# a digit Divisible Number

A Digit Divisible Number is a number that does not contain $$0$$ as a digit, and every consecutive sequence of its digits from right divides the number. For example, the number $$55$$ is a digit divisible number because $$5\mid 55$$ and $$55\mid55$$, and $$55$$ does not contain $$0$$ anywhere.

The number $$n=123$$ is not a digit divisible number because $$3\mid123$$ and $$123\mid123$$ but $$23\nmid123$$

Here is a list from java code: for all integers less than $$2\;147\;483\;647$$, the DDNs are

$$1,2,3,4,5,6,7,8,9,11,12,15,21,22,24,25,31,32,33,35,36,41,42,44,45,48,51,52,55,61,62,63,64,65,66,71,72,75,77,81,82,84,85,88,91,92,93,95,96,99,125,225,312,315,325,375,425,525,612,615,624,625,675,725 ,728,735,816,825,832,912,915,925,936,945,975,1125,2125,3125,3375,4125,5125,5625,6125,6375,7125 ,7875,8125,9125,9225,9375,53125,91125,91875,95625,721875$$

which indicates that this is complete list and no more integer is a digit divisible number.

I want to prove this or find a counterexample, I'm looking for a hint on how to solve it?

• This is sequence A178158, except excluding numbers that contain 0.
– Dan
Commented Jun 13, 2022 at 22:32
• Potentially relevant observation: Every DDN with 4 or more digits ends in 5.
– Dan
Commented Jun 13, 2022 at 22:36
• Note in the list for the OEIS sequence linked by @Dan, there are no entries with no zeroes above $721825$. Commented Jun 14, 2022 at 0:27

Let's take this one digit at a time.

## If $$n$$ is divisible by its own last digit

$$n = 10a + b$$

• If $$b \in \lbrace1, 2, 5 \rbrace$$ (a factor of 10), then the condition is satisfied regardless of what $$a$$ is.
• Otherwise, we must have $$b | 10a$$.

## ...and is also divisible by its last two digits

$$n = 100a + 10b + c$$

• If $$10b + c = 25$$, then the condition is satisfied regardless of what $$a$$ is. In theory, the last two digits could be any factor of 100 (01, 02, 04, 05, 10, 20, 25, or 50), but 25 is the only one without any 0's in it.
• Otherwise, we must have both $$c|10(10a+b)$$ and $$(10b+c)|100a$$.

Of the $$9 \times 9 = 81$$ possible combinations of $$bc$$ (concatenation, not multiplication), there are 10 that can not possibly meet both divisibility requirements: 14, 18, 34, 38, 54, 58, 74, 78, 94, and 98. Or in regex terms, [13579][48]. For these, the last digit requires $$4|n$$, but then the ten's digit has to be even, because 20 is divisible by 4, but 10 isn't.

This leaves 71 valid combinations for the last two digits:

11, 12, 13, 15, 16, 17, 19, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 32, 33, 35, 36, 37, 39, 41, 42, 43, 44, 45, 46, 47, 48, 49, 51, 52, 53, 55, 56, 57, 59, 61, 62, 63, 64, 65, 66, 67, 68, 69, 71, 72, 73, 75, 76, 77, 79, 81, 82, 83, 84, 85, 86, 87, 88, 89, 91, 92, 93, 95, 96, 97, 99

## ...and is also divisible by its last three digits

$$n = 1000a + 100b + 10c + d$$

If $$bcd = 125$$, then the condition is satisfied regardless of $$a$$. (All other factors of 1000 have 0's in them.)

There are 229 valid combinations for the last 3 digits. I may categorize them later, but for now, here's a list:

111, 112, 115, 121, 122, 123, 125, 126, 128, 129, 135, 136, 144, 145, 146, 152, 155, 165, 168, 175, 176, 177, 182, 184, 185, 191, 192, 195, 211, 212, 213, 215, 216, 222, 223, 224, 225, 232, 235, 236, 244, 245, 246, 248, 252, 255, 256, 264, 265, 272, 275, 276, 277, 284, 285, 287, 288, 291, 292, 296, 311, 312, 315, 321, 322, 323, 325, 328, 333, 335, 336, 342, 344, 345, 346, 352, 355, 365, 366, 368, 369, 375, 377, 382, 384, 391, 392, 411, 412, 415, 416, 422, 423, 424, 425, 426, 429, 432, 435, 436, 444, 445, 446, 448, 452, 455, 456, 464, 465, 472, 475, 477, 488, 491, 492, 496, 511, 512, 513, 515, 522, 525, 528, 535, 536, 544, 545, 552, 555, 565, 568, 575, 576, 577, 584, 591, 611, 612, 615, 616, 621, 622, 624, 625, 632, 633, 635, 636, 639, 642, 644, 645, 648, 652, 655, 656, 664, 665, 666, 672, 675, 677, 684, 688, 691, 696, 711, 712, 715, 722, 725, 726, 728, 735, 736, 742, 744, 745, 752, 755, 765, 775, 777, 784, 791, 811, 813, 815, 816, 822, 824, 825, 832, 835, 844, 845, 848, 852, 855, 864, 865, 875, 877, 888, 891, 896, 911, 912, 915, 921, 922, 925, 928, 933, 935, 936, 942, 944, 945, 952, 955, 963, 965, 966, 975, 977, 984, 991, 999

