Let A = $a_1a_2 \cdots a_n$ and let $c_i$ be inverse of $(\frac{A}{a_i})$(mod $a_i$). Now, let k = $\sum_{i=1}^n\frac{A}{a_i} \cdot c_i \cdot b_i$ (mod A). Using this, answer the following:
Find the smallest natural n such that it leaves remainder 0,1,2,3 respectively divided by 2,3,5,7 respectively.
Answer: We ignore the divisibility by 2 for now, and use Chinese Remainder Theorem on the numbers. We get
n = 3$\cdot$5$\cdot$3 + 3$\cdot$7$\cdot$2 + 5$\cdot$7$\cdot$2 (mod 105) $\equiv$ 157 $\equiv$ 52 (mod 105) and we also have 2|52 so that 52 is the answer.
Approach: First take 3$\cdot$5 as take 7 equal to $a_1$ now want to find the inverse of this mod 7 which is just 1 and wanted the congruence mod 7 as 3 (want it congruent to 3 mod 7) and have $\frac{A}{a_i}$ which is 3$\cdot$5 into its inverse $c_i$ which is 1 and then $\cdot$ $b_i$ equal to 3 so have 3$\cdot$5$\cdot$3 + now check for 5, 3$\cdot$7 is already congruent to 1 mod 5 so don't have to add any inverse here to make this 1 and want the congruence to be 2 so + 3$\cdot$7$\cdot$2, and 5$\cdot$7 is congruent to 2 mod 3 and want to find its inverse which is 2 so + 5$\cdot$7$\cdot$2
Question: I don't really know how to get the whole n = 3$\cdot$5$\cdot$3 + 3$\cdot$7$\cdot$2 + 5$\cdot$7$\cdot$2 (mod 105) reasonably by using the definition for k. Could someone please help explain how exactly to come up with it
