Question

Find the residue of \( (67 + 89 + 90 + 87) \pmod{11} \):

• A. 3
• B. 0
• C. 2
• D. 1

Hint:

In modular arithmetic, it's often easier to use negative residues for numbers close to the modulus. For example, \( 87 \equiv 10 \pmod{11} \) is the same as \( 87 \equiv -1 \pmod{11} \). The sum would then be \( 1 + 1 + 2 - 1 = 3 \), giving the answer directly.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept: 

This problem involves modular arithmetic. The residue of a sum modulo \(n\) is the same as the sum of the individual residues modulo \(n\). This property allows us to simplify the calculation by working with smaller numbers.

Step 2: Key Formula or Approach:

The key property is \[ (a + b + c + d) \pmod{n} = \left( (a \pmod{n}) + (b \pmod{n}) + (c \pmod{n}) + (d \pmod{n}) \right) \pmod{n}. \] We will find the residue of each number in the sum modulo 11.

Step 3: Detailed Explanation:

1. \( 67 \div 11 \): \( 67 = 6 \times 11 + 1 \). So, \( 67 \equiv 1 \pmod{11} \).
2. \( 89 \div 11 \): \( 89 = 8 \times 11 + 1 \). So, \( 89 \equiv 1 \pmod{11} \).
3. \( 90 \div 11 \): \( 90 = 8 \times 11 + 2 \). So, \( 90 \equiv 2 \pmod{11} \).
4. \( 87 \div 11 \): \( 87 = 7 \times 11 + 10 \). So, \( 87 \equiv 10 \pmod{11} \).

Now, add the residues: \[ 1 + 1 + 2 + 10 = 14. \] Finally, find the residue of this sum modulo 11: \[ 14 \pmod{11}: \quad 14 = 1 \times 11 + 3. \quad \text{So,} \quad 14 \equiv 3 \pmod{11}. \]

Step 4: Final Answer:

The residue is \( \boxed{3} \).