What is the remainder when 85 x 87 x 89 x 91 x 95 x 96 is divided by 100?

To solve this elegantly, we use the 'Zero Remainder Condition' seen in the 2022 exam [8]: if the divisor is fully contained in the product, the remainder is 0. We know 100 is made of two 5s and two 2s ($5 \\times 5 \\times 2 \\times 2$) [64]. Let's look for these ingredients in the numbers provided: 85 gives the first 5 ($17 \\times 5$). 95 gives the second 5 ($19 \\times 5$). 96 is a large even number ($32 \\times 3$), so it easily provides the two 2s. Since we have found the full set of factors for 100, the product is perfectly divisible by 100. We can completely ignore the other numbers (87, 89, 91) as distractors [35]. The remainder is 0.