Math Problem Statement
Find the last 2 digits of the smallest 5-digit number that when divided by 8, 11, and 12 gives a remainder of 1, 2, and 5 respectively.
Solution
To solve this problem, we need to find the smallest 5-digit number such that:
N &\equiv 1 \ (\text{mod} \ 8), \\ N &\equiv 2 \ (\text{mod} \ 11), \\ N &\equiv 5 \ (\text{mod} \ 12). \end{aligned}$$ We'll use the Chinese Remainder Theorem (CRT) to solve these congruences. ### Step 1: Rewrite the congruences Given: $$\begin{aligned} N &\equiv 1 \ (\text{mod} \ 8), \\ N &\equiv 2 \ (\text{mod} \ 11), \\ N &\equiv 5 \ (\text{mod} \ 12). \end{aligned}$$ Since $$12 = 3 \times 4$$, and 8 is divisible by 4, we can consider the system: $$\begin{aligned} N &\equiv 5 \ (\text{mod} \ 3), \\ N &\equiv 5 \ (\text{mod} \ 4). \end{aligned}$$ The first congruence $$N \equiv 5 \ (\text{mod} \ 3)$$ reduces to $$N \equiv 2 \ (\text{mod} \ 3)$$ since $$5 \equiv 2 \ (\text{mod} \ 3)$$. Thus, the new system is: $$\begin{aligned} N &\equiv 1 \ (\text{mod} \ 8), \\ N &\equiv 2 \ (\text{mod} \ 11), \\ N &\equiv 2 \ (\text{mod} \ 3), \\ N &\equiv 5 \ (\text{mod} \ 4). \end{aligned}$$ ### Step 2: Solve the system of congruences Let's solve $$N$$ modulo 8, 11, and 12 directly. 1. **Congruence 1 and 2**: Combine $$N \equiv 1 \ (\text{mod} \ 8)$$ and $$N \equiv 2 \ (\text{mod} \ 11)$$. Assume $$N = 8k + 1$$. Substitute this into the second congruence: $$8k + 1 \equiv 2 \ (\text{mod} \ 11)$$ Simplifying: $$8k \equiv 1 \ (\text{mod} \ 11)$$ To solve for $$k$$, find the multiplicative inverse of 8 modulo 11. The inverse of 8 modulo 11 is 7 (since $$8 \times 7 = 56 \equiv 1 \ (\text{mod} \ 11)$$). Thus: $$k \equiv 7 \ (\text{mod} \ 11)$$ Let $$k = 11m + 7$$, so: $$N = 8(11m + 7) + 1 = 88m + 57$$ 2. **Congruence 3**: Now use $$N = 88m + 57$$ in $$N \equiv 5 \ (\text{mod} \ 12)$$: $$88m + 57 \equiv 5 \ (\text{mod} \ 12)$$ Since $$88 \equiv 4 \ (\text{mod} \ 12)$$ and $$57 \equiv 9 \ (\text{mod} \ 12)$$, we get: $$4m + 9 \equiv 5 \ (\text{mod} \ 12)$$ $$4m \equiv -4 \equiv 8 \ (\text{mod} \ 12)$$ Since the inverse of 4 modulo 12 is 3: $$m \equiv 8 \times 3 = 24 \equiv 0 \ (\text{mod} \ 3)$$ Let $$m = 3n$$, so: $$N = 88(3n) + 57 = 264n + 57$$ We want the smallest 5-digit number, so: $$264n + 57 \geq 10000$$ Solving for $$n$$: $$264n \geq 9943$$ $$n \geq \left\lceil \frac{9943}{264} \right\rceil = 38$$ Substitute $$n = 38$$ into the equation for $$N$$: $$N = 264(38) + 57 = 10032 + 57 = 10089$$ ### Conclusion The smallest 5-digit number $$N$$ that satisfies the given conditions is 10089. The last two digits of $$N$$ are **89**. --- Would you like more details, or have any questions? Here are some related questions: 1. How does the Chinese Remainder Theorem apply to solving congruences? 2. What are multiplicative inverses in modular arithmetic? 3. Can we solve systems of congruences with non-coprime moduli? 4. How can we check if our solution is correct using direct substitution? 5. How does the step-by-step solution process work for congruences involving large numbers? **Tip:** When solving modular equations, finding the smallest solution often requires combining congruences systematically to simplify the problem.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Modular Arithmetic
Chinese Remainder Theorem
Formulas
-
Theorems
Chinese Remainder Theorem
Suitable Grade Level
Advanced High School and Above
Related Recommendation
Find the Minimum 3-Digit Number Divisible by 7, 5, and 4 with a Remainder of 2
Chinese Remainder Theorem: Finding Remainders for Successive Divisions
Find the Smallest Integer x Satisfying Multiple Congruences
Finding the Largest Three-Digit Number with Specific Remainders
Solve a 4-Digit Number Problem Using the Chinese Remainder Theorem