# 2つの数字には固有の階乗が含まれていますか？

2つの数字を階乗に分解してください。彼らが何かを共有する場合、虚偽の値を返します。それ以外の場合は、真理値を返します。
この最近の質問に触発された）

# 例

``````20 = 3! + 3! + 3! + 2!
49 = 4! + 4! + 1!
``````

``````132 = 5! + 3! + 3!
32 = 4! + 3! + 2!
``````

3！両方の表現に現れるので、偽の値を返します。

# テストケース

If one input is `0`, the answer will always be
truthy. Other truthy テストケース:

``````{6, 3}, {4, 61}, {73, 2}, {12, 1}, {240, 2}, {5, 264}, {2, 91}, {673, 18},
{3, 12}, {72, 10}, {121, 26}, {127, 746}
``````

If both inputs are odd integers, or if both inputs are the same
positive integer, then the output will always be falsey. Other
falsey テストケース:

``````{8, 5}, {7, 5}, {27, 47}, {53, 11}, {13, 123}, {75, 77}, {163, 160}, {148, 53},
{225, 178}, {285, 169}, {39, 51}, {207, 334}, {153, 21}, {390, 128}, {506, 584},
{626, 370}, {819, 354}
``````

This is ,
so fewest bytes wins!

ベストアンサー

# Jelly, 7 bytes

``````Æ!ṠḄ&/¬
``````

Try it
online!

### 使い方

``````Æ!ṠḄ&/¬  Main link. Argument: (x, y) (pair of integers)

Æ!       Convert x and y to factorial base.
Ṡ      Apply the sign function to each digit.
Ḅ     Unbinary; convert each resulting Boolean array from base 2 to integer.
&/   Reduce the resulting pair of integers by bitwise AND.
¬  Take the logical NOT of the result.
``````