Overview

\(N\) gears are arranged in a row. When gear \(1\) is rotated, we need to find the minimum number of rotations \(R\) such that all gears’ marks simultaneously return to their original positions, expressed as an irreducible fraction.

Analysis

Deriving the Number of Rotations for Each Gear

When gear \(i\) and gear \(i+1\) mesh, the number of teeth that pass through is the same. When gear \(i\) makes one full rotation, \(T_i\) teeth advance, so gear \(i+1\) rotates \(T_i / T_{i+1}\) times.

Repeating this, when gear \(1\) rotates \(R\) times: - Gear \(2\) rotates \(R \cdot T_1 / T_2\) times - Gear \(3\) rotates \(R \cdot T_1 / T_3\) times - In general, gear \(k\) rotates \(R \cdot T_1 / T_k\) times

Condition for Marks to Return

A gear’s mark returns to its original position when the number of rotations of that gear is exactly a positive integer. Therefore, for all \(k = 1, \ldots, N\):

\[R \cdot \frac{T_1}{T_k} \in \mathbb{Z}^{+}\]

must hold. This is equivalent to \(R\) being a positive integer multiple of the fraction \(T_k / T_1\).

LCM of Fractions

The minimum positive \(R\) is the least common multiple (LCM) of the set of fractions \(\{T_1/T_1, T_2/T_1, \ldots, T_N/T_1\}\).

We reduce each \(T_i / T_1\) to its lowest terms. Letting \(g_i = \gcd(T_i, T_1)\):

\[\frac{T_i}{T_1} = \frac{T_i / g_i}{T_1 / g_i}\]

The formula for the LCM of fractions is as follows:

\[\text{lcm}\left(\frac{a_1}{b_1}, \frac{a_2}{b_2}, \ldots\right) = \frac{\text{lcm}(a_1, a_2, \ldots)}{\gcd(b_1, b_2, \ldots)}\]

Concrete Example

For \(T = [6, 4, 3]\): - \(T_1/T_1 = 1/1\), \(T_2/T_1 = 4/6 = 2/3\), \(T_3/T_1 = 3/6 = 1/2\) - Set of numerators: \(\{1, 2, 1\}\) → \(\text{lcm} = 2\) - Set of denominators: \(\{1, 3, 2\}\) → \(\gcd = 1\) - Answer: \(R = 2/1\) (all marks return when gear 1 rotates 2 times)

Algorithm

1. For each \(i\), compute \(g_i = \gcd(T_i, T_1)\), and obtain the numerator \(T_i/g_i\) and denominator \(T_1/g_i\)
2. Compute the LCM of all numerators
3. Compute the GCD of all denominators
4. Reduce the result to an irreducible fraction and output it

Complexity

• Time complexity: \(O(N \log(\max T_i))\) (GCD and LCM computation for each element)
• Space complexity: \(O(N)\)

Implementation Notes

• When computing the LCM, calculate in the order a // gcd(a, b) * b to prevent overflow (in Python this is not an issue due to arbitrary-precision integers, but it’s a point to be aware of)

• When \(i = 1\), we have \(T_1/T_1 = 1/1\), so numerator 1 and denominator 1 are included. Therefore, the LCM is always a multiple of the other numerators, and regarding whether the GCD necessarily becomes a divisor of 1 (since denominator 1 is included, one might think the GCD becomes 1, and indeed \(T_1/g_1 = T_1/T_1 = 1\) means denominator 1 is included, so the GCD of all denominators is always 1… but actually, taking the GCD with other denominators means the answer could potentially be non-integer)
  → In practice, since \(i=1\) makes the denominator 1, and 1 is included in the \(\gcd\) computation, the result is always 1… one might think, but in the code all \(i\) are correctly processed, and since \(i=1\) contributes denominator 1, the final GCD of denominators becomes 1, making the answer always an integer.
  → Correction: Due to the problem constraints, the answer can be a fraction. This is because from the condition “the number of rotations of gear \(k\), \(R \cdot T_1/T_k\), must be an integer,” \(R\) is not necessarily an integer (the condition for \(k=1\) requires \(R\) to be an integer, but actually checking the code, since \(i=1\) contributes denominator 1, the GCD becomes 1, so the denominator of the answer is 1). While the output format allows fractions, in this derivation the answer is always an integer \(\text{lcm}(T_1/g_1, T_2/g_2, \ldots) / 1\).

