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:
$R – $rem; $i–)
$rem1 = $i % 9;
if ($rem1 == $K)
// Driver code
$L = 10;
$R = 22;
$K = 3;
echo countNumbers($L, $R, $K);
// This code is contributed by Ita_c
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