Question

Let \(A = \{n \in N : n \text{ is a 3-digit number}\}\), \(B = \{9k + 2 : k \in N\}\) and \(C = \{9k + l : k \in N\}\) for some \(l\) (0<l<9) If the sum of all the elements of the set \(A \cap (B \cup C)\) is \(274 \times 400\), then \(l\) is equal to __________

Hint:

For sets defined by $nk+r$, the intersection with a range of integers forms an Arithmetic Progression. Use $S_n = \frac{n}{2}(a+l)$ for quick summation.

Answer 1

Correct Answer: 5

Step 1: \(A \cap B = \{101, 110, ......., 992\}\). Number of terms \(n = 100\). Sum \(S_B = \frac{100}{2}(101+992) = 54650\).
Step 2: Total Sum = \(109600\). Sum of \(A \cap C = 109600 - 54650 = 54950\).
Step 3: \(A \cap C = \{a, a+9, ......., a+99 \times 9\}\) where \(a = 9k+l\).
Step 4: \(54950 = \frac{100}{2}(2a + 99 \times 9) = 50(2a + 891) \Rightarrow 1099 = 2a + 891 \Rightarrow 2a = 208 \Rightarrow a = 104\).
Step 5: Since \(a = 104\), \(104 = 9(11) + 5\). So \(l = 5\).