If eggs are removed from a basket 2, 3, 4, 5, and 6 at a time, there remain, respectfully, 1, 2, 3, 4, and 5 eggs. But if the eggs are removed 7 at a time, no eggs remain. What is the least number of eggs that could have been in the basket
N mod 2 =1, N mod 3 =2, N mod 4 =3, N mod 5 = 4, N mod 6 =5, N mod 7 =0, solve for N
LCM(2,3,4,5,6,7) = 420
Using Chinese Remainder Theorem + Modular Multiplicative Inverse, we get:
N =119. The least number of eggs in the basket.
The full solution is:
420m + 119, where m=0, 1, 2, 3..........etc.
