What is the least value that should be added to 1812 in order to make the resulting number exactly divisible by 7, 11 and 14?

Introduction / Context:
This problem involves divisibility and least common multiple. We are asked to adjust a given number by adding the smallest possible value so that the new number becomes divisible by three given divisors. Understanding how to use least common multiple to combine divisibility conditions is very important in aptitude exams.

Given Data / Assumptions:
- We start with the number 1812. - We want 1812 plus some least non negative integer k to be divisible by 7, 11 and 14 simultaneously. - The required number must obey the divisibility conditions for all three divisors at once.

Concept / Approach:
- A number divisible by 14 and by 7 and by 11 must be divisible by the least common multiple of these three numbers. - Since 14 already contains the prime factor 7, the combined least common multiple is lcm(7, 11, 14) = 2 * 7 * 11 = 154. - Therefore, we need to find the smallest value k such that 1812 + k is a multiple of 154. - That is, we find the next multiple of 154 greater than or equal to 1812 and compute the difference.

Step-by-Step Solution:
Step 1: Factor the numbers. 14 = 2 * 7, while 7 and 11 are primes. Step 2: The least common multiple is 2 * 7 * 11 = 154. Step 3: We need a number of the form 154 * n that is just greater than or equal to 1812. Step 4: Compute 1812 divided by 154. 154 * 10 = 1540 and 154 * 11 = 1694, 154 * 12 = 1848. Step 5: 1848 is the first multiple of 154 that is greater than 1812. Step 6: The required addition k is 1848 - 1812 = 36. Step 7: Therefore, 1812 + 36 = 1848 is divisible by 7, 11 and 14.

Verification / Alternative check:
Check divisibility of 1848: 1848 / 7 = 264 exactly, so it is divisible by 7. Next, 1848 / 11 = 168 exactly, so it is divisible by 11. Since 1848 is divisible by 14 = 2 * 7 and we already know it is divisible by 7, checking evenness: 1848 is even, so it is divisible by 2 as well. Hence it satisfies all conditions.

Why Other Options Are Wrong:
Option A (12): 1812 + 12 = 1824; 1824 is not a multiple of 154 and fails divisibility by at least one of the three numbers. Option C (72): 1812 + 72 = 1884; 1884 divided by 154 is not an integer, so this is not a common multiple. Option D (154): 1812 + 154 = 1966; this is 154 more than 1848 and is not guaranteed to be the smallest such adjustment, and in fact 1966 is not a multiple of 154. Option E (18): 1812 + 18 = 1830 which is not divisible by 154, so it fails as well.

Common Pitfalls:
- Forgetting that 14 already includes a factor of 7 and incorrectly multiplying 7, 11 and 14 directly without reducing. - Trying to test each divisor separately on 1812 + k without using the least common multiple to simplify the problem. - Failing to choose the smallest positive difference to the next multiple, which can lead to a larger than necessary answer.

Final Answer:
The least value that must be added to 1812 to make it divisible by 7, 11 and 14 is 36.