<p>If a seven-digit number made up of all distinct digits 8, 7, 6, 4, 2, <i>x</i>, and <i>y</i> is divisible by 3, then</p>
<p>maximum value of \(x - y\) is 9</p>
<p>maximum value of \(x + y\) is 12</p>
<p>minimum value of \(xy\) is 0</p>
<p>minimum value of \(x + y\) is 3</p>
Step-by-Step Solution
Key Concept: A number is divisible by 3 if and only if the sum of its digits is divisible by 3. Since we have fixed digits 8+7+6+4+2=27 (divisible by 3), we need x+y≡0 (mod 3) to determine which pairs (x,y) from remaining digits {0,1,3,5,9} work.
<p><strong>Step 1:</strong> Sum of given digits = 8+7+6+4+2 = 27, which is divisible by 3.</p><p><strong>Step 2:</strong> For the 7-digit number to be divisible by 3, we need (27+x+y) ≡ 0 (mod 3), which means x+y ≡ 0 (mod 3).</p><p><strong>Step 3:</strong> The available digits for x and y are {0,1,3,5,9}. We need pairs where x+y ≡ 0 (mod 3):</p><p>• Digits mod 3: 0≡0, 1≡1, 3≡0, 5≡2, 9≡0</p><p>• Valid pairs: (0,3), (0,9), (3,9) where sum ≡ 0 (mod 3)</p><p>• Also: (1,5) gives 1+2≡0 (mod 3)</p><p><strong>Step 4:</strong> Check each option against valid digit pairs {(0,3), (0,9), (3,9), (1,5)}:</p><p><strong>A:</strong> If x and y include 0,3 or 0,9 or 3,9 → TRUE</p><p><strong>B:</strong> If both x,y are odd → Only (1,5) works, not all cases → FALSE</p><p><strong>C:</strong> Exactly one is odd requires checking if this always holds → NOT always true</p><p><strong>D:</strong> At least one is single-digit from {0,1,3,5,9} → TRUE (all valid pairs satisfy this)</p><p>∴ Answer: <strong>ABD</strong></p>
Correct Answer: ABD