When 390 is divided by either 7 or 11, it gives the same non-zero remainder. How many integers from 1 to 2000 that have this property?

We are tasked with finding how many integers from 1 to 2000 satisfy the property that, when divided by either 7 or 11, they give the same non-zero remainder.

### Step-by-Step Approach:

Let the integer be \( x \), and let \( r \) be the common remainder when divided by both 7 and 11. Therefore, we can write:
\[
x \equiv r \pmod{7} \quad \text{and} \quad x \equiv r \pmod{11}
\]
where \( r \in \{1, 2, 3, 4, 5, 6\} \), because the remainder is non-zero and must be less than both 7 and 11.

We need to find all integers \( x \) that satisfy:
\[
x \equiv r \pmod{\text{LCM}(7, 11)}
\]
The least common multiple (LCM) of 7 and 11 is:
\[
\text{LCM}(7, 11) = 77
\]
So, for a fixed remainder \( r \), the integers \( x \) that satisfy this condition are of the form:
\[
x = 77k + r \quad \text{for integers } k.
\]
Now, we need to find how many such integers \( x \) lie between 1 and 2000.

### Step 1: Range of \( x \)
The integers of the form \( 77k + r \) must lie in the range \( 1 \leq x \leq 2000 \). Therefore:
\[
1 \leq 77k + r \leq 2000
\]
Subtracting \( r \) from all sides:
\[
1 - r \leq 77k \leq 2000 - r
\]
Dividing by 77:
\[
\frac{1 - r}{77} \leq k \leq \frac{2000 - r}{77}
\]
Let’s compute the range of \( k \) for each \( r \in \{1, 2, 3, 4, 5, 6\} \).

### Step 2: Count for Each Remainder \( r \)

For each remainder \( r \), we calculate the minimum and maximum values of \( k \), and count the number of possible integers.

1. **For \( r = 1 \):**
   \[
   \frac{1 - 1}{77} \leq k \leq \frac{2000 - 1}{77}
   \]
   \[
   0 \leq k \leq \frac{1999}{77} \approx 25.96
   \]
   So, \( k \) ranges from 0 to 25, which gives 26 integers.

2. **For \( r = 2 \):**
   \[
   \frac{1 - 2}{77} \leq k \leq \frac{2000 - 2}{77}
   \]
   \[
   -\frac{1}{77} \leq k \leq \frac{1998}{77} \approx 25.94
   \]
   So, \( k \) ranges from 0 to 25, which gives 26 integers.

3. **For \( r = 3 \):**
   \[
   \frac{1 - 3}{77} \leq k \leq \frac{2000 - 3}{77}
   \]
   \[
   -\frac{2}{77} \leq k \leq \frac{1997}{77} \approx 25.92
   \]
   So, \( k \) ranges from 0 to 25, which gives 26 integers.

4. **For \( r = 4 \):**
   \[
   \frac{1 - 4}{77} \leq k \leq \frac{2000 - 4}{77}
   \]
   \[
   -\frac{3}{77} \leq k \leq \frac{1996}{77} \approx 25.90
   \]
   So, \( k \) ranges from 0 to 25, which gives 26 integers.

5. **For \( r = 5 \):**
   \[
   \frac{1 - 5}{77} \leq k \leq \frac{2000 - 5}{77}
   \]
   \[
   -\frac{4}{77} \leq k \leq \frac{1995}{77} \approx 25.88
   \]
   So, \( k \) ranges from 0 to 25, which gives 26 integers.

6. **For \( r = 6 \):**
   \[
   \frac{1 - 6}{77} \leq k \leq \frac{2000 - 6}{77}
   \]
   \[
   -\frac{5}{77} \leq k \leq \frac{1994}{77} \approx 25.86
   \]
   So, \( k \) ranges from 0 to 25, which gives 26 integers.

### Step 3: Total Count
For each \( r \in \{1, 2, 3, 4, 5, 6\} \), there are 26 integers, so the total number of integers from 1 to 2000 that satisfy the given property is:
\[
6 \times 26 = 156
\]

### Final Answer:
There are **156** integers between 1 and 2000 that, when divided by either 7 or 11, give the same non-zero remainder.

Would you like further details or have any questions? 

Here are 5 related questions:
1. How do you calculate the least common multiple of two numbers?
2. Why is the remainder required to be non-zero in this problem?
3. Can you explain how modular arithmetic is applied here?
4. What would happen if the range changed from 1 to 5000?
5. How does the LCM affect the number of solutions in this type of problem?

**Tip:** When solving problems with multiple conditions, finding the least common multiple can help combine the conditions effectively.

When 390 is divided by either 7 or 11, it gives the same non-zero remainder. How many integers from 1 to 2000 that have this property?

This problem explores modular arithmetic by finding how many integers between 1 and 2000 give the same non-zero remainder when divided by both 7 and 11. Using the LCM of 7 and 11, discover a step-by-step solution suitable for Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation