# three distinct prime factors $x$, $y$ and $10x+y$, where $x$ and $y$ are each less than 10. What is the largest possible value of $m$?

The number $m$ is a three-digit positive integer and is the product of the three distinct prime factors $x$, $y$ and $10x+y$, where $x$ and $y$ are each less than 10. What is the largest possible value of $m$?

Any hints are greatly appreciated.

• There are only $\binom 42=6$ possible pairs of distinct, single digit primes. Just try each.
– lulu
Commented Oct 28, 2017 at 20:53
• Keep in mind you also need to try these pairs in reverse. Really there are $4 * 3 = 12$ possible choices for $x$ and $y$. Still, very easily brute-forceable. Commented Oct 28, 2017 at 20:55
• So well $10x + y$ can not be y = 2, 5 so $y = 3,7$ and $x = 2,5, 3,7$ with $3\not \mid x+y$. That means $10x + y$ may be $23, 37, 53,73$. Obvious $7*3*73 > 53*3*5 > 23*2*3$ and $7*3*73 > 7*3*37 > 2*3*23$. It's not obvious whether $53*3*5$ is more or less than $7*3*37$ but that's not important. The answer is easily seen to by $7*3*73$. Commented Oct 28, 2017 at 21:16

It isn't too hard here to simply brute force, with some optimization. Note $y \neq 2, 5$, as that would lead to $10x + y$ being nonprime. Now, let's consider the pairs of $(x, y)$ that do work. These pairs are: (2, 3); (3, 7); (5, 3); (7, 3). I assume you can proceed from here.

$\{x,y\} \subset \{2,3,5,7\}$

$10x + 2$ and $10x + 5$ are not prime. $5 + 7 = 12 = 3*4;2+7 = 9=3*3$ so $57,75,27,72$ are not prime. So possible contenders are.

$(x,y, 10x+y): = (7,3,73),(3,7,37),(5,3,53), (2,3,23)$.

$7*3*73 > 21*70 > 1400$ and not three digits.

Clearly $2< 3, 3 < 5 < 7$ and $23< 37< 53$ so $(2,3,23)$ is out.

$7*37$ and $5*53$ and similar in size so it will be either $3*7*37$ or $5*3*53$ depending upon whether $7*37 = 210 + 49$ is more or less than $5*53 = 250 +15$.