How to query for a specific type in type union arguments?

Question

typescript

How to make it so that in the interface 'B', the function '1' takes an argument of type 'A', whose name has the value '1', and so on?

type A = {
    name: '1'
} | {
    name: '2'
} | {
    name: '3'
}

interface B {
    1: (a: A) => void // 'A', где name 1
    2: (a: A) => void // 'A', где name 2
    3: (a: A) => void // 'A', где name 3
}

Answer 1

interface A<T extends number> {
  name: `${T}`;
}

A<1>
A<2>
...

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https://www.typescriptlang.org/play?#code/JYOwLgpg...