Apply Chinese remainder theorem in computer arithmetic with large integers to find the sum of 123684 and 413456.

Solution Using Chinese Remainder Theorem (CRT)

The Chinese Remainder Theorem (CRT) is a method to solve systems of congruences with pairwise coprime moduli. Here, we use it to compute the sum of two large integers efficiently by breaking them into smaller congruences.

Step 1: Choose Moduli

Select two coprime integers (moduli) $m_1=5$ and $m_2=7$. These are small primes that simplify computation.

Step 2: Compute Congruences

Find $123684\mathrm{mod}5$ and $123684\mathrm{mod}7$:

• $123684÷5=24736\times 5=123680$, remainder $4$. So, $123684\equiv 4\mathrm{mod}5$.
• $123684÷7=17669\times 7=123683$, remainder $1$. So, $123684\equiv 1\mathrm{mod}7$.

Similarly, for $413456$:

• $413456÷5=82691\times 5=413455$, remainder $1$. So, $413456\equiv 1\mathrm{mod}5$.
• $413456÷7=59065\times 7=413455$, remainder $1$. So, $413456\equiv 1\mathrm{mod}7$.

Step 3: Sum Congruences

Add the congruences:

• $123684+413456\equiv (4+1)\mathrm{mod}5\Rightarrow 0\mathrm{mod}5$.
• $123684+413456\equiv (1+1)\mathrm{mod}7\Rightarrow 2\mathrm{mod}7$.

Now, solve for $x$ such that: $$\{\begin{array}{l}x\equiv 0\mathrm{mod}5,\\ x\equiv 2\mathrm{mod}7.\end{array}$$

Step 4: Apply CRT

Let $x=5k$. Substitute into the second congruence: $$5k\equiv 2\mathrm{mod}7.$$ Multiply both sides by the modular inverse of $5\mathrm{mod}7$. Since $5\times 3=15\equiv 1\mathrm{mod}7$, the inverse is $3$: $$k\equiv 2\times 3\mathrm{mod}7\Rightarrow k\equiv 6\mathrm{mod}7.$$ Thus, $k=7m+6$ for some integer $m$. Substitute back: $$x=5k=5(7m+6)=35m+30.$$ The smallest positive solution is $x=30$.

Step 5: Verify

Check $30\mathrm{mod}5=0$ and $30\mathrm{mod}7=2$. Correct.

Step 6: Final Sum

The sum modulo $35$ (since $5\times 7=35$) is $30$. To find the actual sum, compute: $$123684+413456=537140.$$ Now, $537140\mathrm{mod}35=30$, confirming our result.

Final Answer: The sum of $123684$ and $413456$ modulo $35$ is 30. The actual sum is 537140.