Crack the Multi‑Modulo Egg‑Counting Puzzle: Three Python Solutions

Introduction

Hello, I am a Python enthusiast. A basic algorithm question was raised in a Python community, and I am sharing the solution here.

Implementation Process

I will share three different ideas that work well for this problem.

Method One

This method was contributed by an experienced member and is implemented as follows:

y = 1
while(True):
    if y % 2 == y % 4 == y % 8 and y % 3 == 0 and y % 7 == 0 and y % 9 == 0 and y % 5 == 4 and y % 6 == 3:
        print("篮子里总共有鸡蛋: %s(个)" % y)
        break
    y += 1

Although this brute‑force approach is slower, it works. The next two methods improve the algorithm.

Method Two

This method, shared by another contributor, uses a larger step size for faster convergence:

a = 9
while True:
    if a%5 == 4 and a%7 == 0 and a%8 == 1 and a%9 == 0:
        print(a)
        break
    a += 9

Increasing the variable by 9 each iteration makes this method noticeably faster than Method One.

Method Three

The third approach, suggested by another expert, modifies the previous logic slightly:

y = 63
while(True):
    if y % 2 == y % 4 == y % 8 and y % 3 == 0 and y % 7 == 0 and y % 9 == 0 and y % 5 == 4 and y % 6 == 3:
        print("篮子里总共有鸡蛋: %s(个)" % y)
        break
    y += 63

Only a single adjustment is needed, as shown in the accompanying illustration.

Conclusion

This article demonstrates how to apply Python programming to solve a specific value‑search problem, progressively optimizing the algorithm across three methods. The solutions deepen understanding of Python coding techniques and algorithmic thinking.