Definition
Position powers
Leftmost digit uses ^1, next ^2, and so on.
Check Disarium Number in Python
Overview
What You’ll Learn
A Disarium number equals the sum of its digits each raised to its left-to-right position. This tutorial covers the definition vs Armstrong, a live preview, worked Python examples, edge cases, and complexity.
Classic Examples
89, 135
Also every one-digit number 1–9 is Disarium.
Right Peel
% 10 // 10
Start exponent at digit count; decrease each peel.
vs Armstrong
Different rule
Armstrong uses one fixed exponent for every digit.
Live Preview
Try any n
Classify positive integers instantly in the browser.
O(d²)
Basic check
Cache powers to bring a single check closer to O(d).
Introduction
Disarium numbers are positive integers equal to the sum of their digits raised to left-to-right position powers. Classic examples: 89 (81 + 92) and 135 (11 + 32 + 53).
Every one-digit number 1–9 is Disarium because d1 = d. Do not confuse this with Armstrong numbers, which use one shared exponent equal to the digit count.
Why it matters?
It trains digit extraction, positional indexing, and careful comparison with similar “digit power” interview problems.
Key Highlights
Left Positions
Positions start at 1 on the leftmost digit.
Rightmost Highest
When peeling right-to-left, start at exponent k.
1–9 Always
All single-digit positives are Disarium.
Not Armstrong
Different exponent rule — say so in interviews.
In short: if digits are d1…dk from the left, check whether d11 + … + dkk equals n.
Foundation
📝 Problem & Approach
Given a positive integer n, decide whether it equals its position-power digit sum.
python
# 89 → 8**1 + 9**2 = 89 → yes
# 135 → 1**1 + 3**2 + 5**3 → yes
# 10 → 1**1 + 0**2 = 1 → noInputs & Outputs
| Item | Type | Description |
|---|---|---|
n | int | Positive integer (this page excludes 0 and negatives). |
| Return / print | bool / text | True if n is Disarium. |
Minimal workflow
Pseudocode
function is_disarium(n): // n > 0
k = number_of_digits(n)
sum = 0
pos = k
while n > 0:
digit = n mod 10
sum += digit ^ pos
pos -= 1
n = floor(n / 10)
return sum == original_nMethod comparison
| Method | Idea | Notes |
|---|---|---|
| Right peel | % 10 / // 10 with descending exponent | Classic interview loop |
| Left-to-right string | Enumerate digits with enumerate(..., 1) | Positions match the definition directly |
| Range scan | Filter 1…N with the same helper | Great for listing 1–9 and 89 |
Cheat Sheet
⚡ Quick Reference
| Goal | Pattern |
|---|---|
| Digit count | len(str(n)) |
| Last digit | n % 10 |
| Drop last digit | n //= 10 |
| Power term | digit ** pos |
| Classic yes | 89, 135, 1…9 |
| Classic no | 10 (→ sum 1) |
Compare
📋 Disarium vs Armstrong vs Digit Sum
Related digit problems — different exponent rules.
Disarium
d_i ^ iExponent = left-to-right position
Armstrong
d_i ^ kSame exponent k for every digit
Digit sum
sum d_iNo powers — just add digits
Interview tip
state the ruleSay positions start at 1 on the left
Context
When This Problem Shows Up
Reach for Disarium checks when positional digit powers matter.
Interview warm-ups
Tests digit loops and careful indexing.
Contrast with Armstrong
Great follow-up after learning narcissistic numbers.
Range listing tasks
Print all Disarium numbers in 1…N for small N.
Teaching powers
Makes position-dependent exponents concrete with 89 and 135.
Not for zero/negatives by default
Most lists use positive integers only — say so up front.
Key benefit: one short boolean check that forces clear thinking about digit order and exponents.
Interactive
🔮 Live Preview
Positive integers only, within JavaScript safe range.
Hands-On
Examples Gallery
Three complete Python programs — single check, range scan, and left-to-right string style. Click View Output to reveal sample console results.
📚 Getting Started
Right-to-left peel with descending exponents.
Example 1 — Single Value Check (89)
Direct and interview-friendly implementation for one input value.
python
def is_disarium(number: int) -> bool:
if number <= 0:
return False
original = number
digits = len(str(number))
total = 0
while number > 0:
digit = number % 10
total += digit ** digits
digits -= 1
number //= 10
return total == original
n = 89
if is_disarium(n):
print(f"{n} is a Disarium number.")
else:
print(f"{n} is not a Disarium number.")How It Works
The first peeled digit is the rightmost one, so we start the exponent from the full digit count and decrease each step.
