Given an integer K and a range of consecutive numbers [L, R]. The task is to count the numbers from the given range which have digital root as K (1 ≤ K ≤ 9). Digital root is sum of digits of a number until it becomes a single digit number. For example, 256 -> 2 + 5 + 6 = 13 -> 1 + 3 = 4.
Input: L = 10, R = 22, K = 3
12 and 21 are the only numbers from the range whose digit sum is 3.
Input: L = 100, R = 200, K = 5
- First thing is to note that for any number Sum of Digits is equal to Number % 9. If remainder is 0, then sum of digits is 9.
- So if K = 9, then replace K with 0.
- Task, now is to find count of numbers in range L to R with modulo 9 equal to K.
- Divide the entire range into the maximum possible groups of 9 starting with L (TotalRange / 9), since in each range there will be exactly one number with modulo 9 equal to K.
- Loop over rest number of elements from R to R – count of rest elements, and check if any number satisfies the condition.
Below is the implementation of the above approach:
- Digital Root (repeated digital sum) of the given large integer
- Print a number containing K digits with digital root D
- Find Nth positive number whose digital root is X
- Sudo Placement[1.7] | Greatest Digital Root
- Smallest root of the equation x^2 + s(x)*x - n = 0, where s(x) is the sum of digits of root x.
- Sum of all even numbers in range L and R
- Sum of all numbers divisible by 6 in a given range
- Sum of all natural numbers in range L to R
- Sum of all odd natural numbers in range L and R
- Sum of all even factors of numbers in the range [l, r]
- Sum of all odd factors of numbers in the range [l, r]
- Count of numbers having only 1 set bit in the range [0, n]
- Sum of Fibonacci Numbers in a range
- Prime numbers in a given range using STL | Set 2
- Sum of all numbers in the given range which are divisible by M
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