Find the remainder when the product 1234 × 1235 × 1237 is divided by 8.

Introduction / Context:
This problem is a remainder question involving multiplication of large numbers. Instead of multiplying the full numbers, we use modular arithmetic to simplify each factor modulo 8 and then multiply the smaller residues. This approach greatly reduces calculation effort and is widely used in aptitude and competitive exams.

Given Data / Assumptions:

• We are given the product 1234 × 1235 × 1237.
• This product is divided by 8.
• We need to find the remainder obtained on division by 8.
• No calculators are assumed; manual simplification through modular arithmetic is expected.

Concept / Approach:
When computing a product modulo a number, we can reduce each factor modulo that number before multiplying. If a ≡ r (mod m) and b ≡ s (mod m), then a * b ≡ r * s (mod m). The same property extends to more factors. Since 8 is a small modulus, it is very easy to find each number's remainder on division by 8 using only the last three bits (or just doing short division). We then multiply the remainders and reduce again modulo 8.

Step-by-Step Solution:
Step 1: Compute 1234 mod 8. We note that 8 * 154 = 1232, so 1234 - 1232 = 2. Thus 1234 ≡ 2 (mod 8).Step 2: Compute 1235 mod 8. Since 1234 ≡ 2 (mod 8), 1235 ≡ 3 (mod 8).Step 3: Compute 1237 mod 8. From 1232 = 8 * 154, 1237 - 1232 = 5, so 1237 ≡ 5 (mod 8).Step 4: Replace each large factor with its remainder: 1234 × 1235 × 1237 ≡ 2 × 3 × 5 (mod 8).Step 5: Multiply the remainders: 2 × 3 = 6, then 6 × 5 = 30.Step 6: Reduce 30 modulo 8. Since 8 * 3 = 24, 30 - 24 = 6.Step 7: Therefore the remainder when 1234 × 1235 × 1237 is divided by 8 is 6.

Verification / Alternative check:
To verify, you could check the intermediate remainders with a different approach, such as using only the last three digits of each number (since 1000 is divisible by 8). The last three digits of 1234, 1235, and 1237 are 234, 235, and 237, which also reduce to remainders 2, 3, and 5 modulo 8. Repeating the multiplication and reduction still gives remainder 6. This confirms that our modular arithmetic steps are correct.

Why Other Options Are Wrong:
Option 4: Would appear if there was a miscalculation in the last modular reduction, such as taking 30 - 26 instead of 30 - 24.Option 2: Could result from mistakenly using one of the remainders incorrectly or mixing up multiplication order.Option 0: This would mean the product is divisible exactly by 8, which is not the case here.Option 1: Represents a random plausible looking remainder but does not match the correct modular arithmetic.

Common Pitfalls:
Typical mistakes include forgetting to reduce numbers modulo 8 before multiplication, leading to unwieldy large products. Another common error is reducing incorrectly, for example miscomputing 1234 mod 8. Some students also forget to reduce the final product modulo 8, stopping at 30 instead of reducing it to 6. Carefully applying modular rules at every step ensures accuracy and speed.

Final Answer:
The remainder when 1234 × 1235 × 1237 is divided by 8 is 6.