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perform arithmetic operations python [closed]

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given the number 12345 = 1x10^4 + 2x10^3 + 3x10^2 + 4x10^1 + 5x10^0, how to perform some arithmetic operations to leave you with just the digit at position 5 (from the left) and output it to the screen? Thanks for the help!

Assuming you want to stick to arithmetic operations (and not strings), use the modulo operator with 10 to get the remainder of division by 10, i.e. the unit:
12345%10
output: 5
For an arbitrary number, you need to compute the position, you can use log10 and ceil:
from math import log10, ceil
N = 5
number = 1234567
number//10**(ceil(log10(number))-N)%10
output: 5

i need more decimal places for pi calculation [closed]

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I'm trying to make a Pi calculator in python but I need more decimal places.
it would help a lot if someone edited my code and carefully explained what they did.
this is the code I'm using.
import math
d = 0
ans = 0
display = 0
while True:
display += 1
d += 1
ans += 1/d**2
if display == 1000000:
print(math.sqrt(ans*6))
display = 0
# displays value calculated every 1m iterations
output after ~85m iterations: (3.14159264498239)
I need more than 15 decimal places (3.14159264498239........)

You’re using a very slowly converging series for π²∕6, so you are not going to get a very precise value this way. Floating point limitations prevent further progress after 3.14159264498239, but you’re not going to get much further in any reasonable amount of time, anyway. You can get around these issues by some combination of
micro-optimising your code,
storing a list of values, reversing it and using math.fsum,
using decimal.Decimal,
using a better series (like this one),
using a method that converges to the value of π quickly, instead of a series (like this one),
using PyPy, or a faster language than Python,
from math import pi.

you could try with a generator:
def oddnumbers():
n = 1
while True:
yield n
n += 2
def pi_series():
odds = oddnumbers()
approximation = 0
while True:
approximation += (4 / next(odds))
yield approximation
approximation -= (4 / next(odds))
yield approximation
approx_pi = pi_series()
for x in range(10000000):
print(next(approx_pi))

Check if the length of the range divided by range element has no division remainder [closed]

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The code below shall show a specific comment if the number of elements in the array isn't integer-type number.
It works for dr=0.5 but not for dr=0.1 because print (4%0.1) returns 0.09999999999999978 instead of 0.
How can I change the code to get 4%0.1==0?
import math
limits_real=(-2,2)
dr=0.1
if (limits_real[1]-limits_real[0])%dr!=0:
print ('Inapropriate limits or elements')

It works for dr=0.5 but not for dr=0.1 because print (4%0.1) returns 0.09999999999999978 instead of 0.
Because floating point numbers have limitations that every programmer should know, see this:
https://docs.python.org/3/tutorial/floatingpoint.html
As showed at the end of above documentation, you can use decimal module, which works exacly right but is slower than normal floating point arihtmetics:
from decimal import Decimal
limits_real=(-2,2)
dr = Decimal("0.1")
if (limits_real[1] - limits_real[0]) % dr != 0:
print ('Inapropriate limits or elements')
Note that you should use a str while constructing the Decimal instance, do not use a float.

I found out only this solution :) At least it works
import math
limits_real=(-2,2)
dr=0.1
if (((limits_real[1] - limits_real[0]) * 1000) % (dr * 1000)) / 1000 != 0:
print ('Inapropriate limits or elements')

How does abs function work with this code : abs(3 + 4j)? [closed]

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I already know how does abs function works.I mean it only shows you how far the number is from zero
The only thing i can't just get is with this example:
print(abs(3 + 4j)) # prints 5 !! Why ?

Because for complex numbers abs(number) will return magnitude of Complex Numbers.
Magnitude value will counted as :
√x2+y2 = √(3² + 4²) = √(9 + 16) = √(25) = 5.0
So abs with complex number will return magnitude of complex numbers.
for further reference you can use https://www.geeksforgeeks.org/abs-in-python/.

As the rest of the answers stated above, 3+4j is a complex number and the formula of calculating the absolute value of a complex number x+yi is sqrt( (x^2) + (y^2) ). In your case it's:
sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5

Because in python, 3+4j is complex number and python calculates the absolute value or magnitude of the complex number when you do abs() on it. Magnitude of 3+4j is 5. Try this :
type(3+4j)
It should give <class 'complex'>.
Note : Magnitude of a complex number a+bj is ((a**2+b**2)**0.5)

Creating a function to generate a random number [closed]

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Note: I cannot use any of the built in functions to solve this.
This is the question:
Given a function that returns a random integer number between 1 and 5, create a function that creates a random integer between 1 and 7.

As it has been answered in Adam Rosenfield's solution a while ago you may try to use an algoritm similar to this one below:
int i;
do
{
i = 5 * (rand5() - 1) + rand5(); // i is now uniformly random between 1 and 25
} while(i > 21);
// i is now uniformly random between 1 and 21
return i % 7 + 1; // result is now uniformly random between 1 and 7
EDIT: Translation to python:
def rand7():
i = 0;
while True:
i = 5 * (rand5() -1) + rand5()
if i > 21:
break
return i % 7 + 1

Apparently the following doesn't work. See the linked answer and the other responses as well.
Well, here is an approach I am thinking of. It could be wrong. I am going with the assumption that the result should be randomly distributed within the larger range.
Generate 7 random numbers using the given random functions (that generates numbers [1,5]) and calculate their sum.
This will result in a value between 7 (1 * 7) and 35 (5 * 7).
Because this value is evenly divisible by the target range and randomly distributed, then it seems like it would be valid to collapse the intermediate range [7,35] back to [1,7] without losing uniformity.
Or maybe that's not quite it - but it seems like exploiting a common multiple between the numbers is key.

import random
def winningFunc():
return originalFuncReturning1to5() + random.randint(0,2)