Computing Trailing Zeros of the Smallest Multiple of Numbers 1 to k in Python

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πŸ’‘ Problem Formulation: We want to find the smallest number x that is divisible by all values from 1 to k, and then count the number of trailing zeros in x. For instance, if k = 5, the smallest number divisible by 1 through 5 is 60, which has one trailing zero.

Method 1: Prime Factorization and Zero Counting

This method involves finding the least common multiple (LCM) of the numbers from 1 to k via their prime factors. The trailing zeros in a number are determined by the power of 10 in its prime factorization, which is in turn determined by the number of 2-5 pairs in its factors. The Python code calculates the LCM and subsequently counts the trailing zeros.

Here’s an example:

import math

def find_lcm(k):
    lcm = 1
    for i in range(1, k+1):
        lcm = lcm * i // math.gcd(lcm, i)
    return lcm

def count_trailing_zeros(x):
    count = 0
    while x % 10 == 0:
        x //= 10
        count += 1
    return count

k = 5
x = find_lcm(k)
trailing_zeros = count_trailing_zeros(x)
print(f"Number of trailing zeros in LCM: {trailing_zeros}")

Output:

Number of trailing zeros in LCM: 1

The provided code snippet first computes the LCM of all numbers from 1 to k using a loop and the math.gcd() function. Once the LCM is found, the count_trailing_zeros() function sequentially divides the number by 10 to tally the zeros at the end of the LCM.

Method 2: Prime Factorization Optimization

This method optimizes the prime factorization process by only focusing on the factors that contribute to trailing zerosβ€”2 and 5. It calculates how many times 2 and 5 appear in the prime factorization of the LCM by counting multiples, thus determining the number of trailing zeros more efficiently.

Here’s an example:

def count_factors(n, k):
    count = 0
    for i in range(1, k+1):
        while i % n == 0:
            count += 1
            i //= n
    return count

k = 5
two_count = count_factors(2, k)
five_count = count_factors(5, k)
trailing_zeros = min(two_count, five_count)
print(f"Number of trailing zeros in LCM: {trailing_zeros}")

Output:

Number of trailing zeros in LCM: 1

The code above determines the number of 2’s and 5’s in the prime factorization of the LCM by counting how often each divides the numbers from 1 to k. The number of trailing zeros is equal to the minimum of these two counts, since a 10 is formed every time we have a pair of 2 and 5.

Method 3: Dynamic Programming to Calculate LCM

Dynamic programming can be employed to compute the LCM of the sequence more efficiently.By storing intermediate results, we can avoid redundant calculations. This technique particularly shines for large values of k where multiple calculations of the LCM elements overlap.

Here’s an example:

from functools import lru_cache

@lru_cache(maxsize=None)
def lcm(a, b):
    return a * b // math.gcd(a, b)

def find_lcm_dp(k):
    if k == 1:
        return 1
    else:
        return lcm(find_lcm_dp(k-1), k)

k = 5
x = find_lcm_dp(k)
trailing_zeros = count_trailing_zeros(x)
print(f"Number of trailing zeros in LCM: {trailing_zeros}")

Output:

Number of trailing zeros in LCM: 1

The find_lcm_dp() function uses dynamic programming (memoization) with the help of Python’s @lru_cache decorator. It avoids redundant GCD calculations by caching the intermediate LCMs, leading to more efficient computation for large k.

Method 4: Mathematical Insight

On inspecting the problem with mathematical insight, we realize that the number of trailing zeros in the LCM up to k is dictated by the power of 5 in its prime factorization since 2 is more abundant. Therefore, we can focus solely on counting the occurrences of 5 to find the number of trailing zeros.

Here’s an example:

def count_trailing_zeros_math(k):
    count = 0
    while k >= 5:
        k //= 5
        count += k
    return count

k = 5
trailing_zeros = count_trailing_zeros_math(k)
print(f"Number of trailing zeros in LCM: {trailing_zeros}")

Output:

Number of trailing zeros in LCM: 1

This approach simplifies the problem by directly calculating the amount of times 5 can divide the numbers from 1 to k. This is more efficient as it eliminates the need to calculate the complete LCM of the numbers.

Bonus One-Liner Method 5: Using Python Libraries

There are several Python libraries available that can handle large integer operations and LCM calculations. Libraries like numpy or sympy can compute the LCM very efficiently without our having to implement the logic manually.

Here’s an example:

import numpy as np

k = 5
x = np.lcm.reduce(range(1, k+1))
trailing_zeros = count_trailing_zeros(x)
print(f"Number of trailing zeros in LCM: {trailing_zeros}")

Output:

Number of trailing zeros in LCM: 1

This one-liner uses numpy‘s lcm.reduce function to find the LCM of all numbers from 1 to k. The function handles array inputs natively and reduces them to their combined LCM in a single line of code.

Summary/Discussion

  • Method 1: Prime Factorization and Zero Counting. Straightforward and comprehensible. Might be inefficient for very large values of k.
  • Method 2: Prime Factorization Optimization. More efficient by focusing on relevant factors. Still not the most efficient for very large k.
  • Method 3: Dynamic Programming to Calculate LCM. Improves performance by avoiding repeated calculations. Best suited for scenarios where k is large.
  • Method 4: Mathematical Insight. Most efficient as it directly counts the power of 5. Requires good insight into the mathematical properties of the problem.
  • Method 5: Using Python Libraries. The simplest and fastest for coding, as it leverages the power of optimized libraries. Dependence on external libraries that might not be specialized in finding trailing zeros.