In [13]: def isAbundant (x) ""Returns whether or not the given number x is abundant. A number is considered to be abundant if the sun of its factors (aside from the number) is greater than the number itself. Example: 12 is abundant since 1+2+3+4+6-16 > 12. However, a number like 15, where the sun of the factors. in 1+3+5 is not abundant. your code here for i in range(2x) if(x1==0) print(x, 'is Abundant Number") elser print(x, 'is not Abundant Number") File "cipython-input-13-42000150fbb5", line 19 else: Syntaxtrror: invalid syntax In [14] abundant numbers [12, 18, 20, 24, 30, 36, 40, 42, 48, 54, 56, 60, 66, 70, 72, 78, 80, 84, 88, 90, 96, 100, 102, 104, 108, 112, 114, 1201 for i in abundant ber assert true (isAbundant (i), str(i) is abundant") not abundant numbers [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 21, 22, 23, 25, 26, 27, 28, 29, 91, 92, 93, 94, 95, 1191 for i is not abundant unbere assert true (not (iaAbundant (i)), str(i) is not abundant") test existence of doestring assert true(len(isAbundant.doce) > 1, "there is no docstring for isbundant") print("Success!")
In [13]: def isAbundant (x) ""Returns whether or not the given number x is abundant. A number is considered to be abundant if the sun of its factors (aside from the number) is greater than the number itself. Example: 12 is abundant since 1+2+3+4+6-16 > 12. However, a number like 15, where the sun of the factors. in 1+3+5 is not abundant. your code here for i in range(2x) if(x1==0) print(x, 'is Abundant Number") elser print(x, 'is not Abundant Number") File "cipython-input-13-42000150fbb5", line 19 else: Syntaxtrror: invalid syntax In [14] abundant numbers [12, 18, 20, 24, 30, 36, 40, 42, 48, 54, 56, 60, 66, 70, 72, 78, 80, 84, 88, 90, 96, 100, 102, 104, 108, 112, 114, 1201 for i in abundant ber assert true (isAbundant (i), str(i) is abundant") not abundant numbers [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 21, 22, 23, 25, 26, 27, 28, 29, 91, 92, 93, 94, 95, 1191 for i is not abundant unbere assert true (not (iaAbundant (i)), str(i) is not abundant") test existence of doestring assert true(len(isAbundant.doce) > 1, "there is no docstring for isbundant") print("Success!")
Computer Networking: A Top-Down Approach (7th Edition)
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Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
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Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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![In [13]: def isAbundant (x):
***Returns whether or not the given number x is abundant.
A number is considered to be abundant if the sum of its factors
(aside from the number) is greater than the number itself.
Example: 12 is abundant since 1+2+3+4+6-16 > 12.
However, a number like 15, where the sum of the factors.
is 1 + 3 + 5 - 9 is not abundant.
your code here
sum-1
for i in range(2,x)
if(x1==0):
Bum-Bum+1
if(sum>) 1
print(x, 'is Abundant Number")
else:
print (x, 'is not Abundant Number')
File "<ipython-input-13-d2000150fbb5>", line 19
else:
SyntaxError: invalid syntax
In [14]: abundant_numbers [12, 18, 20, 24, 30, 36, 40, 42, 48,
54, 56, 60, 66, 70, 72, 78, 80, 84,
88, 90, 96, 100, 102, 104, 108, 112,
114, 1201
for i in abundant numbers
assert true (isAbundant (i), str(i) is abundant")
not abundant_numbers [1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11, 13,
14, 15, 16, 17, 19,
21, 22, 23, 25, 26, 27, 28, 29,
91, 92, 93, 94, 95, 1191
for i in not abundant_numbers:
assert true (not (isAbundant (i)), str(i) is not abundant")
#test existence of docstring
assert true (len (isAbundant._doc_) > 1, "there is no docstring for isAbundant")
print("Success!")](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2d7dabf0-c01d-46c4-af62-b15d43439327%2F3fe4c8ed-7cff-4fe5-9a5b-c5e0fd6027e3%2Flazlmbq_processed.png&w=3840&q=75)
Transcribed Image Text:In [13]: def isAbundant (x):
***Returns whether or not the given number x is abundant.
A number is considered to be abundant if the sum of its factors
(aside from the number) is greater than the number itself.
Example: 12 is abundant since 1+2+3+4+6-16 > 12.
However, a number like 15, where the sum of the factors.
is 1 + 3 + 5 - 9 is not abundant.
your code here
sum-1
for i in range(2,x)
if(x1==0):
Bum-Bum+1
if(sum>) 1
print(x, 'is Abundant Number")
else:
print (x, 'is not Abundant Number')
File "<ipython-input-13-d2000150fbb5>", line 19
else:
SyntaxError: invalid syntax
In [14]: abundant_numbers [12, 18, 20, 24, 30, 36, 40, 42, 48,
54, 56, 60, 66, 70, 72, 78, 80, 84,
88, 90, 96, 100, 102, 104, 108, 112,
114, 1201
for i in abundant numbers
assert true (isAbundant (i), str(i) is abundant")
not abundant_numbers [1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11, 13,
14, 15, 16, 17, 19,
21, 22, 23, 25, 26, 27, 28, 29,
91, 92, 93, 94, 95, 1191
for i in not abundant_numbers:
assert true (not (isAbundant (i)), str(i) is not abundant")
#test existence of docstring
assert true (len (isAbundant._doc_) > 1, "there is no docstring for isAbundant")
print("Success!")
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