Solve..... a number of oranges is shared among 7 pple n 6 remain, the same number is shared is shared among 6 n 5 remain shared among 5 n 4 remain. What is the number?

We need to find a number \( N \) of oranges such that:

1. When \( N \) is divided by 7, the remainder is 6:
   \[
   N \equiv 6 \pmod{7}
   \]

2. When \( N \) is divided by 6, the remainder is 5:
   \[
   N \equiv 5 \pmod{6}
   \]

3. When \( N \) is divided by 5, the remainder is 4:
   \[
   N \equiv 4 \pmod{5}
   \]

This can be solved using the **Chinese Remainder Theorem**.

---

### Step 1: Represent the problem mathematically
The conditions can be rewritten:
- \( N = 7k + 6 \) for some integer \( k \),
- \( N = 6m + 5 \) for some integer \( m \),
- \( N = 5n + 4 \) for some integer \( n \).

---

### Step 2: Solve the system step-by-step
#### Combine \( N \equiv 6 \pmod{7} \) and \( N \equiv 5 \pmod{6} \)
Let \( N = 7k + 6 \), and substitute into \( N \equiv 5 \pmod{6} \):
\[
7k + 6 \equiv 5 \pmod{6}
\]
Simplify \( 7k \mod 6 \) (note: \( 7 \equiv 1 \pmod{6} \)):
\[
k + 6 \equiv 5 \pmod{6} \quad \Rightarrow \quad k \equiv -1 \pmod{6}.
\]
Since \( k \equiv -1 \pmod{6} \), we rewrite as:
\[
k \equiv 5 \pmod{6}.
\]
Thus, \( k = 6j + 5 \) for some integer \( j \). Substitute \( k \) into \( N = 7k + 6 \):
\[
N = 7(6j + 5) + 6 = 42j + 35 + 6 = 42j + 41.
\]
So:
\[
N \equiv 41 \pmod{42}.
\]

---

#### Combine \( N \equiv 41 \pmod{42} \) and \( N \equiv 4 \pmod{5} \)
Let \( N = 42p + 41 \), and substitute into \( N \equiv 4 \pmod{5} \):
\[
42p + 41 \equiv 4 \pmod{5}.
\]
Simplify \( 42 \mod 5 \) (note: \( 42 \equiv 2 \pmod{5} \)):
\[
2p + 41 \equiv 4 \pmod{5}.
\]
Simplify \( 41 \mod 5 \) (note: \( 41 \equiv 1 \pmod{5} \)):
\[
2p + 1 \equiv 4 \pmod{5} \quad \Rightarrow \quad 2p \equiv 3 \pmod{5}.
\]
Multiply through by the modular inverse of 2 modulo 5 (which is 3):
\[
p \equiv 3 \cdot 3 \pmod{5} \quad \Rightarrow \quad p \equiv 9 \pmod{5} \quad \Rightarrow \quad p \equiv 4 \pmod{5}.
\]
Thus, \( p = 5q + 4 \) for some integer \( q \). Substitute \( p \) into \( N = 42p + 41 \):
\[
N = 42(5q + 4) + 41 = 210q + 168 + 41 = 210q + 209.
\]
So:
\[
N \equiv 209 \pmod{210}.
\]

---

### Final Answer:
The number of oranges is:
\[
\boxed{209}.
\]

### Verification:
1. \( 209 \div 7 = 29 \) remainder \( 6 \) ✔️
2. \( 209 \div 6 = 34 \) remainder \( 5 \) ✔️
3. \( 209 \div 5 = 41 \) remainder \( 4 \) ✔️

Let me know if you'd like a detailed breakdown of any step! 😊 

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#### 5 Related Questions:
1. How does the Chinese Remainder Theorem work in modular arithmetic problems?
2. What are other methods to solve simultaneous modular equations?
3. Can you find the general solution for this problem?
4. How would the solution change if there were an additional modulus condition?
5. What is the significance of modular arithmetic in number theory?

#### Tip:
When solving modular arithmetic problems, always verify the solution by substituting it back into the original congruences.

Learn how to solve a modular arithmetic problem using the Chinese Remainder Theorem. The problem involves dividing a number of oranges among 7, 6, and 5 people with specific remainders, and the solution shows that the number of oranges is 209. Detailed step-by-step breakdown included.

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