0767EasyMath

Digits That Divide The Reading

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Trains the technique from

LeetCode 2520Count the Digits That Divide a Number

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 meter shows a positive whole number reading.

Walk the decimal digits of reading from left to right. A digit carries when the reading divides by that digit with no remainder. A digit equal to 0 never carries, since nothing divides by zero.

Return how many digit positions carry. A digit that appears more than once is counted once for each position it occupies.

Examples

Example 1

Input: reading = 4620
Output: 3

4620 divides evenly by 4, by 6 and by 2, so those three positions carry. The final digit is 0 and never carries.

Example 2

Input: reading = 305
Output: 1

305 leaves a remainder of 2 when divided by 3, the middle digit is 0 and cannot carry, and 305 divides evenly by 5. One position carries.

Example 3

Input: reading = 936
Output: 3

936 divides evenly by 9, by 3 and by 6, so all three positions carry.

Constraints

1 <= reading <= 10^9

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 dividing_digits(reading: int) -> int:

Java

public int dividingDigits(int reading)

The intended approach

Time O(L) for a reading with L digits.Space O(L) for the decimal text.

Read the number's decimal text once. Skip any `0` digit outright, and for every other digit test the remainder of the whole reading against it, counting the position when the remainder is zero. Repeated digits are handled naturally because each position is tested on its own.

Pythonentry point: dividing_digits

def dividing_digits(reading):
    # Time O(L) for an L digit reading, space O(L) for its decimal text.
    total = 0
    for ch in str(reading):
        d = ord(ch) - 48
        if d == 0:
            continue
        if reading % d == 0:
            total += 1
    return total

The obvious version

Peel the digits off arithmetically with repeated division by ten, then go through the nine possible non-zero digit values, and for each one that divides the reading add how many times it turned up.

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_dividing_digits(reading):
    digits = []
    x = reading
    while x > 0:
        x, d = divmod(x, 10)
        digits.append(d)
    total = 0
    for d in range(1, 10):
        if reading % d == 0:
            total += digits.count(d)
    return total

How this problem was checked

Reference solution executed against 14 test cases — 14 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

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