Digital Multiplication

We got this question in the mail from a friend two days ago:

What is the largest positive 2 digit number that is divisible by the multiplication of its digits?

A number theory question! Good fun. Usually with this kind of thing I start with messing around.

89 8 × 9

121 × 2

60 6 × 0

Okay so, I notice a few things: prime numbers (except 11 !) will never be divisible by their digital product. Same with numbers with 0 in them.

Next thing I usually do is to write out all the possibilities (or categories) when manageable. There's only 90, in this case.

90919293949596979899
80818283848586878889
70717273747576777879
60616263646566676869
50515253545556575859
40414243444546474849
30313233343536373839
20212223242526272829
10111213141516171819

We can also rule out numbers that have an even tens digit but an odd ones digit.

90919293949596979899
80818283848586878889
70717273747576777879
60616263646566676869
50515253545556575859
40414243444546474849
30313233343536373839
20212223242526272829
10111213141516171819

Similarily, we can rule out numbers with 3, 6, or 9 as a digit but that are not threeven.

90919293949596979899
80818283848586878889
70717273747576777879
60616263646566676869
50515253545556575859
40414243444546474849
30313233343536373839
20212223242526272829
10111213141516171819

Same with numbers containing 4 or 8 but that are quadrodd.

90919293949596979899
80818283848586878889
70717273747576777879
60616263646566676869
50515253545556575859
40414243444546474849
30313233343536373839
20212223242526272829
10111213141516171819

We could keep going, finding optimisations, but at some point we should just check them.


Okay. Just one more. We can get rid of that 22, 33, 44... diagonal by realising theyre all of the form k×11, and thus not divisible by k2.

90919293949596979899
80818283848586878889
70717273747576777879
60616263646566676869
50515253545556575859
40414243444546474849
30313233343536373839
20212223242526272829
10111213141516171819

And after checking the rest manually...

90919293949596979899
80818283848586878889
70717273747576777879
60616263646566676869
50515253545556575859
40414243444546474849
30313233343536373839
20212223242526272829
10111213141516171819

We find that the only valid ones remaining are 11, 12, 15, 24, and 36, which is the highest. We did it!

Prrr, nice work... I found something cool, want to see?

Sure, what is it?

Define y,x as the 10s and 1s digit of the number. 2 digit number that is divisible by the multiplication of its digits means 10y+x = nyx.
Graph that with varying values of n to get:
The intersections with the integer grid are what we are looking for!

Oh I see, so you've put the 10s digit along the y axis and the 1s digit along the x axis, like I did earlier. And then you've graphed that curve, 10y+x = nyx, across it? That is pretty cool.

Miauw