Binary Magic Cards [closed]

Question

There's a cool magic trick that works using the power of binary. The effect of the trick is as follows:

1. An audience member chooses some natural number in the range of 1 to x where x is chosen by the magician.

2. The magician hands the audience member some special cards. Each card contains some numbers from 1 to x.

3. The audience member selects the cards which contain their number.

4. Almost instantly, the magician can determine the original number selected.

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Specification:

The numbers used for the cards are determined based on binary place value. Each card is first labeled with a power of 2. The first card becomes 1, the second becomes 2, the third becomes 4, and so on.

From now on, I will refer to card n as the card labeled with n.

To determine whether a number k is on card n, determine whether k in binary has at 1 at place value n. Consider the numbers k=13 and n=4.

K in binary is 1101. The second digit (n=4) is 1, so k=13, n=4 is a valid combination.

Goal:

Given two natural numbers 0 < n < 128 and 0 < k < 128, determine whether n appears on card k. Any reasonable input and output is allowed. Standard loopholes are banned.

This is code-golf, so the fewest bytes wins.

Test cases

Answer 1

Pyth, 5 bytes

._.&E

Try it here.

Notice the cute face ._.

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How?

._.&EQ  -  Q means input and is implicit.

  .&    - Bitwise AND between:
    E     - The second input and
     Q    - The first input.
._      - Sign. 0 if it equals 0, 1 otherwise.

If inconsistent values are allowed:

Pyth, 3 bytes

No cute face this time ._.

.&E

Try it here.

Answer 2

Jelly, 1 byte

&

Try it online!

Bitwise AND Builtin

Answer 3

05AB1E, 2 bytes

&Ā

Try it online!

Answer 4

Proton, 3 bytes

(&)

Try it online!

(Waiting for pull from TIO before it works on TIO but you can verify it by cloning the repository)

Functional operator of the Bitwise AND operator