While my preparation for exams, came across this question.
"Find the smallest number which leaves remainder 1,2 and 3 when divided by 11,51 and 91"
Find considerable time in solving this. I have also gone through similar questions here.
This site gives answer for this question as 1277 with a detailed strategy. can this be used as a reliable approach or is there better approach available. Please advice so that I can start solving such problems
EDIT:Adding relevant portions of the answer cited for clarity
How (51051/11)y1 ≡ 1 (mod 11) can be written in the following block?
x ≡ 1 (mod 11)
x ≡ 2 (mod 51)
x ≡ 3 (mod 91)
11,51,91 are positive integers and pairwise co-prime. Hence, as per Chinese remainder theorem, there is exactly one number between 1 and 11×51×91 = 51051 that meets all these criteria
N = 11×51×91 = 51051
(51051/11)y1 ≡ 1 (mod 11)
After finding y1, y2 and y3, how the following line can be written for finding x?
x = 1 × (51051/11) × 10 + 2 × (51051/51) × 8 + 3 × (51051/91) × 85 mod 51051
I could relate other concepts from wiki