Q1.
When will two even numbers add up to give a multiple of 4? This problem is similar to the question of identifying when adding two numbers will result in an even number. Can you see this?
Answer
Exactly when both are multiples of 4, or neither is. There are three cases to examine, and algebra settles all three:
| Case | Algebra | Multiple of 4? | Example |
|---|---|---|---|
| both multiples of 4 | 4p + 4q = 4(p + q) | Always | 12 + 16 = 28 = 4 × 7 |
| neither a multiple of 4 | (4p + 2) + (4q + 2) = 4(p + q + 1) | Always | 6 + 10 = 16 = 4 × 4 |
| one of each | 4p + (4q + 2) = 4(p + q) + 2 | Never | 8 + 10 = 18 = 4 × 4 + 2 |
Yes, it is the same problem one level up. Compare the two tables:
even + even = even ↔ 4p + 4q is a multiple of 4
odd + odd = even ↔ (4p + 2) + (4q + 2) is a multiple of 4
even + odd = odd ↔ 4p + (4q + 2) is not a multiple of 4
odd + odd = even ↔ (4p + 2) + (4q + 2) is a multiple of 4
even + odd = odd ↔ 4p + (4q + 2) is not a multiple of 4
Why it happens: in the parity question every whole number leaves remainder 0 or 1 on division by 2, and 'same remainder' wins. Here every even number leaves remainder 0 or 2 on division by 4, and again 'same remainder' wins. The structure is identical — only the divisor has changed.