(he's thinking about it)back to homepage
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
12
60
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.
| 92 | 93 | 94 | 95 | 96 | 98 | 99 | |||
| 81 | 82 | 84 | 85 | 86 | 87 | 88 | |||
| 72 | 74 | 75 | 76 | 77 | 78 | ||||
| 62 | 63 | 64 | 65 | 66 | 68 | 69 | |||
| 51 | 52 | 54 | 55 | 56 | 57 | 58 | |||
| 42 | 44 | 45 | 46 | 48 | 49 | ||||
| 32 | 33 | 34 | 35 | 36 | 38 | 39 | |||
| 21 | 22 | 24 | 25 | 26 | 27 | 28 | |||
| 11 | 12 | 14 | 15 | 16 | 18 |
We can also rule out numbers that have an even tens digit but an odd ones digit.
| 92 | 93 | 94 | 95 | 96 | 98 | 99 | |||
| 82 | 84 | 86 | 88 | ||||||
| 72 | 74 | 75 | 76 | 77 | 78 | ||||
| 62 | 64 | 66 | 68 | ||||||
| 51 | 52 | 54 | 55 | 56 | 57 | 58 | |||
| 42 | 44 | 46 | 48 | ||||||
| 32 | 33 | 34 | 35 | 36 | 38 | 39 | |||
| 22 | 24 | 26 | 28 | ||||||
| 11 | 12 | 14 | 15 | 16 | 18 |
Similarily, we can rule out numbers with 3, 6, or 9 as a digit but that are not threeven.
| 93 | 96 | 99 | |||||||
| 82 | 84 | 88 | |||||||
| 72 | 74 | 75 | 77 | 78 | |||||
| 66 | |||||||||
| 51 | 52 | 54 | 55 | 57 | 58 | ||||
| 42 | 44 | 48 | |||||||
| 33 | 36 | 39 | |||||||
| 22 | 24 | 28 | |||||||
| 11 | 12 | 14 | 15 | 18 |
Same with numbers containing 4 or 8 but that are quadrodd.
| 93 | 96 | 99 | |||||||
| 84 | 88 | ||||||||
| 72 | 75 | 77 | |||||||
| 66 | |||||||||
| 51 | 52 | 55 | 57 | ||||||
| 44 | 48 | ||||||||
| 33 | 36 | 39 | |||||||
| 22 | 24 | 28 | |||||||
| 11 | 12 | 15 |
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.
| 93 | 96 | ||||||||
| 84 | |||||||||
| 72 | 75 | ||||||||
| 51 | 52 | 57 | |||||||
| 48 | |||||||||
| 36 | 39 | ||||||||
| 24 | 28 | ||||||||
| 11 | 12 | 15 |
And after checking the rest manually...
| 36 | |||||||||
| 24 | |||||||||
| 11 | 12 | 15 |
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 digitsmeans 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