S1Q1 · Check If Multiple of 5 but Not 3

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⚡ Quick Reference

Function: is_multiple_of_5_not_3(num: int) -> bool

Core idea: check both divisibility conditions in one expression.

def is_multiple_of_5_not_3(num):
    return num % 5 == 0 and num % 3 != 0

Key rules: - Multiple of 5: num % 5 == 0 - NOT a multiple of 3: num % 3 != 0 - Both must hold - use and - Works for negative numbers too (Python % is always non-negative for positive divisors)

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Problem Statement

Problem

Write a function is_multiple_of_5_not_3(num) that returns True if num is divisible by 5 but not by 3.

Examples:

Input

10

Output

True

Input

15

Output

False

Input

-25

Output

True

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Tracing all examples

num	num % 5	Multiple of 5?	num % 3	Multiple of 3?	Result
10	0 ✅	Yes	1	No	True
25	0 ✅	Yes	1	No	True
-50	0 ✅	Yes	1	No	True
15	0 ✅	Yes	0	Yes ❌	False
9	4 ❌	No	-	-	False
12	2 ❌	No	-	-	False
-25	0 ✅	Yes	2	No	True

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Solution approaches

Pythonic (direct return)ExplanatoryUsing lambda

def is_multiple_of_5_not_3(num):
    return num % 5 == 0 and num % 3 != 0

def is_multiple_of_5_not_3(num):
    is_mult_of_5 = num % 5 == 0
    is_mult_of_3 = num % 3 == 0
    return is_mult_of_5 and not is_mult_of_3

is_multiple_of_5_not_3 = lambda num: num % 5 == 0 and num % 3 != 0

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Key takeaways

01

Two conditions combined with and

num % 5 == 0 and num % 3 != 0 - both must be true. Short-circuit and skips the second check if the first already fails (not a multiple of 5).

02

Works for negatives - Python % is non-negative

In Python, -25 % 5 == 0 and -25 % 3 == 2. The modulo operator always returns a non-negative result when the divisor is positive, so no special handling is needed for negative inputs.

03

15 is the classic trap - multiple of both

15 is divisible by both 5 and 3 (it's a multiple of 15). The num % 3 != 0 condition correctly excludes it. Always check both conditions explicitly.