Print the numbers from 1-100 skipping the numbers divisible by 3 & 5
I want to print numbers from 1-100 skipping the numbers divisible by 3 & 5 and when I use the code-1 I'm not getting the correct output, I am getting full counting 1-100
#CODE1
i=1
a=1
while i<=100:
if (a%3==0 and a%5==0) :
a=a+1
else:
print(a)
a=a+1
i=i+1
but when I use the CODE-2 I am getting the desired result
#CODE2
i=1
a=1
while i<=100:
if ((a%3 and a%5)==0) :
a=a+1
else:
print(a)
a=a+1
i=i+1
notice the fourth line of the code, why is wrong with the 1st code?
python python-3.x algorithm if-statement while-loop
edited Nov 13 '18 at 17:35
jpp
102k2165116
asked Nov 13 '18 at 17:22
aman goyalaman goyal
324
add a comment |
I want to print numbers from 1-100 skipping the numbers divisible by 3 & 5 and when I use the code-1 I'm not getting the correct output, I am getting full counting 1-100
#CODE1
i=1
a=1
while i<=100:
if (a%3==0 and a%5==0) :
a=a+1
else:
print(a)
a=a+1
i=i+1
but when I use the CODE-2 I am getting the desired result
#CODE2
i=1
a=1
while i<=100:
if ((a%3 and a%5)==0) :
a=a+1
else:
print(a)
a=a+1
i=i+1
notice the fourth line of the code, why is wrong with the 1st code?
python python-3.x algorithm if-statement while-loop
edited Nov 13 '18 at 17:35
jpp
102k2165116
asked Nov 13 '18 at 17:22
aman goyalaman goyal
324
add a comment |
I want to print numbers from 1-100 skipping the numbers divisible by 3 & 5 and when I use the code-1 I'm not getting the correct output, I am getting full counting 1-100
#CODE1
i=1
a=1
while i<=100:
if (a%3==0 and a%5==0) :
a=a+1
else:
print(a)
a=a+1
i=i+1
but when I use the CODE-2 I am getting the desired result
#CODE2
i=1
a=1
while i<=100:
if ((a%3 and a%5)==0) :
a=a+1
else:
print(a)
a=a+1
i=i+1
notice the fourth line of the code, why is wrong with the 1st code?
python python-3.x algorithm if-statement while-loop
edited Nov 13 '18 at 17:35
jpp
102k2165116
asked Nov 13 '18 at 17:22
aman goyalaman goyal
324
I want to print numbers from 1-100 skipping the numbers divisible by 3 & 5 and when I use the code-1 I'm not getting the correct output, I am getting full counting 1-100
#CODE1
i=1
a=1
while i<=100:
if (a%3==0 and a%5==0) :
a=a+1
else:
print(a)
a=a+1
i=i+1
but when I use the CODE-2 I am getting the desired result
#CODE2
i=1
a=1
while i<=100:
if ((a%3 and a%5)==0) :
a=a+1
else:
print(a)
a=a+1
i=i+1
notice the fourth line of the code, why is wrong with the 1st code?
python python-3.x algorithm if-statement while-loop
python python-3.x algorithm if-statement while-loop
edited Nov 13 '18 at 17:35
jpp
102k2165116
asked Nov 13 '18 at 17:22
aman goyalaman goyal
324
edited Nov 13 '18 at 17:35
jpp
102k2165116
asked Nov 13 '18 at 17:22
aman goyalaman goyal
324
edited Nov 13 '18 at 17:35
jpp
102k2165116
edited Nov 13 '18 at 17:35
jpp
102k2165116
edited Nov 13 '18 at 17:35
jpp
102k2165116
102k2165116
asked Nov 13 '18 at 17:22
aman goyalaman goyal
324
asked Nov 13 '18 at 17:22
aman goyalaman goyal
324
asked Nov 13 '18 at 17:22
aman goyalaman goyal
324
324
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
The first program is incorrect because you on line 6-7 you increase the counter without checking to see if you need to print the number.
A cleaner way to write this could would be:
for counter in xrange(1, 101):
if not ((counter % 5 == 0) or (counter % 3 == 0)):
print (a)
answered Nov 13 '18 at 17:42
user3614104user3614104
745
If the first program "increase[s] the counter without checking to see if you need to print the number", the second program also does (the only difference between the two is how they check), yet that works.
– Dukeling
Nov 13 '18 at 18:26
add a comment |
Consider this:
a = 10
(a%3 == 0) and (a%5 == 0) # False
(a%3 and a%5) == 0 # True
The first attempt gives False incorrectly because it needs both conditions to be satisfied; you need or instead. If you look carefully, some numbers (e.g. 15) are excluded, coinciding with numbers which have both 3 and 5 as factors.
The second attempt is correct because if a is not divisible by either 3 or 5, the expression evaluates to False, and 0 == False gives True. More idiomatic would be to write:
not (a%3 and a%5)
answered Nov 13 '18 at 17:34
jppjpp
102k2165116
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The first program is incorrect because you on line 6-7 you increase the counter without checking to see if you need to print the number.
