18Root 9
- Chain
1 + 8 = 9
Condense a Number (Digital Root) in Python
What you’ll learn
- How repeated digit sum gives a single-digit answer.
- Iterative method and O(1) closed form with mod 9.
- Important cases like 0, multiples of 9, and negatives.
Overview
Starting from a nonnegative integer, sum its decimal digits; if the result still has multiple digits, repeat. The final single digit is the digital root. Example: 9875 → 29 → 11 → 2.
Two programs
Iterative reduction and closed-form mod 9 method for nonnegative integers.
Live preview
Enter a number and instantly see its condensed value.
Terminology
Clear distinction between a one-time digit sum and the final digital root.
Prerequisites
Modulo, integer division, loops, and basic functions.
- Use
% 10and// 10to extract digits. - Optional: understand why
mod 9gives a shortcut in base 10.
Digital root vs digit sum
The digit sum of 9875 is 29. Condensing continues: 29 → 11 → 2.
Digit sum is one pass. Digital root repeats digit sum until one digit remains.
One passdigit sum
Repeatdigital root
mod 9closed form
Congruence mod 9
In base ten, n and the sum of its digits are congruent modulo 9. Repeating this preserves the same remainder, which yields the digital root shortcut.
98759875 % 9 = 2, matching the iterative chain ending at 2.
Intuition
49Root 4
- Chain
4 + 9 = 13 → 1 + 3 = 4
Takeaway: repeated digit sums keep the same value modulo 9.
Live preview
Enter a nonnegative integer and get digital root.
Algorithm (iterative)
Goal: reduce n to one digit by repeated digit sum.
While n > 9
Compute the sum of digits using % 10 and // 10.
Assign and repeat
Set n = digit_sum until n is a single digit.
Closed form (for n ≥ 0)
If n == 0, return 0; else r = n % 9, return 9 when r == 0, otherwise r.
📜 Pseudocode
Pseudocode
function condense(n):
while n > 9:
s = 0
while n > 0:
s += n % 10
n //= 10
n = s
return n1
Iterative condensation
Reference-style loop: keep reducing until the working value is a single digit.
python
def condense_number(number: int) -> int:
n = number
while n > 9:
digit_sum = 0
while n > 0:
digit_sum += n % 10
n //= 10
n = digit_sum
return n
number = 9875
print(f"The condensed form of {number} is: {condense_number(number)}")Explanation
The inner loop accumulates digits into digit_sum; the outer loop repeats until the value is within 0..9.
2
Closed form using mod 9
Same answer for nonnegative inputs in O(1) time.
python
def digital_root_nonnegative(n: int) -> int:
if n == 0:
return 0
r = n % 9
return 9 if r == 0 else r
print(f"dr(9875) = {digital_root_nonnegative(9875)} (closed form)")
print(f"dr(999999999999999999) = {digital_root_nonnegative(999999999999999999)} (closed form)")Explanation
Special-case 0. For positive values divisible by 9, return 9 instead of 0.
Optimization
Use formula. For nonnegative n, mod 9 gives the answer in O(1).
Huge inputs. For numbers beyond integer range, compute mod 9 over the number string.
Interview tip: explain iterative method first, then derive the shortcut and edge cases.
❓ FAQ
It means repeatedly adding digits until only one digit remains (digital root).
No. Digit sum is one pass. Condensing repeats until result is a single digit.
For n > 0: if n % 9 == 0 then 9 else n % 9. For n == 0, answer is 0.
In base 10, number and sum of digits are congruent modulo 9.
Usually we define this for nonnegative integers; you can use abs(n) if needed.
Iterative method is small digit-processing loops; closed form is O(1).
🔄 Input / output examples
Change the input literal in either sample and verify these expected roots.
| Input | Digital root | Notes |
|---|---|---|
9875 | 2 | Reference walkthrough |
0 | 0 | Special case |
9 | 9 | Already one digit |
18 | 9 | 1 + 8 = 9 |
Edge cases and pitfalls
Define behavior for negatives up front; most implementations assume nonnegative integers.
Zero
n = 0
Digital root is 0.
Multiples of 9
n % 9 = 0 and n > 0
Root is 9, not 0.
Negative input
Define policy
Use abs(n) or reject input explicitly.
Huge values
Beyond integer range
Use string-based digit processing if values exceed numeric limits.
⏱️ Time and space complexity
| Approach | Time | Extra space |
|---|---|---|
| Iterative digit reduction | Small digit loops | O(1) |
| Closed form | O(1) | O(1) |
For string-based very large numbers, each pass is linear in the number of digits.
Summary
- Condense means repeated digit sums until one digit remains.
- Methods: iterative reduction and
mod 9shortcut for nonnegative inputs. - Important: handle
0, multiples of9, and negative-input policy.
Did you know?
For positive numbers, digital root follows 1 + (n - 1) % 9.
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