0505EasyMath

Stamped Cloakroom Ticket

Tracked in this browser only

Trains the technique from

LeetCode 3622Check Divisibility by Digit Sum and Product

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A cloakroom prints a positive whole number n on every ticket and stamps the lucky ones. Take the decimal digits of n as they are written, with no leading zeros. Add up those digits to get their total, multiply them together to get their product, and add the total to the product to get the ticket's key.

A ticket is stamped when n is divisible by its key. Return true if the ticket printed with n is stamped and false otherwise.

A ticket whose number contains the digit 0 has a digit product of 0, and 0 counts as divisible by any positive key, so such a ticket is stamped exactly when n is divisible by its digit total.

Examples

Example 1

Input: n = 4536
Output: true

The digits 4, 5, 3 and 6 add to 18 and multiply to 360, so the key is 378. Since 4536 is 12 times 378, the ticket is stamped.

Example 2

Input: n = 4537
Output: false

The digits 4, 5, 3 and 7 add to 19 and multiply to 420, so the key is 439, and 4537 leaves a remainder of 159 against it.

Example 3

Input: n = 1000000
Output: true

The digits add to 1 and multiply to 0, so the key is 1, and every whole number is divisible by 1.

Constraints

1 <= n <= 10^6

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def is_stamped(n: int) -> bool:

Java

public boolean isStamped(int n)

The intended approach

Time O(d) where d is the number of digits of n, so O(log n).Space O(1): two running values.

Strip the digits with repeated remainder and division by ten, building the digit total and the digit product in the same loop, then report whether `n` leaves no remainder against their sum. The key is at least one for every allowed `n`, since a digit total of zero is impossible, so the division is always defined.

Pythonentry point: is_stamped

def is_stamped(n):
    # Peel the digits off with repeated division by ten, accumulating their total
    # and their product in the same walk, then test the single divisibility.
    # A zero digit sends the product to zero, and zero is a multiple of every
    # positive integer, so such a ticket is judged on its digit total alone.
    # O(d) time in the number of digits, O(1) space.
    total = 0
    product = 1
    rest = n
    while rest > 0:
        digit = rest % 10
        total += digit
        product *= digit
        rest //= 10
    return n % (total + product) == 0

The obvious version

Read the digits off the decimal spelling of `n`, add and multiply them to form the key, then step through the multiples of the key one at a time until reaching or passing `n` and check whether `n` was landed on. O(n / key) steps.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_is_stamped(n):
    digits = [int(ch) for ch in str(n)]
    total = sum(digits)
    product = 1
    for d in digits:
        product *= d
    divisor = total + product
    step = divisor
    while step < n:
        step += divisor
    return step == n

How this problem was checked

Reference solution executed against 12 test cases — 12 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in