Conditionals_Lab_1

Question

Write a program that randomly generates an integer between 1 and 499. 



1. If the random number square root has no decimal values (eg. 64 is 8, 9 is 3). Output the random number, its square root, and a statement indicating it has no decimal values.
2. If the random number square root has decimal values (eg. 65 is 8.06225). Output the random number, its square root rounded to two decimal places and a statement indicating it has decimal values.

Answer 1

#########################################

# Here is the new one....

############################

import math

import random

x=random.randint(1,499)

print("The random number is " + str(x))

perfect_square_flag = False

done_flag = False

t=1

while (done_flag==False):

tSqr=t*t

if (tSqr==x):

perfect_square_flag=True

done_flag=True

break

if (tSqr<x):

t=t+1

else:

done_flag=True

if (perfect_square_flag):

print(str(x) + " is a perfect square with square root "+ str(int(math.sqrt(x))))

else:

print(str(x) + " is not a perfect square with square root " + str(round(math.sqrt(x),2)))

##################################################################

#------------------------- Here is the OLD one

############################################################

import math

import random

perfect_squares=[1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400,441,484]

x=random.randint(1,499)

perfect_square_flag=False

for n in perfect_squares:

if int(n)==int(x):

perfect_square_flag=True

break

if (perfect_square_flag):

print(str(x) + " is a perfect square with square root "+ str(int(math.sqrt(x))))

else:

print(str(x) + " is not a perfect square with square root " + str(round(math.sqrt(x),2)))