IAS 2023 — CSAT Maths

Correct Option (1)

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

The divisor is $100$100, which can be factorized as $100 = 2^2 \times 5^2 = 4 \times 25$100=22×52=4×25.

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

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

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

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

Rearranging the terms to highlight the factors of 100:

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

This clearly shows that the product contains $5 \times 5 \times 4 = 100$5×5×4=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$85×87×89×91×95×96 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$22 and $5^2$52), confirming that it is a multiple of 100 and thus leaves a remainder of 0.