What is the remainder when $85\times 87\times 89\times 91\times 95\times 96$is divided by 100?

Correct Option (1)

To determine the remainder when $85\times 87\times 89\times 91\times 95\times 96$ is divided by 100, we analyze the prime factorization of both the divisor and the dividend.

The divisor is $100$, which can be factorized as $100=2^2\times 5^2=4\times 25$.

Next, we examine the prime factors present in the terms of the product $85\times 87\times 89\times 91\times 95\times 96$:

• The term $85$ contains a factor of $5$ ($85=5\times 17$).
• The term $95$ contains another factor of $5$ ($95=5\times 19$).
• The term $96$ contains a factor of $4$ ($96=4\times 24$).

Therefore, the product $85\times 87\times 89\times 91\times 95\times 96$ can be expressed as:

$(5\times 17)\times 87\times 89\times 91\times (5\times 19)\times (4\times 24)$

Rearranging the terms to highlight the factors of 100:

$(5\times 5\times 4)\times (17\times 87\times 89\times 91\times 19\times 24)$

This clearly shows that the product contains $5\times 5\times 4=100$ as a factor. Consequently, the entire product is a multiple of 100 and is perfectly divisible by 100.

When a number is perfectly divisible by another number, the remainder is 0.

Incorrect Options:

Options 2 (1), 3 (2), and 4 (4) are incorrect. A non-zero remainder would imply that the product $85\times 87\times 89\times 91\times 95\times 96$ is not a perfect multiple of 100. However, as demonstrated, the product explicitly contains all the prime factors required for divisibility by 100 ($2^2$ and $5^2$), confirming that it is a multiple of 100 and thus leaves a remainder of 0.