This python3 code generalizes rounding. It includes a "base" argument, which can be a float greater than 0. The "number of places" argument may also be a float. When the base is 10 Round uses builtins.round.

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import builtins
def Round(a,base_place=0,base=10):
'''
implement Rounding to bases other than 10.
known deficiencies:
fails with decimal input
fails with fractions input
Beware of 0 < base < 1 since b**(-p) == (1/b)**(p)
>>> [Round(x,-1,5) for x in (142,143,147,)]
[140, 145, 145]
>>>
>>> [(p,Round(12.345,p,2)) for p in range(-3,3)]
[(-3, 16.0), (-2, 12.0), (-1, 12.0), (0, 12.0), (1, 12.5), (2, 12.25)]
>>>
>>> # Use for base < 1
>>> # one might want decimal input
>>> # Round to the nearest US nickel
>>> no_pennies = Round(12.07,-1,0.05)
>>> print('{0:.2f}'.format(no_pennies))
12.05
>>> # The advertising game
>>> price_per_gallon = Round(12.379999999999,-1,1/100)
>>> print('{0:.2f}'.format(price_per_gallon))
12.38
>>>
>>> # round to nearest multiple of square root of two
>>> x = Round(12.379999999999,1/2,2)
>>> print (x / 2**(1/2))
9.0
>>>
'''
# consider using sign transfer for negative a
if base == 10:
return builtins.round(a,base_place)
if base <= 0:
raise ValueError('base too complex')
b = base**base_place
return type(a)(int(a*b+0.5)/b)
``` |

The doctests show how to use the program when you're trying to fill your gas tank, or as a shop owner tired of pennies.

Rounding to binary is not only fun but might be useful for platform independent conversions between float and fractions.

And of course rounding to the nearest multiple of *n* meets a grade school challenge.