C Program
/* C - Neon Number (short & working) */ #include <stdio.h> int main() { int n; if (scanf("%d", &n)!=1) return 0; long sq = 1L*n*n, sum = 0; while (sq) { sum += sq % 10; sq /= 10; } if (n==0) sum = 0; /* handle n=0 correctly */ printf(sum == n ? "Neon\n" : "Not Neon\n"); return 0; }
C Output
Input:
9Output:
Neon
C++ Program
// C++ - Neon Number (short & working) #include <bits/stdc++.h> using namespace std; int main() { long long n; if(!(cin>>n)) return 0; long long sq = n*n, sum = 0; while (sq) { sum += sq%10; sq/=10; } if (n==0) sum = 0; cout << (sum==n ? "Neon" : "Not Neon") << '\n'; }
C++ Output
Input:
10Output:
Not Neon
JAVA Program
// Java - Neon Number (short & working) import java.util.*; public class Main { public static void main(String[] args){ Scanner s = new Scanner(System.in); long n = s.nextLong(); long sq = n*n, sum = 0; while (sq > 0) { sum += sq % 10; sq /= 10; } if (n==0) sum = 0; System.out.println(sum==n ? "Neon" : "Not Neon"); } }
JAVA Output
Input:
1Output:
Neon
Python Program
# Python - Neon Number (short & working) n = int(input().strip()) sq = n*n sum_ = sum(int(d) for d in str(sq)) if sq else 0 print("Neon" if sum_ == n else "Not Neon")
Python Output
Input:
12Output:
Not Neon
Example
A number is called a Neon number if the sum of the digits of its square equals the number itself. Consider n = 9. Its square is 81. Adding the digits gives 8 + 1 = 9, which is exactly the original number, so 9 is Neon. The same happens for n = 1 because 1² = 1 and the digit sum is 1. 0 is also Neon since 0² = 0 and the digit sum is 0. On the other hand, n = 10 gives 10² = 100, whose digit sum is 1, which does not equal 10, so it is not Neon.
Real-Life Analogy
Suppose you had a machine which initially "amplifies" a number by squaring it and then compresses this output by adding its digits together. A Neon number is unique in that after undergoing this amplify-and-compress operation, you get back the original number. It's similar to verifying the authenticity of a product by subjecting it to a process pipeline and obtaining the same identifier afterwards.
Logic and Steps
The reasoning is simple: read n, calculate sq = n * n, peel off the last digit of sq by repeatedly taking modulus ten, add to running sum, and discard that last digit by integer division by ten. Once all digits are done, check if sum equals n. If yes, then print "Neon", else print "Not Neon". The base case n = 0 is a small corner where the loop wouldn't execute since sq is zero; doing it explicitly keeps the sum at zero.
Learning Insights
This exercise reinforces the ability of digit extraction and summation using % 10 and / 10 (or // 10 in Python). It teaches one to think in terms of integer operations instead of strings, which is good in resource-constrained settings and in appreciating how numbers work at the bit and digit levels. It also introduces one to the concept of derived properties of numbers where we take a value and check a simple predicate, a pattern that occurs in checksums and digital signature verifications.
Why It Matters
Neon numbers themselves are a delightful number theory curiosity, but the exercise teaches fundamental techniques applied extensively in programming. Digit sum is a foundation for operations such as digital root computation, number format checking, and producing lightweight hashes. With experience in exact integer manipulation, you are more at ease dealing with input, protecting yourself against edge cases, and authoring dense, reliable code.
Interview Use Case
Interviewers prefer problems in which the statement is straightforward but challenge whether you can take a definition and implement it in clean, correct code with appropriate edge handling. You might be required to generalize to a range, count Neon numbers between bounds, evade string manipulations, or assess complexity. A clean description such as "compute n², add its digits, compare to n; it's O(d) where d is the number of digits in n²" demonstrates both clarity and discipline. Pointing out the n = 0 edge case scores extra marks for completeness.
Extensions and Variations
A natural variation is "Neon numbers in a range," where you iterate from L to R and perform the same test on each number. Another spin is trying bases other than base ten, where "digit sum" is a function of representation; in base b, you take % b and / b over and over again. You can also relate this to other digit-power patterns such as Armstrong numbers, or to digital root properties with repeated digit sums settling into small invariant helpful for rapid tests of divisibility.
How It Connects to Real Projects
Reducing digits of an altered value reflects basic checksum concepts and fingerprinting methods in which a considerable value is mapped to a small-sized signature for rapid comparison. Many embedded systems, telemetry compression, and ID checks rely on such reductions as they are inexpensive, deterministic, and easy to code without the overhead of heavy libraries.
Common Pitfalls
Common errors include comparing the digit sum of n rather than the digit sum of n², which quietly passes some but not all cases but misses the definition. Another is inadvertently making floating-point operations or string conversions that incur overhead or interfere with performance constraints. Lastly, omitting the n = 0 case can result in incorrect answers if the loop never executes and an uninitialized sum is compared.
SEO-Optimized Closing Paragraph
A Neon number is a program in C, C++, Java, and Python that identifies if the sum of digits of a number's square is the same as the number, and hence is a great beginners' exercise to learn digit extraction, integer math, and control flow. This topic reinforces problem-solving capabilities for coding interview problems, competitive programming, and actual-world checksum and numeric authentication applications. By getting a clean digit-sum implementation and dealing with edge conditions such as zero, you establish trust with loops, modular arithmetic, and effective number manipulation that applies very directly to many fundamental programming problems.
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