If {X: 2 d- x d- 19; X integer} and 7 + x = 4 (mod 9), find the highest value of x

The question text came through garbled. It most likely reads: "If {x : 2 ≤ x ≤ 19; x an integer} and 7 + x ≡ 4 (mod 9), find the highest value of x."

Step 1: Analyze the question We need integers x between 2 and 19 inclusive that satisfy 7 + x ≡ 4 (mod 9), and then pick the largest.

Step 2: The concept "a ≡ b (mod 9)" means a and b leave the same remainder when divided by 9. Equivalently, a − b is a multiple of 9.

Step 3: Solve 7 + x ≡ 4 (mod 9) x ≡ 4 − 7 (mod 9) x ≡ −3 (mod 9) x ≡ 6 (mod 9)

So x = 6, 15, 24, ... Within 2 ≤ x ≤ 19, the possible values are 6 and 15.

Check x = 15: 7 + 15 = 22, and 22 = 2(9) + 4, so the remainder is 4 ✓.

Check the other options:

• x = 18: 7 + 18 = 25, remainder 7 ✗
• x = 5: 12, remainder 3 ✗
• x = 2: 9, remainder 0 ✗

Step 4: Answer The highest value is 15, which is option C.

Exam tip: For "a + x ≡ b (mod n)", find x ≡ b − a (mod n), then add multiples of n to list the values within the given range. Always check the largest candidate by substituting it back.