q12: Alternative solution

[CHAMP-123]

[CHAMP-123]

l=[] for i in range(1000,3001): t=[int(j)%2 for j in list(str(i))] tp=list(set(t)) if tp.count(1)==0: l.append(str(i)) print(','.join(l))

how about this?

[venkatasai1432]

a='' for i in range(1000,3001): if i%2==0: a=a+str(i) a=a+str(',') print(a) it works

[venkatasai1432]

a=[] for i in range(1000,3001): if i%2==0: a=a+[str(i)] print(','.join(a))

Is it ok OR can you want another one ?

[Manish-Ranjan]

You can try this

Result = [ i for in range(1000, 3001) if i%2 == 0] print(Result)

[venkatasai1432]

Hi, It returns syntax error because i is not declared there the below one return the output as expected

Result = [ i for i in range(1000, 3001) if i%2 == 0] print(Result)

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On Sun, Sep 8, 2019 at 1:36 PM Manish-Ranjan [email protected] wrote:

You can try this

Result = [ i for in range(1000, 3001) if i%2 == 0] print(Result)

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[Manish-Ranjan]

Hi, i didn't get any type of error. you don't have to declare variable i .Range function always gives you integer but if you want a list of string values then you simply typecast int into string ( str(i) for i in range(1000, 3001) if i%2 == 0]