An old woman went to the market and a horse stepped on her basket and smashed her eggs. The rider offered to pay for the eggs and asked her how many there were. She did not remember the exact number, but when she had taken them two at a time there was one egg left, and the same happened when she took three, four, five, and six at a time. But when she took them seven at a time, they came out even. What is the smallest number of eggs she could have had?

(In reply to

more formally by Steven Lord)

The problem essentially asks to find integer x to solve a system of congruences:

x = 1 mod 60

x = 0 mod 7

With 60 and 7 being coprime, solving this is basically applying the Chinese Remainder Theorem. The wikipedia page on the theorem has a lot of useful info. https://en.wikipedia.org/wiki/Chinese_remainder_theorem

An explicit solution can be found by first calculating a pair of values S and T such that 60*S + 7*T = 1. Then a solution x = 0*60*S + 1*7*T. To solve for S and T this without guessing an extended Euclidean algorithm can be applied. https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm

index | quotient | remainder | S value | T value

0 | | 60 | 1 | 0

1 | | 7 | 0 | 1

2 | 60/7=8 | 60-7*8=4 | 1-0*8=1 | 0-1*8=-8

3 | 7/4=1 | 7-4*1=3 | 0-1*1=-1 | 1-(-8)*1=9

4 | 4/3=1 | 4-3*1=1 | 1-(-1)*1=2 | -8-9*1=-17

5 | 3/1=3 | 3-1*3=0 | Stop here

The S and T values from the last nonzero remainder are the values needed. Then 60*2 + 7*(-17) = 1. Then back to the Chinese Remainder Theorem. If we already have 60*2 + 7*(-17) = 1 then x = 0*60*2 + 1*7*(-17) = -119. Then the smallest positive integer solution is -119+60*7 = **301**.