#math trivia for #August24: #236 has the form abc where a*b = c. How many three-digit numbers are like this?

— Burt Kaliski Jr. (@modulomathy) August 24, 2013

The answer is 32.

There are two main ways to figure it out.

Assume that a is between 1 and 9 in the following, i.e., leading 0s are not allowed. In other words, as in usual usage, a three-digit number is a number between 100 and 999.

One way is to count the number of pairs (*a*, *b*) where *a* is between 1 and 9, *b* is between 0 and 9, and *c* = *a***b* is between 0 and 9. Thus, we need to know, for each value of a, how many of its multiples (including 0 and itself) are less than 10. The number of such multiples is as follows:

*a*= 1: 10 multiples (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)*a*= 2: 5 multiples (0, 2, 4, 6, 8)*a*= 3: 4 multiples (0, 3, 6, 9)*a*= 4: 3 multiples (0, 4, 8)*a*= 5, 6, 7, 8, 9: 2 multiple each (0 and*a*)

The total number of multiples is 10 + 5 + 4 + 3 + 5*2 = 32.

Another way is to count the number of ways to factor a digit *c* between 0 and 9 into two digits *a* and *b* where *a* is between 1 and 9, and *b* is between 0 and 9. This produces the following values:

*c*= 0: 9 factorizations (1*0, 2*0, 3*0, 4*0, 5*0, 6*0, 7*0, 8*0, 9*0)*c*= 1: 1 factorization (1*1)*c*= 2, 3, 5, 7 (primes): 2 factorizations each (1**c*and*c**1)*c*= 4, 9 (prime squares): 3 factorizations each (1**c*, sqrt(*c*)*sqrt(*c*),*c**1)*c*= 8 (prime cube): 4 factorizations (1*8, 2*4, 4*2, 8*1)*c*= 6 (bicomposite): 4 factorizations (1*6, 2*3, 3*2, 6*1)

The total is again 32 = 9 + 1 + 4*2 + 2*3 + 4 + 4.

I’ve found that it’s often helpful to try to solve a problem in two (or more) ways, in case one of the ways doesn’t work out, or in case it seems to work out, but actually produces an error. Having another way available provides a good check and balance.

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