Find the smallest positive integer which, when decreased by 7, is exactly divisible by 12, 16, 18, 21 and 28.

Introduction / Context:
This problem is a standard application of least common multiple (L.C.M.) in which a number, after being decreased by a certain constant, becomes divisible by several integers. Such questions test whether you can recognise that the adjusted number must be a common multiple and then find the smallest original number that satisfies the condition.

Given Data / Assumptions:

• We seek a smallest number N.
• N, when diminished by 7 (N - 7), is divisible by 12, 16, 18, 21, and 28.
• We must find this smallest positive integer N.

Concept / Approach:
If N - 7 is divisible by several numbers, then N - 7 must be a common multiple of those numbers. To get the smallest such N, we take N - 7 as the least common multiple of 12, 16, 18, 21, and 28. Then we add 7 back to obtain N. Therefore the key step is computing the L.C.M. of the given set of integers.

Step-by-Step Solution:
Step 1: Let N be the required number.Step 2: Given that N - 7 is divisible by 12, 16, 18, 21, and 28.Step 3: Therefore, N - 7 must be a common multiple of these five numbers.Step 4: To get the smallest possible N, take N - 7 equal to the L.C.M. of 12, 16, 18, 21, and 28.Step 5: Compute prime factors: 12 = 2^2 * 3, 16 = 2^4, 18 = 2 * 3^2, 21 = 3 * 7, 28 = 2^2 * 7.Step 6: Take highest powers: 2^4, 3^2, and 7^1.Step 7: L.C.M. = 2^4 * 3^2 * 7 = 16 * 9 * 7 = 144 * 7 = 1008.Step 8: So N - 7 = 1008, giving N = 1008 + 7 = 1015.

Verification / Alternative check:
Check N - 7 = 1015 - 7 = 1008.Now verify divisibility: 1008 ÷ 12 = 84, 1008 ÷ 16 = 63, 1008 ÷ 18 = 56, 1008 ÷ 21 = 48, 1008 ÷ 28 = 36. All quotients are integers, so N = 1015 satisfies all the conditions, and since 1008 is the L.C.M., no smaller N will work.

Why Other Options Are Wrong:
Option a (1008) corresponds to N - 7, not N itself. Option c (1022) and option d (1032) do not yield N - 7 values that are common multiples of all five numbers. Testing them will show at least one divisor fails in each case.

Common Pitfalls:
A common mistake is to stop at the L.C.M. and forget to add the 7 back, leading to the wrong answer 1008. Others miscalculate the L.C.M. because they do not consistently take the highest power of each prime across all numbers. Being systematic in prime factorization and remembering to undo the subtraction is crucial.

Final Answer:
The smallest required number is 1015.