Given a string, turn each letter into its ASCII character code and join them together to create a number - let's call this number `total1`:

```
'ABC' --> 'A' = 65, 'B' = 66, 'C' = 67 --> 656667
```

Then replace any incidence of the number `7` with the number `1`, and call this number 'total2':
```
total1 = 656667
              ^
total2 = 656661
              ^
```

Then return the difference between the sum of the digits in `total1` and `total2`:

```
  (6 + 5 + 6 + 6 + 6 + 7)
- (6 + 5 + 6 + 6 + 6 + 1)
-------------------------
                       6
```

Difficulty: introductory

Starter code:

def calc(x):
	

Solutions:
def calc(s):
    total1 = ''.join(map(lambda c: str(ord(c)), s))
    total2 = total1.replace('7', '1')
    return sum(map(int, total1)) - sum(map(int, total2))

def calc(x):
    return ''.join(str(ord(ch)) for ch in x).count('7') * 6

def calc(x):
    return ''.join([str(ord(x[i])) for i in range(len(x)) ]).count('7') * 6

def calc(x):
    total1 = "".join(str(ord(char)) for char in x)
    total2 = total1.replace("7","1")
    return sum(int(x) for x in total1) - sum(int(x) for x in total2)

def calc(x):
    return sum(sum(6 for i in str(ord(i)) if i == '7') for i in x)

def calc(string):
    return "".join(str(ord(char)) for char in string).count("7") * 6

def calc(s):
    return sum(2 if c == 'M' else c in '%/9CFGHIJKLNOWaku' for c in s) * 6

def calc(x):
    return sum(6 for c in ''.join(map(str, map(ord, x))) if c== '7')

def calc(x):
    n = ''.join(map(lambda y: str(ord(y)), x))
    return n.count('7') * 6

def calc(x):
    n1 = "".join([str(ord(c)) for c in x])
    n2 = n1.replace('7', '1')
    return sum(map(int, n1))-sum(map(int, n2))