JavaScript (Node.js), (削除) 43 (削除ここまで) (削除) 41 (削除ここまで) 37(削除) 37 (削除ここまで) 36 bytes
Thanks @Dennis Dennis for the idea of using string interpolation in this answer and save 4 bytes!
Thanks @ØrjanJohansen for -1!
a=>b=>eval(`0x${a}*0x${b}==0x$<0x${a*b}`)
Take input as 2 numbers in currying syntax (a)(b)Of course when the destination base is less than the original base (string also workslike in my Jelly answer, because Javascriptthe base is very weakly typed2), and output true for easy, the false< for not easy to multiplymust be flipped.
JavaScript (Node.js), (削除) 43 (削除ここまで) (削除) 41 (削除ここまで) 37 bytes
Thanks @Dennis for the idea of using string interpolation in this answer and save 4 bytes!
a=>b=>eval(`0x${a}*0x${b}==0x${a*b}`)
Take input as 2 numbers in currying syntax (a)(b) (string also works, because Javascript is very weakly typed), and output true for easy, false for not easy to multiply.
JavaScript (Node.js), (削除) 43 (削除ここまで) (削除) 41 (削除ここまで) (削除) 37 (削除ここまで) 36 bytes
Thanks @Dennis for the idea of using string interpolation in this answer and save 4 bytes!
Thanks @ØrjanJohansen for -1!
a=>b=>eval(`0x${a}*0x${b}<0x${a*b}`)
Of course when the destination base is less than the original base (like in my Jelly answer, the base is 2) the < must be flipped.
JavaScript (Node.js), (削除) 43 (削除ここまで) 41(削除) 41 (削除ここまで) 37 bytes
Thanks @Dennis for the idea of using string interpolation in this answer and save 4 bytes!
a=>b=>(g=n=>eval('0x'+n))a=>b=>eval(`0x${a)*g(}*0x${b)==g(}==0x${a*b}`)
The algorithm should be correctTake input as 2 numbers in currying syntax (a)(b) (string also works, because Javascript is very weakly typed), and output true for easy, false for not easy to multiply.
JavaScript (Node.js), (削除) 43 (削除ここまで) 41 bytes
a=>b=>(g=n=>eval('0x'+n))(a)*g(b)==g(a*b)
The algorithm should be correct.
JavaScript (Node.js), (削除) 43 (削除ここまで) (削除) 41 (削除ここまで) 37 bytes
Thanks @Dennis for the idea of using string interpolation in this answer and save 4 bytes!
a=>b=>eval(`0x${a}*0x${b}==0x${a*b}`)
Take input as 2 numbers in currying syntax (a)(b) (string also works, because Javascript is very weakly typed), and output true for easy, false for not easy to multiply.
JavaScript (Node.js), 43(削除) 43 (削除ここまで) 41 bytes
a=>b=>(g=n=>parseIntg=n=>eval(n,36'0x'+n))(a)*g(b)==g(a*b)
I think thisThe algorithm should be correct. @LeakyNun ?
JavaScript (Node.js), 43 bytes
a=>b=>(g=n=>parseInt(n,36))(a)*g(b)==g(a*b)
I think this should be correct. @LeakyNun ?
JavaScript (Node.js), (削除) 43 (削除ここまで) 41 bytes
a=>b=>(g=n=>eval('0x'+n))(a)*g(b)==g(a*b)
The algorithm should be correct.