CSS  Grid with perfect squares
By : trial
Date : March 29 2020, 07:55 AM
wish help you to fix your issue You need to combine these style rules to get what you need. The float property ensures they stack in a horizontal row, the block rule allows you to set the height and width of the element and the overflow hidden rule stops it from expanding with the content. code :
.square {
float: left;
width:200px;
height:200px;
display:block;
overflow:hidden;
}

Double Squares: counting numbers which are sums of two perfect squares
By : user2617018
Date : March 29 2020, 07:55 AM
this one helps. Factor the number n, and check if it has a prime factor p with odd valuation, such that p = 3 (mod 4). It does if and only if n is not a sum of two squares. The number of solutions has a closed form expression involving the number of divisors of n. See this, Theorem 3 for a precise statement.

Find perfect squares
By : Sheeraz Mughal
Date : March 29 2020, 07:55 AM
will help you I am trying to write a loop that calls a method to determine if a number entered is a perfect square. It compiles just fine so I must have a logic error, for the life of me though I can't find it. No matter the number I put in it seems to always return false, leading me to believe the problem lies inside the isPerfect() method. I however, do not know enough about java yet to know where to go from here. Here's the code I've got so far: , Two potential issues. code :
int num = Math.round(Math.sqrt(input));

Calculating the number of perfect squares, perfect cubes,etc in a given range?
By : LdoZ
Date : March 29 2020, 07:55 AM
wish of those help Float numbers calculations are not exact. 64**1/3 could have value like ~3.99999975, so floor gives 3. Or 4.000000016, so ceil gives 5 (I did not check real value). You must take numerical errors into account.

Sum of all the perfect squares
By : KoltonLandry
Date : March 29 2020, 07:55 AM
Does that help If I understood your question right, you are asking how to write a function to loop through all the values in a list, sum the numbers that are perfect squares, and ignore the others. Here is my code with comments explaining what is going on. code :
# We need to use the math module in this program.
import math
# Function declaration
def sumSquares(numbers):
# Start by defining the sum as 0
sum = 0
# Loop through each number in the list
for num in numbers:
# Check if number is a square by:
# 1. Taking the integer square root of that number
# 2. Squaring it
# 3. And checking if that is equal to the original number
if num == int(math.sqrt(num)) ** 2:
# If it is a perfect square, add it to the total sum.
sum += num
sumSquares([1, 4, 9, 30])

