Modular Arithmetic Problems

A Step-by-Step Guide to Finding the Least Positive Value of x

Question 1.

Find the least positive value of \(x\) that satisfies each of the following congruences:

  1. \( 71 \equiv x \pmod{8} \)
  2. \( 78 + x \equiv 3 \pmod{5} \)
  3. \( 89 \equiv (x + 3) \pmod{4} \)
  4. \( 96 \equiv \frac{x}{7} \pmod{5} \)
  5. \( 5x \equiv 4 \pmod{6} \)

Solutions

(i) \( 71 \equiv x \pmod{8} \)

This congruence asks for the remainder when 71 is divided by 8. We can write 71 as a multiple of 8 plus a remainder: \[ 71 = 8 \times 8 + 7 \] The remainder is 7. Therefore, \( 71 \equiv 7 \pmod{8} \).

The least positive value of \(x\) is 7.

(ii) \( 78 + x \equiv 3 \pmod{5} \)

First, let's simplify the number 78 in modulo 5. The remainder of 78 divided by 5 is 3, so \( 78 \equiv 3 \pmod{5} \).
Substitute this into the congruence: \[ 3 + x \equiv 3 \pmod{5} \] Subtract 3 from both sides: \[ x \equiv 0 \pmod{5} \] This means \(x\) must be a multiple of 5. The problem asks for the least positive value of \(x\).

The least positive multiple of 5 is 5.

(iii) \( 89 \equiv (x + 3) \pmod{4} \)

First, simplify 89 in modulo 4. The remainder of 89 divided by 4 is 1, since \( 89 = 4 \times 22 + 1 \). So, \( 89 \equiv 1 \pmod{4} \).
Our congruence becomes: \[ 1 \equiv x + 3 \pmod{4} \] Subtract 3 from both sides: \[ 1 - 3 \equiv x \pmod{4} \] \[ -2 \equiv x \pmod{4} \] To find the least positive value, we can add multiples of 4 to -2 until we get a positive number: \( -2 + 4 = 2 \).

Thus, \( x \equiv 2 \pmod{4} \). The least positive value is 2.

(iv) \( 96 \equiv \frac{x}{7} \pmod{5} \)

This congruence implies that \(\frac{x}{7}\) is an integer, so \(x\) must be a multiple of 7. Let's say \( x = 7k \) for some integer \(k\).
Substitute this into the congruence: \[ 96 \equiv \frac{7k}{7} \pmod{5} \] \[ 96 \equiv k \pmod{5} \] Now, we find the value of 96 in modulo 5. Since \( 96 = 5 \times 19 + 1 \), we have \( 96 \equiv 1 \pmod{5} \). So, \( k \equiv 1 \pmod{5} \).
We need the least positive value for \(x\). This means we need the least positive integer \(k\), which is \(k=1\). Now we find \(x\): \[ x = 7k = 7(1) = 7 \]

The least positive value for \(x\) is 7.

(v) \( 5x \equiv 4 \pmod{6} \)

We can simplify the term \(5x\). In modulo 6, 5 is the same as -1 (since \(5 - 6 = -1\)). This makes the calculation easier. \[ -1x \equiv 4 \pmod{6} \] \[ -x \equiv 4 \pmod{6} \] Multiply both sides by -1: \[ x \equiv -4 \pmod{6} \] To find the least positive equivalent value, we add 6: \( x \equiv -4 + 6 \pmod{6} \). \[ x \equiv 2 \pmod{6} \] The solutions are 2, 8, 14, and so on.

The least positive value for \(x\) is 2.