We are given an integer N. We need to write a program to find the least positive integer X made up of only digits 9’s and 0’s, such that, X is a multiple of N.
Note: It is assumed that the value of X will not exceed 106.
Input : N = 5 Output : X = 90 Exaplanation: 90 is the smallest number made up of 9's and 0's which is divisible by 5. Input : N = 7 Output : X = 9009 Exaplanation: 9009 is smallest number made up of 9's and 0's which is divisible by 7.
The idea to solve this problem is to generate and store all of the numbers which can be formed using digits 0 & 9. Then find the smallest number among these generated number which is divisible by N.
We will use the method of generating binary numbers to generate all numbers which can be formed by using digits 0 & 9.
Below is the implementation of above idea:
# Python3 program to find smallest multiple of
# a given number made of digits 0 and 9 only
from queue import Queue
# Preprocessing function to generate
# all possible numbers formed by 0 and 9
# Create an empty queue of strings
q = Queue()
# enque the first number
# This loops is like BFS of a tree
# with 9 as root, 0 as left child
# and 9 as right child and so on
for count in range(MAX_COUNT, -1, -1):
s1 = q.queue
# storing the front of queue
# in the vector
s2 = s1
# Append “0” to s1 and enqueue it
s1 += “0”
# Append “9” to s2 and enqueue it. Note
# that s2 contains the previous front
s2 += “9”
# function to find smallest number made
# up of only digits 9’s and 0’s, which
# is a multiple of n.
# traverse the vector to find
# the smallest multiple of n
for i in range(len(vec)):
# int is used for string to
if (int(vec[i]) % n == 0):
# Driver Code
# Maximum number of numbers
# made of 0 and 9
MAX_COUNT = 10000
# stack to store all numbers that
# can be formed using digits 0 and
# 9 and are less than 10^5
vec = 
n = 7
# This code is contributed by PranchalK
Time Complexity: O(n)