=0
n=0
for i in a:
   for j in a:
     for k in a:
         s=i+j*10+k*100
         n=n+1
         if (math.fmod(s,k)+1)*(math.fmod(s,j)+1)*(math.fmod(s,i)+1)==1:
           w=w+1
           print(s,w)
print (w,n)

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Xuming Liang
Jun 11, 2016

My code: 

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def self_digit_divisible(n):
    nstring=str(n)
    for i in range(len(str(n))):
        if int(nstring[i])==0:
            return 0
        elif n%int(str(n)[i]) != 0:
            return 0
    else:
        return 1

counting=0

for n in range(100,1000):
    if self_digit_divisible(n)==1:
        counting+=1

print(counting)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Standard solution. Can you think of a way to cut down on code in writing your function?

Viki Zeta
Jun 3, 2016 
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'POSSIBILITY ESTIMATOR'

nos = range(100, 1000)
possible = list()

for i in nos:
    a = list(map(int, list(str(i))))
    div = True
    if 0 in a:
        continue
    for j in a:
        if i % j != 0:
            div = False
            break

    if div : 
        possible.append(i)

print len(possible)

Simple, that aids 56

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Mehdi K.
May 23, 2016

Python 

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>>> from math import gcd
>>> def lcm(a,b): return a*b//gcd(a,b)
... 
>>> sum(1 for i in range(1, 10) for j in range(1, 10) for k in range(1, 10) if (100 * i + 10 * j + k) % lcm(i, lcm(j, k)) == 0)
56

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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