a positive integer when divided by q leaves remainder 21 when double that number is divided by q the remainder is 13 then q is

Let's solve the problem step by step.

Let the positive integer be \( n \), and let \( q \) be the divisor.

1. The first condition is that when \( n \) is divided by \( q \), the remainder is 21:
   \[
   n = kq + 21
   \]
   for some integer \( k \).

2. The second condition is that when \( 2n \) is divided by \( q \), the remainder is 13:
   \[
   2n = mq + 13
   \]
   for some integer \( m \).

Substituting \( n = kq + 21 \) into \( 2n \), we get:
\[
2n = 2(kq + 21) = 2kq + 42
\]
When \( 2n \) is divided by \( q \), the remainder is 13, so:
\[
2kq + 42 \equiv 13 \pmod{q}
\]
This simplifies to:
\[
42 \equiv 13 \pmod{q}
\]
Subtracting 13 from both sides:
\[
29 \equiv 0 \pmod{q}
\]
This implies \( q = 29 \).

Thus, the value of \( q \) is **29**.

Would you like more details or have any questions?

Here are some related questions:
1. What is the general method for solving remainder problems in modular arithmetic?
2. Can you find other integer values for \( q \) that satisfy similar conditions?
3. How can you determine the value of \( n \) given \( q \)?
4. What is the significance of modulus in division problems?
5. How would the problem change if the remainder for \( 2n \) was larger than \( q \)?

**Tip:** When solving problems involving remainders, consider using modular arithmetic to simplify calculations and find patterns.

Explore how to solve a modular arithmetic problem involving divisibility and remainders. Learn step-by-step solutions using modular congruences and understand the implications of the Chinese Remainder Theorem. Suitable for high school students studying number theory and modular arithmetic.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation