Given number of digits ‘N’ and base ‘B’, the task is to count all the ‘N’ digit numbers without leading zeros that are in base ‘B’.
Input: N = 2, B = 2 Output: 2 All possible numbers without leading zeros are 10 and 11. Input: N = 5, B = 8 Output: 28672
- If the base is ‘B’ then every digit of the number can take any value within the range [0, B-1].
- So, B ‘N’ digit numbers are possible with base ‘B’ (including the numbers with leading zeros).
- And, if we fix the first digit as ‘0’ then the rest of the ‘N-1’ digits can form a total of B numbers.
- So, total number of ‘N’ digit numbers with base ‘B’ possible without leading zeros are B – B.
Below is the implementation of above Approach:
- Count numbers having N 0's and and M 1's with no leading zeros
- Count of N-bit binary numbers without leading zeros
- Given a number N in decimal base, find number of its digits in any base (base b)
- C++ program to find all numbers less than n, which are palindromic in base 10 and base 2.
- Check if a given number can be represented in given a no. of digits in any base
- Check if the number is even or odd whose digits and base (radix) is given
- Write a program to add two numbers in base 14
- Count of N-digit numbers in base K with no two consecutive zeroes
- Count of numbers between range having only non-zero digits whose sum of digits is N and number is divisible by M
- Count numbers in given range such that sum of even digits is greater than sum of odd digits
- Numbers with sum of digits equal to the sum of digits of its all prime factor
- Numbers of Length N having digits A and B and whose sum of digits contain only digits A and B
- Count unique numbers that can be generated from N by adding one and removing trailing zeros
- Compute sum of digits in all numbers from 1 to n
- Print numbers with digits 0 and 1 only such that their sum is N
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