Question

Let A = {6, 10, 14...1006} and B is a set of positive divisors of the integer 360, then sum of elements of A$\cap$B is

Answer 1

154

Answer 2

Solution

1. The set $A = \{6, 10, 14, \dots, 1006\}$ is an arithmetic progression with first term $a=6$ and common difference $d=4$. Note that every term in $A$ is of the form $4k+2$.

2. The divisors of $360$ (where $360=2^3\times3^2\times5$) are:

   $$\{1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360\}.$$

3. To find $A\cap B$, select those divisors which are of the form $4k+2$:

   • $6 \equiv 2 \pmod{4}$
   • $10 \equiv 2 \pmod{4}$
   • $18 \equiv 2 \pmod{4}$
   • $30 \equiv 2 \pmod{4}$
   • $90 \equiv 2 \pmod{4}$

4. Sum these numbers:

   $$6 + 10 + 18 + 30 + 90 = 154.$$

Explanation (Core Minimal):

• Identify numbers in $A$ are $4k+2$.
• List divisors of $360$ and select those congruent to $2 \pmod{4}$: $6, 10, 18, 30, 90$.
• Their sum is $154$.

Answer: $154$