Confounding Factors

September 24, 2010 at 6:07 pm 3 comments

My colleague Julia is preparing a talk about factoring for an elementary audience, and she created the following problem to use as a warm-up:

Take a two‑digit number ab, and find the least common multiple of a, b, and ab. For example, if you take the number 35, then LCM(3, 5, 35) = 105. For which two‑digit number ab is LCM(abab) the greatest? (The notation ab is used to indicate the two‑digit number with tens digit a and units digit b, which is equal to 10a + b. This notation is used to distinguish the two‑digit number ab from the product ab.)

Here are some math jokes about factors:

What do you call an amount that exactly divides a recipe for a sweet confection?
A fudge factor.

What do algebra equations and British television have in common?
An X Factor.

Sadly, both of those are my original jokes. Sorry. To cleanse your palate, check out one of Randall Munroe’s original jokes about factoring:

Entry filed under: Uncategorized. Tags: , , , , .

NCTM Regional Conference in Denver Rhyme Time

3 Comments Add your own

  • 1. xander  |  September 25, 2010 at 12:29 am

    The answer to the warm up is 89.

    For simplicity, let c = MAX(10a+b,10b+a).

    In general, LCM(a,b,c) = a*b*c/GCD(a,b,c). Intuitively, we want to maximize the numerator while minimizing the denominator.

    The smallest possible denominator occurs when a, b, and c are relatively prime, so it would be great to find three numbers that are relatively prime.

    To maximize the numerator, we can start by examining the maximum possible values for a and b. So take a=9 and b=8. 9 and 8 are relatively prime, so we are in business thus far. 98 and 8 share common factor (i.e. 2), but 89 is relatively prime to both 9 and 8, so 89 is a good candidate. LCM(9,8,89) = 8*9*89 = 6408.

    a=9 and b=7 is another possible candidate. 97 is relatively prime to 9 and 7, thus LCM(9,7,97) = 9*7*97 = 6111. As 6111<6408, we can dismiss 97 as a candidate number.

    All other candidates are inferior to those listed above. At best, we would have LCM(a,b,c) = a*b*c, where a ≤9, b<7 and c<97. Hence a*b*c<6111<6408.

    Therefore 89 is the two digit number which maximizes LCM(a,b,c).

    I don’t find this solution to be entirely satisfactory, as it ultimately relies upon some brute forcing or guess-and-checking (i.e. several candidate numbers need to be selected and tested), but I can’t come up with anything better/more generalizable.

    As an added note, brute forcing in C gave the same result.


  • 2. sfgss  |  November 20, 2017 at 2:47 pm



Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

Trackback this post  |  Subscribe to the comments via RSS Feed

About MJ4MF

The Math Jokes 4 Mathy Folks blog is an online extension to the book Math Jokes 4 Mathy Folks. The blog contains jokes submitted by readers, new jokes discovered by the author, details about speaking appearances and workshops, and other random bits of information that might be interesting to the strange folks who like math jokes.

MJ4MF (offline version)

Math Jokes 4 Mathy Folks is available from Amazon, Borders, Barnes & Noble, NCTM, Robert D. Reed Publishers, and other purveyors of exceptional literature.

Past Posts

September 2010

Enter your email address to subscribe to the MJ4MF blog and receive new posts via email.

Join 474 other followers

Visitor Locations

free counters

%d bloggers like this: