Q1.
Let us try to find a few more pairs of numbers from their sums and differences: (a) Sum = 27, Difference = 9 (b) Sum = 4, Difference = 12 (c) Sum = 0, Difference = 10 (d) Sum = 0, Difference = –10 (e) Sum = –7, Difference = –1 (f) Sum = –7, Difference = –13
Answer
Use the same two half-steps every time.
First number = (sum + difference) ÷ 2
Second number = (sum – difference) ÷ 2
Second number = (sum – difference) ÷ 2
| Sum | Difference | First number | Second number | Check | |
|---|---|---|---|---|---|
| (a) | 27 | 9 | 18 | 9 | 18 + 9 = 27, 18 – 9 = 9 |
| (b) | 4 | 12 | 8 | –4 | 8 + (–4) = 4, 8 – (–4) = 12 |
| (c) | 0 | 10 | 5 | –5 | 5 + (–5) = 0, 5 – (–5) = 10 |
| (d) | 0 | –10 | –5 | 5 | –5 + 5 = 0, –5 – 5 = –10 |
| (e) | –7 | –1 | –4 | –3 | –4 + (–3) = –7, –4 – (–3) = –1 |
| (f) | –7 | –13 | –10 | 3 | –10 + 3 = –7, –10 – 3 = –13 |
Two of them are worth walking through slowly.
(e) First = (–7 + (–1)) ÷ 2 = (–8) ÷ 2 = –4
Second = (–7 – (–1)) ÷ 2 = (–6) ÷ 2 = –3
(f) First = (–7 + (–13)) ÷ 2 = (–20) ÷ 2 = –10
Second = (–7 – (–13)) ÷ 2 = 6 ÷ 2 = 3
Second = (–7 – (–1)) ÷ 2 = (–6) ÷ 2 = –3
(f) First = (–7 + (–13)) ÷ 2 = (–20) ÷ 2 = –10
Second = (–7 – (–13)) ÷ 2 = 6 ÷ 2 = 3
Why it happens: add the two conditions and the second number cancels out. (first + second) + (first – second) = 2 × first, so first = (sum + difference) ÷ 2. Subtract them instead and the first number cancels, giving second = (sum – difference) ÷ 2. That is why the trick never fails.
Tip: notice (c) and (d). They are the same two numbers written in the two possible orders, so their differences are 10 and –10.