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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

Pang
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Kiran
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1 Answers1

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Using CRT (as it seems you have done) you should find the solution set given by $$x = 51051k + 1277$$ where $k \in \mathbb{Z}$. So, to find the smallest possible solution take $k = 0$ and we have that $1277$ is the smallest solution.

It looks like you found that $$x = (51051/11) \cdot 10 + 2 \cdot (51051/51) \cdot 8 + 3 \cdot (51051/91)\cdot 85 \pmod {51051}$$

so you only need to reduce $\pmod {51051}$:

\begin{align} x &= 205481 \pmod {51051}\\ &= 1277 \pmod {51051} \end{align}

Clyde Kertzer
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