Starting with a four-digit integer that is not a multiple of 1000 , an integer with fewer digits can be obtained by removing the leading digit and ignoring leading zeros. For example, removing the leading digit from 1023 gives the integer 23, and removing the leading digit from 2165 gives 165. How many integers from 1001 to 4999, inclusive, other than multiples of 1000, have the property that the integer obtained by removing the leading digit is a factor of the original integer?
We will count the integers with the desired property by considering four cases, one for each possible leading digit.

The leading digit is 1: In this situation, we are looking for integers of the form $1000+n$ where $n$ is an integer from 1 to 999 inclusive so that $1000+n$ is a multiple of $n$.

For $1000+n$ to be a multiple of $n$, there must be some integer $k$ so that $k n=1000+n$, which can be rearranged to get $n(k-1)=1000$.

Thus, $1000+n$ will have the desired property exactly when $n$ is a positive factor of 1000 that is less than 1000 .

The positive factors of 1000 that are less than 1000 are

$$
1,2,4,5,8,10,20,25,40,50,100,125,200,250,500
$$

There are 15 integers in the list above, so we get 15 integers in this case.

The leading digit is 2: Similar to case 1, we need to find all integers $n$ for which $1 \leq n \leq 999$ and $2000+n$ is a multiple of $n$.

The latter condition means there is an integer $k$ with the property that $2000+n=k n$ which can be rearranged to get $2000=n(k-1)$.

The positive factors of 2000 that are less than 1000 are

$$
1,2,4,5,8,10,16,20,25,40,50,80,100,125,200,250,400,500
$$

of which there are 18. Therefore, there are 18 integers in this case.

The leading digit is 3: As in cases 1 and 2 , we need to find all integers $n$ for which $1 \leq n \leq 999$ and $3000+n$ is a multiple of $n$.

Again, this means we need to count the positive factors of 3000 that are less than 1000.

If $m$ is a factor of 3000 , then it is either a multiple of 3 or it is not.

If it is not a multiple of 3 , then it is one of the 15 factors of 1000 from case 1.

Otherwise, it is 3 times one of the factors of 1000 from case 1.

Among the factors from case 1, all except 500 remain less than 1000 when multiplied by 3 .

Therefore, there are $15+14=29$ integers in this case.

The leading digit is 4: In this case, we need to count the positive factors of 4000 that are smaller than 1000 .

The prime factorization of 4000 is $2^{5} 5^{3}$.

Each factor of 4000 is either a power of 2, 5 times a power of 2, 25 times a power of 2, or 125 times a power of 2 .

Each of $1=2^{0}, 2=2^{1}, 4=2^{2}, 8=2^{3}, 16=2^{4}$, and $32=2^{5}$ is less than 1000 , so 6 of the factors are powers of 2 .

Multiplying each of the powers of 2 in the previous line by 5 gives $5,10,20,40,80$, and 160, each of which is less than 1000.

Multiplying 5 by each of the factors in the previous line gives 25, 50, 100, 200, 400, and 800, each of which is less than 1000.

Multiplying 5 by each of the factors in the previous line gives 125, 250, 500, 1000, 2000, 4000, three of which are less than 1000.

This gives a total of $6+6+6+3=21$ factors of 4000 that are less than 1000 , so we get 21 integers in this case.

Thus, the number of integers with the desired property is $15+18+29+21=83$.

ANSWER: 83

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