How could I go about finding the division remainder of a number in Python?
For example:
If the number is 26 and divided number is 7, then the division remainder is 5.
(since 7+7+7=21 and 26-21=5.)
For simple divisibility testing, see How do you check whether a number is divisible by another number?.
you are looking for the modulo operator:
a % b
for example:
>>> 26 % 7
5
Of course, maybe they wanted you to implement it yourself, which wouldn't be too difficult either.
The remainder of a division can be discovered using the operator %:
>>> 26%7
5
In case you need both the quotient and the modulo, there's the builtin divmod function:
>>> seconds= 137
>>> minutes, seconds= divmod(seconds, 60)
26 % 7 (you will get remainder)
26 / 7 (you will get divisor, can be float value)
26 // 7 (you will get divisor, only integer value)
If you want to get quotient and remainder in one line of code (more general usecase), use:
quotient, remainder = divmod(dividend, divisor)
#or
divmod(26, 7)
From Python 3.7, there is a new math.remainder() function:
from math import remainder
print(remainder(26,7))
Output:
-2.0 # not 5
Note, as above, it's not the same as %.
Quoting the documentation:
math.remainder(x, y)
Return the IEEE 754-style remainder of x with respect to y. For finite x and finite nonzero y, this is the difference x - n*y, where n is the closest integer to the exact value of the quotient x / y. If x / y is exactly halfway between two consecutive integers, the nearest even integer is used for n. The remainder r = remainder(x, y) thus always satisfies abs(r) <= 0.5 * abs(y).
Special cases follow IEEE 754: in particular, remainder(x, math.inf) is x for any finite x, and remainder(x, 0) and remainder(math.inf, x) raise ValueError for any non-NaN x. If the result of the remainder operation is zero, that zero will have the same sign as x.
On platforms using IEEE 754 binary floating-point, the result of this operation is always exactly representable: no rounding error is introduced.
Issue29962 describes the rationale for creating the new function.
If you want to avoid modulo, you can also use a combination of the four basic operations :)
26 - (26 // 7 * 7) = 5
We can solve this by using modulus operator (%)
26 % 7 = 5;
but 26 / 7 = 3 because it will give quotient but % operator will give remainder.
Use the % instead of the / when you divide. This will return the remainder for you. So in your case
26 % 7 = 5
Modulo would be the correct answer, but if you're doing it manually this should work.
num = input("Enter a number: ")
div = input("Enter a divisor: ")
while num >= div:
num -= div
print num
You can find remainder using modulo operator Example
a=14
b=10
print(a%b)
It will print 4
If you want the remainder of your division problem, just use the actual remainder rules, just like in mathematics. Granted this won't give you a decimal output.
valone = 8
valtwo = 3
x = valone / valtwo
r = valone - (valtwo * x)
print "Answer: %s with a remainder of %s" % (x, r)
If you want to make this in a calculator format, just substitute valone = 8
with valone = int(input("Value One")). Do the same with valtwo = 3, but different vairables obviously.
Here's an integer version of remainder in Python, which should give the same results as C's "%" operator:
def remainder(n, d):
return (-1 if n < 0 else 1) * (abs(n) % abs(d))
Expected results:
remainder(123, 10) == 3
remainder(123, -10) == 3
remainder(-123, 10) == -3
remainder(-123, -10) == -3
you can define a function and call it remainder with 2 values like rem(number1,number2) that returns number1%number2 then create a while and set it to true then print out two inputs for your function holding number 1 and 2 then print(rem(number1,number2)
