How to find the remainder of division that satisfies a condition?

I am looking for a simple way to find the number c that making the (a+c) is divisibility b. For example, if a=12, b =4, then c =0. If a=13, b=4 then c=3

This is my code but it does not work well

a =12

b = 4

if (a % b):

  c =0

else:

  c = b - a % b

print (c)

Sorry if my question is too simple.

asked Nov 21 '18 at 23:12

John

96041333

With what numbers does it not work well?

– usr2564301
Nov 21 '18 at 23:13

Result of the first case is not correct. This is output repl.it/repls/TiredGratefulInformation

– John
Nov 21 '18 at 23:14

You got the comparison the wrong way around.

– usr2564301
Nov 21 '18 at 23:15

Thanks, Actually, I want to make it without the condition if. Is it possible?

– John
Nov 21 '18 at 23:18

1

(b - a % b) % b

– Peter Wood
Nov 21 '18 at 23:20

add a comment |

I am looking for a simple way to find the number c that making the (a+c) is divisibility b. For example, if a=12, b =4, then c =0. If a=13, b=4 then c=3

This is my code but it does not work well

a =12

b = 4

if (a % b):

  c =0

else:

  c = b - a % b

print (c)

Sorry if my question is too simple.

asked Nov 21 '18 at 23:12

John

96041333

With what numbers does it not work well?

– usr2564301
Nov 21 '18 at 23:13

Result of the first case is not correct. This is output repl.it/repls/TiredGratefulInformation

– John
Nov 21 '18 at 23:14

You got the comparison the wrong way around.

– usr2564301
Nov 21 '18 at 23:15

Thanks, Actually, I want to make it without the condition if. Is it possible?

– John
Nov 21 '18 at 23:18

1

(b - a % b) % b

– Peter Wood
Nov 21 '18 at 23:20

add a comment |

I am looking for a simple way to find the number c that making the (a+c) is divisibility b. For example, if a=12, b =4, then c =0. If a=13, b=4 then c=3

This is my code but it does not work well

a =12

b = 4

if (a % b):

  c =0

else:

  c = b - a % b

print (c)

Sorry if my question is too simple.

asked Nov 21 '18 at 23:12

John

96041333

I am looking for a simple way to find the number c that making the (a+c) is divisibility b. For example, if a=12, b =4, then c =0. If a=13, b=4 then c=3

This is my code but it does not work well

a =12

b = 4

if (a % b):

  c =0

else:

  c = b - a % b

print (c)

Sorry if my question is too simple.

python python-3.x

asked Nov 21 '18 at 23:12

John

96041333

asked Nov 21 '18 at 23:12

John

96041333

asked Nov 21 '18 at 23:12

John

96041333

asked Nov 21 '18 at 23:12

John

96041333

asked Nov 21 '18 at 23:12

John

96041333

With what numbers does it not work well?

– usr2564301
Nov 21 '18 at 23:13

Result of the first case is not correct. This is output repl.it/repls/TiredGratefulInformation

– John
Nov 21 '18 at 23:14

You got the comparison the wrong way around.

– usr2564301
Nov 21 '18 at 23:15

Thanks, Actually, I want to make it without the condition if. Is it possible?

– John
Nov 21 '18 at 23:18

1

(b - a % b) % b

– Peter Wood
Nov 21 '18 at 23:20

add a comment |

With what numbers does it not work well?

– usr2564301
Nov 21 '18 at 23:13

Result of the first case is not correct. This is output repl.it/repls/TiredGratefulInformation

– John
Nov 21 '18 at 23:14

You got the comparison the wrong way around.

– usr2564301
Nov 21 '18 at 23:15

Thanks, Actually, I want to make it without the condition if. Is it possible?

– John
Nov 21 '18 at 23:18

1

(b - a % b) % b

– Peter Wood
Nov 21 '18 at 23:20

With what numbers does it not work well?

– usr2564301
Nov 21 '18 at 23:13

Result of the first case is not correct. This is output repl.it/repls/TiredGratefulInformation

– John
Nov 21 '18 at 23:14

You got the comparison the wrong way around.

– usr2564301
Nov 21 '18 at 23:15

Thanks, Actually, I want to make it without the condition if. Is it possible?

– John
Nov 21 '18 at 23:18

(b - a % b) % b

– Peter Wood
Nov 21 '18 at 23:20

add a comment |

2 Answers
2

active

oldest

votes

There's probably a simpler way, but this works:

c = (b - a % b) % b

answered Nov 21 '18 at 23:19

zvone

9,65112346

add a comment |

c = -a % b

You'd need something more complicated in languages like C, where % matches the sign of the left operand instead of the right.

answered Nov 21 '18 at 23:24

user2357112

155k12167260

This is also tricky with decimal.Decimal, where the same problem happens.

– zvone
Nov 22 '18 at 19:06

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

There's probably a simpler way, but this works:

c = (b - a % b) % b

answered Nov 21 '18 at 23:19

zvone

9,65112346

add a comment |

There's probably a simpler way, but this works:

c = (b - a % b) % b

answered Nov 21 '18 at 23:19

zvone

9,65112346

add a comment |

There's probably a simpler way, but this works:

c = (b - a % b) % b

answered Nov 21 '18 at 23:19

zvone

9,65112346

There's probably a simpler way, but this works:

c = (b - a % b) % b

answered Nov 21 '18 at 23:19

zvone

9,65112346

answered Nov 21 '18 at 23:19

zvone

9,65112346

answered Nov 21 '18 at 23:19

zvone

9,65112346

answered Nov 21 '18 at 23:19

zvone

9,65112346

add a comment |

c = -a % b

You'd need something more complicated in languages like C, where % matches the sign of the left operand instead of the right.

answered Nov 21 '18 at 23:24

user2357112

155k12167260

This is also tricky with decimal.Decimal, where the same problem happens.

– zvone
Nov 22 '18 at 19:06

add a comment |

c = -a % b

You'd need something more complicated in languages like C, where % matches the sign of the left operand instead of the right.

answered Nov 21 '18 at 23:24

user2357112

155k12167260

This is also tricky with decimal.Decimal, where the same problem happens.

– zvone
Nov 22 '18 at 19:06

add a comment |

c = -a % b

You'd need something more complicated in languages like C, where % matches the sign of the left operand instead of the right.

answered Nov 21 '18 at 23:24

user2357112

155k12167260

c = -a % b

You'd need something more complicated in languages like C, where % matches the sign of the left operand instead of the right.

answered Nov 21 '18 at 23:24

user2357112

155k12167260

answered Nov 21 '18 at 23:24

user2357112

155k12167260

answered Nov 21 '18 at 23:24

user2357112

155k12167260

answered Nov 21 '18 at 23:24

user2357112

155k12167260

This is also tricky with decimal.Decimal, where the same problem happens.

– zvone
Nov 22 '18 at 19:06

add a comment |

This is also tricky with decimal.Decimal, where the same problem happens.

– zvone
Nov 22 '18 at 19:06

This is also tricky with decimal.Decimal, where the same problem happens.

– zvone
Nov 22 '18 at 19:06

add a comment |

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