⚡ Range Output
Reuse the helper to filter a beginner interval.
Example 2 — Disarium Numbers from 1 to 100
Prints all Disarium numbers in this range.
python
def is_disarium(num: int) -> bool:
if num <= 0:
return False
original = num
digits = len(str(num))
total = 0
while num > 0:
digit = num % 10
total += digit ** digits
digits -= 1
num //= 10
return total == original
print("Disarium numbers in the range 1 to 100:")
for i in range(1, 101):
if is_disarium(i):
print(i, end=" ")How It Works
All one-digit positive numbers are Disarium. In two digits up to 100, only 89 matches.
🔁 Left-to-Right Style
Match the definition directly with string enumeration.
Example 3 — Enumerate Digits from the Left
Positions start at 1 — no descending exponent bookkeeping.
python
def is_disarium_ltr(number: int) -> bool:
if number <= 0:
return False
total = 0
for pos, ch in enumerate(str(number), start=1):
total += int(ch) ** pos
return total == number
for n in (89, 135, 10, 7):
label = "Disarium" if is_disarium_ltr(n) else "not Disarium"
print(f"{n}: {label}")How It Works
enumerate(..., start=1) assigns each digit its left-to-right position, which matches the mathematical definition without reversing exponents.
🧠 How the Algorithm Decides
1
Count digits
Set exponent to total digit count (or walk left-to-right with pos = 1…k).
Setup
2
Peel and add
Raise each digit to its position power and accumulate.
Sum
3
Compare
If the sum equals the original number, it is Disarium.
Decide
=
Yes or no
Equality → Disarium; otherwise not.
Trace
🔎 Worked Walkthrough — n = 89
Trace the right-peel method. Digit count k = 2.
| Digit peeled | Exponent | Term | Running sum |
|---|---|---|---|
9 (rightmost) | 2 | 81 | 81 |
8 | 1 | 8 | 89 |
Final sum 89 equals n → Disarium.
Use Cases
Where Disarium checks show up beyond the interview prompt.
1. Interview Warm-Ups
Digit peeling plus positional powers in one problem.
Example: write is_disarium(n).
2. Teaching Powers
Makes 81 + 92 memorable with a famous 89.
Example: chalkboard 89 and 135.
3. Armstrong Contrast
Clarify fixed vs positional exponents.
Example: compare 153 vs 135.
4. Range Filters
List Disarium numbers in a classroom interval.
Example: 1 to 100 → 1…9, 89.
5. Indexing Practice
enumerate(start=1) matches the definition cleanly.
Example: left-to-right string style.
6. Complexity Follow-Ups
Discuss O(d²) vs caching digit powers.
Example: precompute for large ranges.
Pro Tip: say “positions start at 1 on the left” before coding — it prevents exponent off-by-ones.
Advantages
Why this pattern works well in interviews and classwork.
1. Clear Definition
Position powers are easy to state on a whiteboard.
2. Two Valid Styles
Right peel or left-to-right enumerate both work.
3. Famous Test Cases
89 and 135 make verification quick.
4. Rich Follow-Ups
Armstrong contrast and complexity caching are natural next questions.
Pro Tip: prefer left-to-right enumeration if you want the code to read exactly like the definition.
Usage Tips
Small habits that keep Disarium solutions interview-ready.
1. State Positions First
Say left-to-right indexing starts at 1 before coding.
2. Keep Original n
Save a copy before the peel loop mutates the working value.
3. Spot-Check 89 and 10
Yes and no cases catch exponent mistakes fast.
4. Contrast Armstrong
Mention the fixed-exponent rule so interviewers know you know the difference.
5. Restrict to Positives
Document that 0 / negatives are out of scope unless asked.
Pro Tip: if peeling from the right, whisper “start at k” — the rightmost digit always gets the largest exponent.
Common Pitfalls
Mistakes that commonly break Disarium solutions.
1. Wrong Exponent Direction
Giving the rightmost digit power 1 when peeling from the right.
→ Start at digit count and decrease.
2. Confusing with Armstrong
Using a fixed exponent equal to digit count for every digit.
→ Use position i for digit i from the left.