A cleaner way to write this could would be:
for counter in xrange(1, 101):
if not ((counter % 5 == 0) or (counter % 3 == 0)):
print (a)
answered Nov 13 '18 at 17:42
user3614104user3614104
745
If the first program "increase[s] the counter without checking to see if you need to print the number", the second program also does (the only difference between the two is how they check), yet that works.
– Dukeling
Nov 13 '18 at 18:26
add a comment |
The first program is incorrect because you on line 6-7 you increase the counter without checking to see if you need to print the number.
A cleaner way to write this could would be:
for counter in xrange(1, 101):
if not ((counter % 5 == 0) or (counter % 3 == 0)):
print (a)
answered Nov 13 '18 at 17:42
user3614104user3614104
745
If the first program "increase[s] the counter without checking to see if you need to print the number", the second program also does (the only difference between the two is how they check), yet that works.
– Dukeling
Nov 13 '18 at 18:26
add a comment |
The first program is incorrect because you on line 6-7 you increase the counter without checking to see if you need to print the number.
A cleaner way to write this could would be:
for counter in xrange(1, 101):
if not ((counter % 5 == 0) or (counter % 3 == 0)):
print (a)
answered Nov 13 '18 at 17:42
user3614104user3614104
745
The first program is incorrect because you on line 6-7 you increase the counter without checking to see if you need to print the number.
A cleaner way to write this could would be:
for counter in xrange(1, 101):
if not ((counter % 5 == 0) or (counter % 3 == 0)):
print (a)
answered Nov 13 '18 at 17:42
user3614104user3614104
745
answered Nov 13 '18 at 17:42
user3614104user3614104
745
answered Nov 13 '18 at 17:42
user3614104user3614104
745
answered Nov 13 '18 at 17:42
user3614104user3614104
745
745
If the first program "increase[s] the counter without checking to see if you need to print the number", the second program also does (the only difference between the two is how they check), yet that works.
– Dukeling
Nov 13 '18 at 18:26
add a comment |
If the first program "increase[s] the counter without checking to see if you need to print the number", the second program also does (the only difference between the two is how they check), yet that works.
– Dukeling
Nov 13 '18 at 18:26
If the first program "increase[s] the counter without checking to see if you need to print the number", the second program also does (the only difference between the two is how they check), yet that works.
– Dukeling
Nov 13 '18 at 18:26
If the first program "increase[s] the counter without checking to see if you need to print the number", the second program also does (the only difference between the two is how they check), yet that works.
– Dukeling
Nov 13 '18 at 18:26
add a comment |
Consider this:
a = 10
(a%3 == 0) and (a%5 == 0) # False
(a%3 and a%5) == 0 # True
The first attempt gives False incorrectly because it needs both conditions to be satisfied; you need or instead. If you look carefully, some numbers (e.g. 15) are excluded, coinciding with numbers which have both 3 and 5 as factors.
The second attempt is correct because if a is not divisible by either 3 or 5, the expression evaluates to False, and 0 == False gives True. More idiomatic would be to write:
not (a%3 and a%5)
answered Nov 13 '18 at 17:34
jppjpp
102k2165116
add a comment |
Consider this:
a = 10
(a%3 == 0) and (a%5 == 0) # False
(a%3 and a%5) == 0 # True
The first attempt gives False incorrectly because it needs both conditions to be satisfied; you need or instead. If you look carefully, some numbers (e.g. 15) are excluded, coinciding with numbers which have both 3 and 5 as factors.
The second attempt is correct because if a is not divisible by either 3 or 5, the expression evaluates to False, and 0 == False gives True. More idiomatic would be to write:
not (a%3 and a%5)
answered Nov 13 '18 at 17:34
jppjpp
102k2165116
add a comment |
Consider this:
a = 10
(a%3 == 0) and (a%5 == 0) # False
(a%3 and a%5) == 0 # True
The first attempt gives False incorrectly because it needs both conditions to be satisfied; you need or instead. If you look carefully, some numbers (e.g. 15) are excluded, coinciding with numbers which have both 3 and 5 as factors.
The second attempt is correct because if a is not divisible by either 3 or 5, the expression evaluates to False, and 0 == False gives True. More idiomatic would be to write:
not (a%3 and a%5)
answered Nov 13 '18 at 17:34
jppjpp
102k2165116
Consider this:
a = 10
(a%3 == 0) and (a%5 == 0) # False
(a%3 and a%5) == 0 # True
The first attempt gives False incorrectly because it needs both conditions to be satisfied; you need or instead. If you look carefully, some numbers (e.g. 15) are excluded, coinciding with numbers which have both 3 and 5 as factors.
The second attempt is correct because if a is not divisible by either 3 or 5, the expression evaluates to False, and 0 == False gives True. More idiomatic would be to write:
not (a%3 and a%5)
answered Nov 13 '18 at 17:34
jppjpp
102k2165116
answered Nov 13 '18 at 17:34
jppjpp
102k2165116
answered Nov 13 '18 at 17:34
jppjpp
102k2165116
answered Nov 13 '18 at 17:34
jppjpp
102k2165116
102k2165116
add a comment |
add a comment |
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