Python says it has a syntax error in this line " elif (i%7==0) or str.count(str(i),'7')>0: " and I cannot figure it out. I'm new to Python so it must be something simple.
k=int(input("enter the value for k:")) n=int(input("enter the value for n:")) if k>=1 and k<=9: for i in range(1,n+1): if (i%7==0) and str.count(str(i),'7')>0: print("boom-boom!") elif (i%7==0) or str.count(str(i),'7')>0: print("boom") else: print(i) The issue is with your identation:
Make sure the "elif" is inline with your "if" and also your "else" statement. Python is sensitive to indentations and spaces.
if (i%7==0) and str.count(str(i),'7')>0: print("boom-boom!") elif (i%7==0) or str.count(str(i),'7')>0: print("boom") else: print(i) Here is an improved solution:
k = int(input("enter the value for k:")) n = int(input("enter the value for n:")) if 1 <= k <= 9: for i in range(1, n + 1): text = str(i) if i % 7 == 0 and text.count('7'): print("boom-boom!") elif i % 7 == 0 or text.count('7'): print("boom") else: print(i) 
elifandelseshould be indented at the same level asif