How do u write a python function with parameter n , which prints square numbers between 0 and n inclusive

Answer 1

Thanks now I can see where I missed it

Answer 2

break it down. Start with importing Math.

import math

Now, define your function with the argument n

import math
def squares(n):

since you want to check numbers between 0 and n, I would use a for loop.

import math
def squares(n):
for i in range(0, n+1):

notice I went n+1.. range starts at 0, thus is ends 1 step before. so range(10) would run 0-9 .. adding 1 to n makes the range complete thru the actual n you desire.

Now add a check in each run of the loop that checks if the current number is a square number. To check, I used the math function that returns the square root of the number. I then check if it is an integer.

import math
def square(n):
for i in range(0,n+1):
if (math.sqrt(i)).is_integer():

if it is a perfect square, We print the number

import math
def square(n):
for i in range(0,n+1):
if (math.sqrt(i)).is_integer():
print(i)

Now you call it with the argument of any number greater than 0.

squares(1000)

Answer 3

When you take your problem and break it down to simple steps, it is only 5 lines of code. Take things 1 step at a time.

With some practice you can get it to be more efficient (square numbers can't end in 2,3,7 or 8 for example meaning you can skip checks on those numbers)

Eventually you will learn about List Comprehension and be able to take those 5 lines of code down to 3 lines

import math
def square(n):
print([i for i in range(0,n+1) if math.sqrt(i).is_integer()])

Answer 4

Thanks now I can see where I missed it