You are running into the old problem with floating point numbers that not all numbers can be represented exactly. The command line is just showing you the full floating point form from memory.

With floating point representation, your rounded version is the same number. Since computers are binary, they store floating point numbers as an integer and then divide it by a power of two so 13.95 will be represented in a similar fashion to 125650429603636838/(2**53).

Double precision numbers have 53 bits (16 digits) of precision and regular floats have 24 bits (8 digits) of precision. The floating point type in Python uses double precision to store the values.

For example,

>>> 125650429603636838/(2**53)
13.949999999999999

>>> 234042163/(2**24)
13.949999988079071

>>> a = 13.946
>>> print(a)
13.946
>>> print("%.2f" % a)
13.95
>>> round(a,2)
13.949999999999999
>>> print("%.2f" % round(a, 2))
13.95
>>> print("{:.2f}".format(a))
13.95
>>> print("{:.2f}".format(round(a, 2)))
13.95
>>> print("{:.15f}".format(round(a, 2)))
13.949999999999999

If you are after only two decimal places (to display a currency value, for example), then you have a couple of better choices:

1. Use integers and store values in cents, not dollars and then divide by 100 to convert to dollars.
2. Or use a fixed point number like decimal.

Original result value: 15.5884572681

Use floor to round down to get 15.58:

import math
def solution(side): 
    result = (side**2)*(3**(0.5))/4 
    return math.floor(result*100)/100
print(solution(6))  # prints 15.58

Use round with precision 2 to get 15.59:

def solution(side): 
    result = (side**2)*(3**(0.5))/4 
    return round(result,2)
print(solution(6))  # prints 15.59

You have to convert the int to a string because int does not have a length property. Also You were checking if the digit was longer than 1 for a twice so I switched the SECOND NUMBER check to b

print('Please enter your 10 digit number')
a = raw_input("FIRST NUMBER: ")
if len(a) > 1:
    print ("Error. Only 1 digit allowed!")
a = int(a)
aa = (a*10)

b = raw_input("SECOND NUMBER: ")
if len(b) > 1:
    print ("Error. Only 1 digit allowed!")
b = int(b)
bb = (b*10)

Or more simply:

You could ask for the number and keep asking until the length is 10 and the input is a number

num = raw_input('Please enter your 10 digit number:')
while len(num) != 10 or (not num.isdigit()):
    print 'Not a 10 digit number'
    num = raw_input('Please enter your 10 digit number:')
num = int(num)
print 'The final number is: ', num

The problem you're having is confusion between a number, and a string representation of a number. These are different things.

When you do this:

nCost = '%.2f'%(nCost*1.25)

… you're multiplying your number by 1.25, then replacing it with a string that represents the number up to two decimal points. So, when you do this:

    nCost = nCost * 1.25

… you're trying to multiply a string by 1.25, which makes no sense.

I suspect you didn't want a string here at all; you just wanted to round the number as a number. To do that, use the round function:

nCost = round(nCost*1.25, 2)

Of course as soon as you multiply by 1.25 again, you may "unround" it by another decimal place or two:

>>> nCost = 1.33333
>>> nCost = round(nCost*1.25, 2)
>>> nCost
1.33
>>> nCost = nCost * 1.25
>>> nCost
1.6625

And really, it's almost always a bad idea to round values before doing arithmetic on them; that just multiplies rounding errors unnecessarily. It's better to just keep the highest-precision value around as long as possible, and round at the last possible second—ideally by using %.2f/{:.2f} formatting in the printing/writing-to-disk/etc. stage.

Another thing to keep in mind is that not all real numbers can be represented exactly as floats. So, for example, round(13.4999, 2) is actually 13.949999999999999289457264. A simple repr or str or %f will print this out as 13.95, so you may not notice… but it's still not exactly 13.95.

The best way to solve this problem, and your other problems, is to use the Decimal type instead of the float type. Then you can round the value to exactly 2 decimal places, do math in a 2-decimal-places context, etc.

The formating method for both integer and decimal is

>>> '{:06.2f}'.format(3.141592653589793)
'003.14'

the part before the . (6 here) denotes the total length of padding (including the .), and after the . (2fhere) denotes digits after decimal point. Hope it helps. checkout the link.

Let:

>>> num = 0.0012345

For up to 3 significant figures:

>>> f'{num:.3}'
'0.00123'

For 3 decimal places:

>>> f'{num:.3f}'
'0.001'

See the "presentation types for floating point and decimal" table at the bottom of this section for any additional requirements provided by e, E, f, F, g, G, n, %, None.

If i got this right this should work:

a=input()
if len(a)==9 and a[0] != '0' and a.isdigit():
  #if you want it to be integer
  a=int(a)
  do something

Hope it helps.