How many integers between 1 and 100 leave a remainder of 3 when divided by 5 and a remainder of 5 when
divided by 7?

We are tasked with finding how many integers between 1 and 100 leave:

- a remainder of 3 when divided by 5, and
- a remainder of 5 when divided by 7.

Let’s break this down step by step.

### Step 1: Represent the conditions as congruences
The conditions can be written as two congruences:
1. \( x \equiv 3 \pmod{5} \) (i.e., \( x \) leaves a remainder of 3 when divided by 5)
2. \( x \equiv 5 \pmod{7} \) (i.e., \( x \) leaves a remainder of 5 when divided by 7)

### Step 2: Solve the system of congruences using the Chinese Remainder Theorem
The system of congruences is:
\[
x \equiv 3 \pmod{5}
\]
\[
x \equiv 5 \pmod{7}
\]

We will solve this system step by step.

### Step 3: Use the first congruence \( x \equiv 3 \pmod{5} \)
Let \( x = 5k + 3 \) for some integer \( k \). We now substitute this into the second congruence.

### Step 4: Substitute into the second congruence
Substitute \( x = 5k + 3 \) into \( x \equiv 5 \pmod{7} \):
\[
5k + 3 \equiv 5 \pmod{7}
\]
Simplifying:
\[
5k \equiv 5 - 3 \pmod{7}
\]
\[
5k \equiv 2 \pmod{7}
\]

### Step 5: Solve \( 5k \equiv 2 \pmod{7} \)
Now, we need to solve \( 5k \equiv 2 \pmod{7} \). To do this, we can multiply both sides by the modular inverse of 5 modulo 7.

The inverse of 5 modulo 7 can be found using trial and error or extended Euclidean algorithm. The inverse of 5 modulo 7 is 3, because:
\[
5 \times 3 = 15 \equiv 1 \pmod{7}
\]

Now multiply both sides of the equation \( 5k \equiv 2 \pmod{7} \) by 3:
\[
3 \times 5k \equiv 3 \times 2 \pmod{7}
\]
\[
k \equiv 6 \pmod{7}
\]

Thus, \( k = 7m + 6 \) for some integer \( m \).

### Step 6: Substitute back into \( x = 5k + 3 \)
Substitute \( k = 7m + 6 \) into \( x = 5k + 3 \):
\[
x = 5(7m + 6) + 3
\]
\[
x = 35m + 30 + 3
\]
\[
x = 35m + 33
\]

### Step 7: Find values of \( x \) between 1 and 100
We need to find the integers \( x = 35m + 33 \) that are between 1 and 100. Solving for \( m \):
\[
1 \leq 35m + 33 \leq 100
\]
Subtract 33 from all parts:
\[
-32 \leq 35m \leq 67
\]
Divide by 35:
\[
-\frac{32}{35} \leq m \leq \frac{67}{35}
\]
\[
m \in \{0, 1\}
\]

### Step 8: Calculate the corresponding \( x \) values
For \( m = 0 \):
\[
x = 35(0) + 33 = 33
\]
For \( m = 1 \):
\[
x = 35(1) + 33 = 68
\]

### Step 9: Conclusion
The integers between 1 and 100 that satisfy both conditions are \( 33 \) and \( 68 \). Therefore, there are **2 integers** that satisfy the given conditions.

---

Do you need more details or have any questions? Here are some related problems you might find interesting:

1. How many integers between 1 and 200 leave a remainder of 4 when divided by 6 and 3 when divided by 8?
2. What are the smallest three integers that satisfy \( x \equiv 2 \pmod{4} \) and \( x \equiv 1 \pmod{3} \)?
3. How many integers between 1 and 500 satisfy \( x \equiv 5 \pmod{7} \) and \( x \equiv 2 \pmod{9} \)?
4. Solve the system \( x \equiv 1 \pmod{2} \) and \( x \equiv 3 \pmod{5} \).
5. How many integers between 1 and 1000 satisfy \( x \equiv 1 \pmod{11} \) and \( x \equiv 5 \pmod{13} \)?

**Tip**: The Chinese Remainder Theorem is a powerful tool for solving systems of simultaneous congruences.

How many integers between 1 and 100 leave a remainder of 3 when divided by 5 and a remainder of 5 when divided by 7?

This problem shows how to find integers between 1 and 100 that leave a remainder of 3 when divided by 5 and a remainder of 5 when divided by 7. The solution involves using the Chinese Remainder Theorem, breaking down step-by-step how to solve the system of congruences.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation