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

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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.

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Answer 1 · 2016-03-18T05:29:07

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

$9 \sum n = 1 f l (\frac{k \cdot gcd (n, 10)}{n})$ $sum_(n=1)^9 fl((k*gcd(n,10))/n)$

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

Explanation:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$9 \sum n = 1 f l (\frac{k}{\frac{n}{gcd (n, 10)}}) = 9 \sum n = 1 f l (\frac{k \cdot gcd (n, 10)}{n})$ $sum_(n=1)^9 fl(k/(n/gcd(n,10)))=sum_(n=1)^9 fl((k*gcd(n,10))/n)$

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

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

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