CF 456B - Fedya and Maths

CF 456B - Fedya and Maths

Rating: 1200
Tags: math, number theory
Solve time: 1m 6s
Verified: yes

Solution

Problem Understanding

The task asks for the remainder when the sum of the first four multiples of a very large number n is divided by 5. In other words, you need to compute (1*n + 2*n + 3*n + 4*n) mod 5. The input n is a single integer that can have up to 10,105 digits, which is far beyond what a standard 64-bit integer can store. This means you cannot directly use typical integer arithmetic or multiply n by 10^4 in memory and then take the modulo, because the number itself could be enormous.

The output is a single integer between 0 and 4, representing the remainder after dividing the sum of these four multiples by 5. The problem is essentially asking for a pattern in modular arithmetic, because the sum 1 + 2 + 3 + 4 is constant, so the only thing that matters is n modulo 5. Edge cases to consider include n being zero, a one-digit number, or having tens of thousands of digits. A naive implementation that tries to store 4*n explicitly will fail for large n, either by crashing due to memory overflow or by taking far too long.

Approaches

A brute-force solution would involve first calculating each multiple 1*n, 2*n, 3*n, and 4*n, summing them, and then taking the modulo 5. This approach is correct for small numbers because multiplication and addition are straightforward. However, if n has 10,000 digits, computing 4*n explicitly is impractical. Even using a big integer library, this would take significant time and memory, far exceeding the intended 1-second limit.

The key observation is that modular arithmetic is distributive over addition and multiplication. That is, (a + b) mod m = ((a mod m) + (b mod m)) mod m and (a * b) mod m = ((a mod m) * (b mod m)) mod m. The sum of coefficients 1 + 2 + 3 + 4 is 10, so (1*n + 2*n + 3*n + 4*n) mod 5 = (10*n) mod 5. Since 10 is divisible by 5, (10*n) mod 5 is always 0, regardless of n. This dramatically simplifies the problem: the output is always 0. There is no need to read the digits of n or perform any arithmetic with the actual number.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(len(n)) or worse depending on big integer arithmetic	O(len(n))	Too slow for n with 10^5 digits
Optimal	O(1)	O(1)	Accepted

Algorithm Walkthrough

1. Recognize that the sum of the first four multiples of n can be factored: 1*n + 2*n + 3*n + 4*n = (1+2+3+4)*n = 10*n.
2. Realize that (10*n) mod 5 simplifies because 10 is divisible by 5, giving (10*n) mod 5 = 0.
3. Return 0 directly, since the expression always evaluates to 0 modulo 5 for any non-negative integer n, regardless of its size.

Why it works: The distributive property of modular arithmetic guarantees correctness. Since 10 is a multiple of 5, multiplying it by any integer n will still be divisible by 5. The modulo operation then produces a remainder of zero. There are no edge cases, because zero multiplied by 10 also yields zero, and the formula is valid for all allowed values of n.

Python Solution

import sys
input = sys.stdin.readline

n = input().strip()
print(0)

This solution reads the input as a string, which is necessary because n can be very large. We do not convert it to an integer because the size of n is irrelevant after realizing that the result is always zero. The strip() method removes any trailing newline characters. Printing 0 produces the correct output without performing any arithmetic operations on n.

Worked Examples

Sample Input 1:

4

Step	Expression	Value
Factor sum	1_4 + 2_4 + 3_4 + 4_4	40
Modulo 5	40 mod 5	0
Output	0	0

Sample Input 2:

123456789012345678901234567890

Step	Expression	Value
Factor sum	1_n + 2_n + 3_n + 4_n	10*n
Modulo 5	10*n mod 5	0
Output	0	0

These examples demonstrate that the algorithm works for both small and extremely large numbers. The modulo calculation depends only on the factor 10, not on n.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	The calculation does not depend on the number of digits in n.
Space	O(1)	No extra memory is used apart from reading the input as a string.

The solution easily fits within the 1-second time limit and 256 MB memory limit because it avoids any arithmetic on large numbers.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n = sys.stdin.readline().strip()
    return str(0)

# provided sample
assert run("4\n") == "0", "sample 1"

# custom cases
assert run("0\n") == "0", "minimum n"
assert run("5\n") == "0", "small n divisible by 5"
assert run("1\n") == "0", "small n not divisible by 5"
assert run("9"*10105 + "\n") == "0", "maximum-size n"

Test input	Expected output	What it validates
0	0	Minimum input
5	0	Small input divisible by 5
1	0	Small input not divisible by 5
9...9 (10^5 digits)	0	Maximum input size

Edge Cases

For n = 0, the sum of the first four multiples is 0 + 0 + 0 + 0 = 0, which modulo 5 is 0. For n with 10,105 digits, the algorithm correctly prints 0 without performing any arithmetic. Even for single-digit numbers not divisible by 5, the sum 1*n + 2*n + 3*n + 4*n is a multiple of 10, and the modulo 5 result is still 0. There are no edge cases that violate this pattern.