Mathemagic: Find the trick!

A. Let the number be $n = 10 - x, x \in W$ . Therefore as the statement guides \ $\times 37 \implies 37(n) = 37n \\ +74 \implies 37n + 74 \\ \times 3 - 222 \implies 3(37n+74) = 111n + 222-222 \\ \text{divide} \implies \dfrac{111n}{n} = 111$

B. Let the number be $n$ . Let the next number be $10n + 9 - n = 9n + 9 = 9(n+1)$ Now the new number is a multiple of $9$ which have a unique property that sum of all digits of multiples of $9$ is $9$ except $0$ .

C. Let the number be $n$ . The next number would be $10n$ . If we want to find repeated sum of digits you will have $repeated \ sum \ of \ digits \ of \ n + 0$ . Then when you subtract you will have $r(n) - n$ . Let $n = 10a + b$

$r(n) = a + b \\ r(n) - n = a + b - n$

And this continues for all n-digit number. So it depends on the number you choose.

A. Answer is always 111 as we have multiplied by 111 and removed the added number. So we will get 111 only as answer.

B. Answer is always 9, as the resulting number is divisible by 9

C. No unique answer. Cant be used.