1. Dropping Two and More Digits

When one digit is dropped from a d-digit integer, there are d integers in the sum and each is d-1 digits long. In general, when n digits are dropped from a d-digit integer, the number of integers in the sum is equal to d! / (n!(d - n)!), or the combination of n digits taken 2, 3, 4... at a time. For example, when d = 20 and n = 3, there are 20! / (3!(20 - 3)!) = 1,140 integers in the sum. In base ten, the integers31171093 01159390698 and 47989422298181591480943 are equal to their missing-digit sums when dropping two, three and four digits, respectively. Here is the delete-2 sum, containing 12! / (2!(12 - 2)!) = 66 integers:

183477122641 = 3477122641 (missing 1 and 8) + 8477122641 (missing 1 and 3) + 8377122641 (missing 1 and 4) + 8347122641 (missing 1 and first 7) + 8347122641 (missing 1 and second 7) + 8347722641 (missing 1 and second 1) + 8347712641 + 8347712641 + 8347712241 + 8347712261 + 8347712264 + 1477122641 + 1377122641 + 1347122641 + 1347122641 + 1347722641 + 1347712641 + 1347712641 + 1347712241 + 1347712261 + 1347712264 + 1877122641 + 1847122641 + 1847122641 + 1847722641 + 1847712641 + 1847712641 + 1847712241 + 1847712261 + 1847712264 + 1837122641 + 1837122641 + 1837722641 + 1837712641 + 1837712641 + 1837712241 + 1837712261 + 1837712264 + 1834122641 + 1834722641 + 1834712641 + 1834712641 + 1834712241 + 1834712261 + 1834712264 + 1834722641 + 1834712641 + 1834712641 + 1834712241 + 1834712261 + 1834712264 + 1834772641 + 1834772641 + 1834772241 + 1834772261 + 1834772264 + 1834771641 + 1834771241 + 1834771261 + 1834771264 + 1834771241 + 1834771261 + 1834771264 + 1834771221 + 1834771224 + 1834771226

2. Trivial Missing-Digit Sums

In any base b, there will be a set of delete-1 missing-digit sums with b+1 digits and first digits in the range 1...b-1, followed by b zeroes. Consider base-2 and base-3:

100_b=2 = 4_b=10 = 00_b=2 (deleting 1) + 10_b=2 (deleting first 0) + 10_b=2 (deleting second 0)

= 0 + 10_b=2 + 10_b=2

= 0 + 10_b=2 x 10_b=2 = 100_b=2

= 0 + 2 x 2 = 4_b=10

1000_b=3 = 27_b=10 = 000_b=3 (deleting 1) + 100_b=3 (deleting first 0) + 100_b=3 (deleting second 0) + 100_b=3 (deleting third 0)

= 0 + 100_b=3 + 100_b=3 + 100_b=3

= 0 + 10_b=3 x 100_b=3 = 1000_b=3

= 0 + 3 x 9 = 27_b=10

2000_b=3 = 54_b=10 = 000_b=3 (deleting 2) + 200_b=3 (deleting first 0) + 200_b=3 (deleting second 0) + 200_b=3 (deleting third 0)

= 0 + 200_b=3 + 200_b=3 + 200_b=3

= 0 + 10_b=3 x 200_b=3 = 2000_b=3

= 0 + 3 x 18 = 54_b=10

Accordingly, the eleven-digit numbers 10,000,000,000 through 90,000,000,000 are trivial missing-digit sums in base ten, because their sums take this form:

10,000,000,000 = 0,000,000,000 (deleting 1) + 1,000,000,000 (deleting first 0) + 1,000,000,000 (deleting second 0) + 1,000,000,000 (deleting third 0) + 1,000,000,000 + 1,000,000,000 + 1,000,000,000 + 1,000,000,000 + 1,000,000,000 + 1,000,000,000 + 1,000,000,000

= 0 + (10 x 1,000,000,000) = 10,000,000,000

And in hexadecimal or base sixteen, the seventeen-digit numbers 10,000,000,000,000,000 through F0,000,000,000,000,000 are missing-digit sums of the same form.

3. Searching for Missing-Digit Sums

Searching for delete-1 missing-digit sums is simplified when one notes that the final two digits of n determine the final digit of its missing-digit sum. One can therefore test simply the final two digits of a given n to determine whether or not it is a potential missing-digit sum. In this way, the search-space is considerably reduced. For example, consider the set of seven-digit base-ten numbers ending in ...01. For these numbers, the final digit of the sum is equal to (digit-0 x 1 + digit-1 x 6) modulo 10 = (0 + 6) mod 10 = 6 mod 10 = 6. Therefore, no seven-digit number ending in ...01 is equal to its own missing-digit-sum in base ten.

Now consider the set of seven-digit numbers ending in ...04. For these numbers, the final digit of the sum is equal to (0 x 1 + 4 x 6) modulo 10 = (0 + 24) mod 10 = 24 mod 10 = 4. This set may therefore contain one or more missing-digit sums. Next consider seven-digit numbers ending ...404. The penultimate (last-but-one) digit of the sum is equal to (2 + 4 x 2 + 0 x 4) modulo 10 = (2 + 8 + 0) mod 10 = 10 mod 10 = 0 (where the 2 is the tens digit of 24 from the sum for the final digit). This set of numbers ending ...404 may therefore contain one or more missing-digit sums. Similar reasoning can be applied to sums in which two, three and more digits are deleted from the original number.

4. Incomplete List of Missing-dDigit Sums in Base Ten

4.1. Delete-1 Sums

14358846, 14400000, 15000000, 28758846, 28800000, 29358846, 29400000, 1107488889, 1107489042, 1111088889, 1111089042, 3277800000, 3281400000, 4388888889, 4388889042, 4392488889, 4392489042, 4500000000,^[1] 5607488889, 5607489042, 5611088889, 5611089042, 7777800000, 7781400000, 8888888889, 8888889042, 8892488889, 8892489042, 10000000000, 20000000000, 30000000000, 40000000000, 50000000000, 60000000000, 70000000000, 80000000000, 90000000000

4.2. Delete-2 Sums

167564622641, 174977122641, 175543159858, 175543162247, 183477122641, 183518142444, 191500000000, 2779888721787, 2784986175699, 212148288981849, 212148288982006, 315131893491390, 321400000000000, 417586822240846, 417586822241003, 418112649991390, 424299754499265, 424341665637682, 526796569137682, 527322398999265, 533548288981849, 533548288982006, 636493411120423, 636531893491390, 642800000000000, 650000000000000, 738986822240846, 738986822241003, 739474144481849, 739474144482006, 739474144500000, 739512649991390, 745699754499265, 745741665637682, 746186822240846, 746186822241003, 751967555620423, 848722398999265, 849167555620423, 854948288981849, 854948288982006, 855396569137682, 862148288981849, 862148288982006, 957893411120423, 957931893491390, 965131893491390, 971400000000000

4.3. Delete-3 Sums

4.4. Delete-4 Sums

1523163197662495253514, 47989422298181591480943, 423579919359414921365511, 737978887988727574986738