Given a positive integer N whose unit’s digit is 3. Find the number of 1s in the smallest repunit which is divisible by the given number N. Every number whose unit’s digit is 3 has a repunit as its multiple. A repunit is a number which has only ones. It is of the form (10n – 1)/9.
As 3 divides 111 which is the smallest repunit
multiple of the number. So the number of ones in 111 is 3.
The repunits are 1, 11, 111, 1111, …. the next repunit to x will always be x*10+1. If the remainder left by x repunit is r then remainder left by the next repunit will always be (r*10+1)%n. Since the repunit can be very large, there is no need to find the repunit number. Simply counting the number of ones will give us the answer.
So, find out the remainders of all repunit numbers until the remainder becomes 0. Once it does, then the count of iterations done to make remainder 0 will be the number of 1’s.
Below is the implementation of above approach :
- Find the smallest number whose digits multiply to a given number n
- Immediate smallest number after re-arranging the digits of a given number
- Smallest number by rearranging digits of a given number
- Get the kth smallest number using the digits of the given number
- Smallest odd number with N digits
- Find smallest number K such that K % p = 0 and q % K = 0
- Smallest Even number with N digits
- Smallest odd digits number not less than N
- Smallest even digits number not less than N
- Smallest N digit number which is a multiple of 5
- Find the kth smallest number with sum of digits as m
- Smallest number divisible by first n numbers
- Smallest number S such that N is a factor of S factorial or S!
- Smallest number with sum of digits as N and divisible by 10^N
- Find smallest permutation of given number
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