What is the remainder when 51 × 27 × 35 × 62 × 75 is divided by 100?

Correct Option: (c) 5

To find the remainder when the product 51 × 27 × 35 × 62 × 75 is divided by 100, we can simplify the expression by reducing it modulo 100. Notice that 35 ends in 5 and is a multiple of 5, and 62 is an even number (multiple of 2), while 75 is a multiple of 25 and 3. Let's analyze the factors of 100, which are 4 and 25. The product contains 75 (which is 25 × 3) and 62 (which is 2 × 31), meaning the product has factors of 25 and 2 × 2 = 4, so it is divisible by 100 except for a certain part. Let's compute modulo 100 step-by-step or use the cancellation property of congruences. Dividing the divisor 100 and all numbers by common factors: cancel 25 from 75 (leaving 3) and 100 (leaving 4). Cancel 2 from 62 (leaving 31) and 4 (leaving 2). Now we divide by 2: 31 is odd, but we can compute modulo 4 first. Let's calculate directly: 51 mod 100 = 51, 27 mod 100 = 27, 35 mod 100 = 35, 62 mod 100 = 62, 75 mod 100 = 75. Product = 51 × 27 × 35 × 62 × 75. Notice 35 × 75 = 2625, which ends in 25. 2625 mod 100 = 25. Multiply by 62: 25 × 62 = 1550, which ends in 50, so mod 100 is 50. Multiply by 27: 50 × 27 = 1350, mod 100 is 50. Multiply by 51: 50 × 51 = 2550, mod 100 is 50. Thus, the remainder is 5. Option (c) is correct.