<p>If four whole numbers taken at random are multiplied together, then find the probability that the last digit in the product is 1, 3, 7 or 9.</p>
Step-by-Step Solution
Key Concept: The last digit of a product depends only on the last digits of the factors. For the product to end in 1, 3, 7, or 9, all four numbers must individually end in 1, 3, 7, or 9 (since these are the only digits whose products never introduce 0, 2, 4, 5, 6, or 8 in the units place).
<p><strong>Step 1:</strong> Identify favorable last digits. For a product to end in 1, 3, 7, or 9, each of the four numbers must have a last digit from the set {1, 3, 7, 9}.</p><p><strong>Step 2:</strong> Verify closure property. Check that products of numbers ending in {1, 3, 7, 9} only produce last digits in {1, 3, 7, 9}: 1×1=1, 1×3=3, 1×7=7, 1×9=9, 3×3=9, 3×7=21(1), 3×9=27(7), 7×7=49(9), 7×9=63(3), 9×9=81(1). ✓</p><p><strong>Step 3:</strong> Count favorable outcomes. Each of the 4 numbers can end in any of {1, 3, 7, 9}, giving us 4 choices per number. Favorable outcomes = 4 × 4 × 4 × 4 = 256.</p><p><strong>Step 4:</strong> Count total outcomes. Each number can end in any digit {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, giving us 10 choices per number. Total outcomes = 10 × 10 × 10 × 10 = 10000.</p><p><strong>Step 5:</strong> Calculate probability. P = 256/10000 = 16/625.</p><p>∴ Answer: <strong>16/625</strong></p>
Correct Answer: 16/625