  Source Code

import math
from functools import reduce

def solve():
    N = int(input())
    T = list(map(int, input().split()))
    
    # Gear 1 rotates R times. Gear i rotates R * T1 / Ti times.
    # For gear i's mark to return, R * T1 / Ti must be a positive integer.
    # So R must be a multiple of Ti / T1 for each i.
    # More precisely, R must be a multiple of Ti / gcd(Ti, T1) divided by T1 / gcd(Ti, T1).
    # Wait, let me think more carefully.
    
    # Gear 1 rotates R times. Gear k rotates R * (T1/T2) * (T2/T3) * ... * (T_{k-1}/T_k) = R * T1 / Tk times.
    # Wait, that's not right either. Let me re-derive.
    
    # Gear i and gear i+1 mesh: when gear i rotates once (Ti teeth), gear i+1 rotates Ti/T_{i+1} times.
    # So if gear 1 rotates R times, gear 2 rotates R * T1/T2 times.
    # Gear 3 rotates (R * T1/T2) * T2/T3 = R * T1/T3 times.
    # In general, gear k rotates R * T1 / Tk times.
    
    # For all marks to return: R * T1 / Tk must be a positive integer for all k = 1, ..., N.
    # For k=1: R must be a positive integer. Wait, R * T1/T1 = R must be integer.
    # For k=i: R * T1 / Ti must be integer.
    
    # So R must be a positive rational number such that:
    # R is a positive integer (from k=1)
    # R * T1 / Ti is a positive integer for all i
    
    # Wait, R being integer already comes from k=1. But R * T1 / Ti integer means
    # R must be a multiple of Ti / gcd(T1, Ti) ... let me think again.
    
    # R * T1 / Ti ∈ Z+  ⟺  R ∈ (Ti / T1) * Z+  ⟺  R = Ti * m / T1 for some positive integer m.
    # But R must satisfy this for ALL i simultaneously.
    
    # Let's write R = P/Q in lowest terms. Then P/Q * T1/Ti must be integer for all i.
    # That means T1 * P / (Ti * Q) must be integer, i.e., Ti * Q | T1 * P.
    
    # Actually, let's think of it differently.
    # R must be such that R * T1 / Ti is a positive integer for each i.
    # Equivalently, R = Ti * n_i / T1 for some positive integer n_i, for each i.
    # The minimum positive R is the LCM of the set {Ti/T1 : i=1,...,N} as fractions.
    
    # LCM of fractions: lcm(a1/b1, a2/b2, ...) = lcm(a1,a2,...) / gcd(b1,b2,...)
    # where each ai/bi is in lowest terms.
    
    # For each i, Ti/T1 in lowest terms: let g_i = gcd(Ti, T1), then numerator = Ti/g_i, denominator = T1/g_i.
    
    T1 = T[0]
    
    numerators = []
    denominators = []
    
    for i in range(N):
        Ti = T[i]
        g = math.gcd(Ti, T1)
        num = Ti // g
        den = T1 // g
        numerators.append(num)
        denominators.append(den)
    
    # LCM of all fractions = lcm(numerators) / gcd(denominators)
    
    def lcm(a, b):
        return a // math.gcd(a, b) * b
    
    result_num = reduce(lcm, numerators)
    result_den = reduce(math.gcd, denominators)
    
    # Simplify
    g = math.gcd(result_num, result_den)
    P = result_num // g
    Q = result_den // g
    
    print(f"{P}/{Q}")

solve()

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