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S1Q1 · Check Even Number and Second Last Digit is Two

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Function: is_even_and_second_last_digit_is_two(num: int) -> bool

Two conditions, both must be true:

  • num is even → num % 2 == 0
  • Second last digit is 2 → (num // 10) % 10 == 2
def is_even_and_second_last_digit_is_two(num: int) -> bool:
    return num % 2 == 0 and (num // 10) % 10 == 2

Key rules: - Both conditions must hold - use and - // 10 removes the last digit, % 10 then isolates the new last digit (which was the second last) - Works correctly for any positive integer including single-digit numbers (second last digit = 0)

Problem Statement

Problem

Write a function is_even_and_second_last_digit_is_two(num: int) -> bool that returns True if both of the following are satisfied, and False otherwise:

  1. num is even
  2. The second last digit of num is 2

Examples:

Input 
232
Output 
False
Input 
120
Output 
True
Input 
25
Output 
False
Input 
13
Output 
False

Understanding the problem

There are exactly two things to check. Both must be True for the function to return True.

Condition How to check
Number is even num % 2 == 0
Second last digit is 2 (num // 10) % 10 == 2

The second condition is the interesting one. Let's break down how to extract the second last digit.

How to extract the second last digit

Think of it in two steps:

  • Step 1 - drop the last digit: integer divide by 10. 120 // 10 → 12
  • Step 2 - isolate the last digit of the result: modulo 10. 12 % 10 → 2

Combined: (num // 10) % 10

Tracing all four examples

num Even? num // 10 (num // 10) % 10 Second last = 2? Result
232 ✅ Yes (232 % 2 = 0) 23 3 ❌ No False
120 ✅ Yes (120 % 2 = 0) 12 2 ✅ Yes True
25 ❌ No (25 % 2 = 1) 2 2 ✅ Yes False
13 ❌ No (13 % 2 = 1) 1 1 ❌ No False

Notice 232 - it is even, and the digit 2 appears in it, but the second last digit is 3, not 2. Both conditions must hold simultaneously.

Notice 25 - the second last digit is 2, but the number is odd. Again, one condition failing is enough to return False.

Solution approaches

def is_even_and_second_last_digit_is_two(num: int) -> bool:
    is_even = (num % 2 == 0)
    second_last = (num // 10) % 10
    if is_even and second_last == 2:
        return True
    else:
        return False

Most readable for beginners - each check is named and explicit.

def is_even_and_second_last_digit_is_two(num: int) -> bool:
    return num % 2 == 0 and (num // 10) % 10 == 2

Since both conditions are boolean expressions, we can return their and directly. No if needed - the result of and is already True or False.

def is_even_and_second_last_digit_is_two(num: int) -> bool:
    s = str(num)
    if len(s) < 2:
        return False          # single-digit: no second last digit
    return num % 2 == 0 and s[-2] == '2'

Convert to string and index s[-2] for the second last character. Clean and readable, but needs an explicit length guard for single-digit numbers. The arithmetic approach handles this automatically ((5 // 10) % 10 = 0, which is never 2).

Why and short-circuits

Python's and operator is short-circuit evaluated:

num % 2 == 0 and (num // 10) % 10 == 2

If the first condition (num % 2 == 0) is already False, Python skips evaluating the second condition entirely and returns False immediately. This means for odd numbers, the second last digit is never computed - a small efficiency bonus.

Key takeaways

01

Extracting the nth digit

General pattern: (num // 10**n) % 10 gives the digit at position n from the right (0-indexed). Last digit: n=0. Second last: n=1.

02

Return boolean expressions directly

If a function returns True or False, you rarely need an if. Just return condition_a and condition_b.

03

and vs or

Use and when all conditions must hold. Use or when any one is enough. Here both must be true → and.