Question

Is there a systematic way to determine the number of numbers between 10 and, say, 50, divisible by their units digits?

I saw a variation on this recently... somewhere. Not allowed to say where. But I ended up writing out some numbers and finding a pattern.

For example...

• anything divisible by 11 is divisible by the units digit, so that includes 11, 22, 33, and 44.
• anything that is a multiple of 12 counts, so 12, 24, 36, and 48, but of course not 60.
• anything ending in 1 counts, so 21, 31, and 41
• anything ending in 2 counts, so 32 and 42.
• only 33 ends in 3 and is also divisible by 3, so that doesn't count.
• numbers ending in 4 count if I didn't say them, nothing really.
• anything ending in 5 counts, so 15, 25, 35, and 45.
• the only number that counts that ends in 6 is 36, but I already said it.
• nothing ending in 7 counts until 77, which is larger than 50.
• nothing ending in 8 counts since I already said 48.
• nothing ending in 9 counts until 99, which is larger than 50.
• and nothing ending in 0 counts.

And I think that's all of them, so 17 numbers. Which is so abnormal, really.

Answer 1

The number of numbers between `10` and `10k` divisible by their units digit can be represented as

`sum_(n=1)^9 fl((k*gcd(n,10))/n)`

where `fl(x)` represents the floor function, mapping `x` to the greatest integer less than or equal to `x`.

Explanation:

This is equivalent to asking how many integers `a` and `b` exist where `1<=b<5` and `1<=a<=9` and `a` divides `10b+a`

Note that `a` divides `10b + a` if and only if `a` divides `10b`. Thus, it suffices to find how many such `b`s exist for each `a`. Also, note that `a` divides `10b` if and only if each prime factor of `a` also is a prime factor of `10b` with appropriate multiplicity.

All that remains, then, is to go through each `a`.

`a = 1`: As all integers are divisible by `1`, all four values for `b` work.

`a=2`: As `10` is divisible by `2`, all four values for `b` work.

`a=3`: As `10` is not divisible by `3`, we must have `b` being divisible by `3`, that is, `b=3`.

`a=4`: As `10` is divisible by `2`, we must have `b` as divisible by `2` to have the appropriate multiplicity. Thus, `b=2` or `b=4`.

`a=5`: As `10` is divisible by `5`, all four values for `b` work.

`a=6`: As `10` is divisible by `2`, we must have `b` as divisible by `3`, that is, `b=3`.

`a=7`: As `10` is not divisible by `7`, we must have `b` as divisible by `7`. But `b<5`, and so no value for `b` works.

`a=8`: As `10` is divisible by `2`, we must have `b` as divisible by `4`, that is, `b=4`

`a=9:` As `10` is not divisible by `3`, we must have `b` as divisible by `3^2`. But `b<5`, and so no value for `b` works.

This concludes each case, and so, adding them up, we get, as concluded in the question, `17` values. This method can easily be extended to greater values, however. For example, if we wanted to go from `10` to `1000`, we would restrict `1<=b<100`. Then, looking at `a=6`, say, we would have `2` divides `10` and thus `6` divides `10b` if and only if `3` divides `b`. There are `33` multiples of `3` in the range for `b`, and thus `33` numbers which end in `6` and are divisible by `6` between `10` and `1000`.

In a shorter, easier to calculate notation, using the observations above, we can write the number of integers between `10` and `10k` as

`sum_(n=1)^9 fl(k/(n/gcd(n,10)))=sum_(n=1)^9 fl((k*gcd(n,10))/n)`

where `fl(x)` represents the floor function, mapping `x` to the greatest integer less than or equal to `x`.