3. Overwriting n Before Compare
Comparing the sum to a mutated working variable.
→ Save
original = nfirst.4. Zero-Based Positions
Using enumerate without
start=1.→ Positions must begin at 1.
5. Accepting Zero Silently
Many lists exclude 0; treat it as non-Disarium here.
→ Return False for n ≤ 0 unless the prompt says otherwise.
Edge Cases
Most lists use positive numbers only; be explicit about zero and negatives.
Zero
n = 0
Commonly excluded by definition; this page checks positive numbers only.
Position
Left-to-right count
Do not use right-to-left positions directly without adjusting exponent order.
Armstrong
Different formula
Armstrong uses fixed exponent equal to digit count for all digits.
One digit
1–9
Always Disarium because d1 = d.
Two digits
Only 89 in 10–99
A quick check when listing up to 100.
Large range
Power growth
Use caching or careful bounds if scanning very large intervals.
Math
⚖️ Facts Worth Knowing
Handy follow-ups interviewers sometimes ask.
- Formal sum. For k-digit n with digits d1…dk, Disarium means n = Σ dii.
- Single digits. Every n in 1…9 is Disarium.
- Classics. 89 and 135 are the standard multi-digit examples.
- Rarity. Disarium numbers become scarce as digit counts grow.
Drill
🎯 Practice Problems
Try these variations to lock in the pattern.
1. Verify classics
- 89, 135, 7 → yes
- 10, 100 → no
2. Match both styles
- Right peel vs left-to-right
- Assert identical booleans
3. Range 1 to 200
- List all Disarium numbers
- Expect 1…9, 89, 135
4. Show the breakdown
- Print each digit^pos term
- Then the total sum
Important
Notes
- Rule: digit powers depend on left position.
- Code: peel from right, decrease exponent each step — or enumerate left-to-right.
- Remember: Disarium is different from Armstrong.
- Basic single check is about O(d²); caching can improve range scans.
Quick Takeaway: sum each digit raised to its left-to-right position; if that equals n, it is Disarium.
Analysis
⏱️ Time and Space Complexity
| Operation | Time | Extra space |
|---|---|---|
| Single check (basic) | O(d²) | O(1) |
| Single check (cached powers) | O(d) | O(d) |
Range [1..N] | About O(N · d²) | O(1) |
🎉 Conclusion
A Disarium number equals the sum of its digits raised to left-to-right position powers. Peel from the right with a descending exponent, or enumerate digits from the left — then compare the sum to n.
Practice the three examples above, then continue to even numbers for a simpler divisibility warm-up.
Always save the original n, start left positions at 1, and distinguish Disarium from Armstrong.
Pro Tips
💡 Best Practices
✅ Do
- State left-position powers first
- Save original before peeling
- Test 89, 135, and 10
- Mention Armstrong difference
- Restrict to positive n unless asked
❌ Don’t
- Use Armstrong’s fixed exponent
- Start positions at 0
- Forget descending exponents on right peel
- Compare against a mutated n
- Treat 0 as Disarium by default
Summary
Key Takeaways
Knowledge Unlocked
Five things to remember about Disarium numbers
Check positional digit powers the interview-friendly way.
5
Core concepts 01
Rule
d_i ^ i
Definition 02
Index
Left starts at 1
Math 03
Peel
Right starts at k
Code 04
Contrast
Not Armstrong
Guard 05
Complexity
~O(d²)
Analysis❓ Frequently Asked Questions
A positive integer n is Disarium if the sum of its digits, each raised to the power of its left-to-right position (starting at 1), equals n.
Using % 10 and // 10 is simple. To keep powers correct, we start exponent from total digit count and decrease it each step.
Yes. Python integer arithmetic is exact for big integers, so ** is safe for this kind of interview program.
Most common definitions focus on positive numbers. In this page we check positive integers only.
Yes for 1 to 9, because d^1 = d.
For d digits, checking one number is O(d^2) in the straightforward version with repeated powers, and can be improved with caching.
Armstrong raises every digit to the same power (digit count). Disarium raises each digit to its left-to-right position index.
89 and 135 are classic. In 1 to 100 you also get all of 1 through 9.
Did you Know? 🔊
Besides 89, 135 is a classic Disarium number because 11 + 32 + 53 = 135. Every one-digit positive number 1 to 9 is Disarium.
Continue to Even Number
Learn how to check whether an integer is even using modulo and bit tricks.
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