Products of Digits: Challenges for Everyone...

[You may also want to look at the preview of the interview with Alec Klein, author of A Class Apart, to be hosted on MathNotations. Alec has agreed to answer my questions about Stuyvesant HS in NYC, other specialized schools and gifted education.]


One never knows where the inspiration for a math challenge might come from. This one came from a book I devoured recently, entitled The Righteous Men by Sam Bourne. If you enjoyed the Da Vinci Code and solving the number puzzles and other codes embedded in the story, you will definitely enjoy this novel. One of the most important clues to unraveling the mystery in the story was stated as a riddle:
Just men we are, our number few
Describable in digits two
We're halved if these do multiply
If we few perish then all must die.

Ok, now some of you are going to ignore the math problem and run out to find this book, but for those who would like a more prosaic version of the math problem, here is our first conundrum:
Note: Students from middle school on can attempt some of these although the proofs are a reach.

(1) Determine a 2-digit positive integer the product of whose digits is one-half the integer.
Now, it won't take you long to find such a number (once you get past the elliptical phrasing), but that's just to whet your appetite. The real challenge begins:

(2) Prove that your answer to (1) is unique, i.e., there is only one solution to the problem.

Comment: We're looking for more than an exhaustive search through all ninety 2-digit numbers or a programmed solution. The key to this and all of the remaining questions is to find an approach to solving a single equation which has 2 or more variables whose domain is the set of positive integers. Students are usually not introduced to solving such equations but they appear frequently on SATs and Math Contests. Because we are looking only for positive integer solutions, a standard algebraic approach must be supplemented with arithmetic concepts and testing of several possibilities. Number theorists refer to these as Diophantine equations.

(3) How about 3-digit integers? If 12 were permitted as the hundreds' digit, then (12)96 would satisfy the problem since (12)(9)(6) = 648, which is one-half of 1296. Unfortunately we can't allow that, so your challenge is to prove that there are NO 3-digit numbers (meaning positive integers) the product of whose digits is one-half the number.

Comments: Again, you're confronted with an equation but this time there are 3 variables. One approach is to solve for h in terms of t and u (I'll let you guess the symbolism) but there surely must be numerous possibilities! Or are there...

(4) Since we couldn't find a 3-digit number with the property, we'll rework the ratio:
Determine a 3-digit number the product of whose digits is one-fourth the number.

(5) Of course, I can't let you off the hook that easily. Prove that your solution in (4) is unique!

Standard disclaimer: These results have not been independently verified. Translation: I devised problems (2)-(5) and therefore there could be errors. If you find other solutions, let me know!

Reminder: Don't forget to give Proper Attribution when using these original problems. See the instructions in the sidebar. I appreciate your understanding and acknowledgment.