Puzzle of the Week for 3 May 1999: Solution

The probability that a randomly chosen integer A, between 10 and 1000, will be divisible by a second integer B, formed by all but the last digit of A, is 123/991 (almost 1 in 8).

There are 991 possible values for A (1000 - 10 + 1 = 991). Those that end with 0 will yield a B that exactly divides A, yielding a quotient of 10 (for example, 230 / 23 = 10). There are 100 such "good" values for A.

If the last digit of A is not zero, the result of dividing A by B will be a quotient greater than 10. For example, if A is 22, B is 2, and the quotient is 11. All two-digit multiples of 11 (11, 22, 33, 44, etc.) have this property; there are 9 such multiples.

If A contains more than two digits, it is easy to see that A/B must be less than 11 (the maximum quotient in this case is 10.9, corresponding to A = 109). Hence there can be no "good" values for A having three or more digits except for the multiples of 10 already counted. It thus remains necessary only to consider two-digit values for A.

In addition to the first 9 multiples of 11, the first 5 multiples of 12 (12, 24, 36, 48, and 60) are evenly divisible by their first digits (but 60 has already been counted, since it ends in 0). Similarly, the first 3 multiples of 13 (13, 26, and 39), and the first 2 multiples of 14 (14 and 28) are "good" values. Finally, the 5 remaining two-digit numbers beginning with 1 (15, 16, 17, 18, and 19), but not their multiples, are "good" numbers. In all, there are 100 + 9 + 4 + 3 + 2 + 5 = 123 "good" values among the 991 possible values of A, so the probability of selecting a "good" A at random is 123/991, or about 12.4%.