## ...and is also divisible by its last four digits

There are 73 valid combinations for the last 4 digits:

1115, 1125, 1225, 1232, 1248, 1275, 1325, 1375, 1425, 1435, 1575, 1675, 1725, 1825, 1875, 2112, 2115, 2125, 2128, 2145, 2175, 2225, 2244, 2275, 2325, 2375, 2448, 2496, 2575, 2725, 2875, 3115, 3125, 3225, 3345, 3375, 3525, 3575, 3675, 3725, 3875, 4115, 4125, 4128, 4224, 4225, 4375, 4575, 4725, 4875, 4896, 5125, 5225, 5375, 5625, 5875, 6125, 6225, 6375, 6675, 6875, 7125, 7175, 7225, 7375, 7875, 8125, 8225, 8375, 9125, 9225, 9375, 9675

## ...and is also divisible by its last 5 digits

We're down to 54 valid combinations:

11125, 11375, 11875, 12375, 14125, 14375, 15625, 16125, 18125, 21125, 21375, 21875, 24125, 24375, 25625, 26125, 28125, 31125, 31875, 33375, 34375, 35625, 36125, 38125, 41125, 41875, 44375, 45625, 48125, 51875, 53125, 54375, 55625, 58125, 59375, 61125, 61875, 64375, 65625, 68125, 71125, 71875, 74375, 77875, 78125, 81125, 84375, 85625, 88125, 91125, 91875, 93375, 95625, 98125

## ...and is also divisible by its last 6 digits

Back up to 141 combinations.

111875, 114375, 116125, 118125, 121875, 124375, 125625, 128125, 131875, 134375, 135625, 138125, 144375, 148125, 153125, 155625, 159375, 161875, 165625, 171875, 174375, 184375, 185625, 191875, 195625, 211875, 214375, 218125, 221375, 221875, 224375, 228125, 231875, 234375, 235625, 245625, 253125, 259375, 261875, 265625, 268125, 284375, 291875, 318125, 321875, 324375, 325625, 328125, 334375, 335625, 353125, 354375, 358125, 359375, 361125, 361875, 365625, 371875, 384375, 391875, 395625, 398125, 414375, 421875, 424375, 428125, 434375, 435625, 453125, 455625, 459375, 465625, 484375, 495625, 511875, 515625, 521875, 528125, 534375, 535625, 545625, 553125, 559375, 561875, 564375, 565625, 571875, 578125, 584375, 591875, 595625, 621875, 624375, 628125, 634375, 635625, 653125, 658125, 659375, 665625, 671875, 684375, 691875, 695625, 721875, 728125, 734375, 735625, 753125, 759375, 761875, 765625, 771875, 774375, 784375, 791875, 811875, 821875, 828125, 834375, 835625, 853125, 859375, 861875, 865625, 871875, 884375, 895625, 914375, 918125, 921875, 924375, 928125, 931875, 934375, 953125, 959375, 965625, 984375, 991875, 995625

## ...and is also divisible by its last 7 digits

Down to 92 combinations:

1121875, 1265625, 1284375, 1328125, 1359375, 1365625, 1421875, 1453125, 1484375, 1515625, 1721875, 1734375, 1828125, 1884375, 1921875, 1953125, 1984375, 2121875, 2165625, 2265625, 2328125, 2365625, 2371875, 2421875, 2453125, 2515625, 2578125, 2721875, 2765625, 3121875, 3328125, 3359375, 3421875, 3453125, 3515625, 3553125, 3765625, 3828125, 3984375, 4265625, 4359375, 4421875, 4453125, 4484375, 4515625, 4721875, 4765625, 4921875, 4984375, 5153125, 5234375, 5421875, 5453125, 5484375, 5515625, 5534375, 5721875, 5765625, 5859375, 5984375, 6121875, 6328125, 6421875, 6515625, 6721875, 6853125, 6953125, 6984375, 7265625, 7284375, 7328125, 7421875, 7515625, 7578125, 7721875, 7734375, 7765625, 7984375, 8328125, 8421875, 8515625, 8721875, 8765625, 8828125, 8984375, 9121875, 9365625, 9421875, 9515625, 9553125, 9765625, 9984375

## ...and is also divisible by its last 8 digits

Only 19 possibilities:

12421875, 14453125, 14765625, 21328125, 33984375, 37578125, 42578125, 47734375, 48515625, 53515625, 55234375, 55859375, 57421875, 59765625, 63984375, 66515625, 78421875, 81421875, 92578125

## ...and is also divisible by its last 9 digits

Well, it looks like there just aren't any numbers that have one of the aforementioned 8-digit numbers as their suffix, and are also divisible by their own last 9 digits. So, this is where our search ends.

• all of the numbers that are bigger than $721875$ fail to hold for all suffix for example $8125 \not |728125$
– user411780
Commented Jun 14, 2022 at 6:13