Divisibility by 10^k-1, 10^k+1 (2nd explanation besides Vivian Loh's is mine)

Geometric proof that the series 1*2 + 2*3 + 3*4 + 4*5 + ... + n*(n+1) equals (1/3)*n*(n+1)*(n+2)

Decimal to octal and duodecimal

We multiply the numbers of each diagonal of the table ( the red and yellow series of numbers, bottom left to top right) one by one with each of the digits of the decimal number to be converted (the blue digits in the leftmost column up to the bar) starting from the bottom in both cases and moving upwards, so we multiply 5 by 1, 9 by 2, 6 by 4 etc. up to 4 times 128 and we add the products. We then remove the greatest multiple of 8 less than the sum from the sum and keep the remainder as a digit (the last first) of the octal number to be found. We keep how many times 8 goes into the sum we found as the carry for the next addition of products. Then we begin at the next blue digit from the end in the leftmost column (9 in this case) and we multiply up to the beginning of the blue digits each digit with each of the numbers of the next diagonal (a red square one in this case) and again we make the sum of the products, convert it as before, add the carry from the previous addition if needed  and convert again if necessary and we get the next digit (from the end) of the octal number to be found. We move along in this way, until only one of the blue digits (4 in this case) is multiplied by 1. The octal number might have more digits from the decimal number because of the carry. The coefficients (numbers in yellow or red squares) always stay the same and we can add more rows to the top one, taking into account that each row starts with 1 and the next number is the sum of twice the number below it and to the left plus the number immediately below it (so 672 in the top row is twice 240 plus 192, 448 is twice 192 plus 64, and 128 is just twice 64 because there is no number immediately below it). Every decimal number can be converted to octal thanks to this table. To convert a decimal number to a base-12 number we use the same table and the same process only that in every second number (second, fourth, sixth etc.) of a diagonal we attach a negative sign. If the sum we end up getting from multiplication with the numbers of a diagonal is now negative, we keep adding multiples of twelve until it turns out positive and we subtract the number of multiples we added from the next sum, once that is complete. The base-12 number can also have 10 and 11 as digits, so to avoid confusion we can also denote them as A and B. The base-12 number can have less digits than the decimal number. By converting to octal we can add the digits of the octal number as if they were decimals and divide the sum by 7 (normal, decimal division). It's casting out 7's, it's the same as casting out 9's. The remainder of the division would be the remainder of the decimal number we started with when that number is divided by 7. On the other hand, we can add pairs of digits of the base-12 number starting from the end as if they were two-digit base-12 numbers (base-12 addition). If the number of the digits is odd we will only add one digit at the end. Repeating the process in the base-12 sum that we got and keeping it up until we get a two-digit base-12 number we calculate its difference from the next lesser base-12 number with the same two digits (the difference could well be ten, eleven or twelve - we need to make a base-12 subtraction). That difference is the remainder of the decimal number we started with when divided by thirteen. The numbers on each row of the table starting from the left row from the bottom give the number of features (vertices, edges, faces, spaces etc.) of rectangular figures (square, cube, hypercube and their analogues in higher dimensions).

Base 10 to base 20

By converting a decimal number to base 20 we can easily check divisibility by 21 (and therefore 7) and 19. The conversion is achieved by subtracting multiples of powers of 2 in descending order (64, 32, 16, 8, 4, 2 etc.) from the decimal number starting at the beginning of the number. 

Example:

Take the decimal number 1453. As there is no point in subtracting a multiple of 8 from 1 we take the first two digits, 14 and subtract the greatest multiple of 4 less than 14, which is 12. We are left with 253. The next power of 2 in descending order after 4 is 2. The greatest multiple of 2 after 25 is 24. Notice that vigesimal (base 20) numbers can have two decimal digits, so we get 12 as the next base 20 “digit". We are left with 1 which carries over to 3 as the vigesimal “digit" 13. Thus the vigesimal number we have found is (in two-decimal-digit notation) 03 12 13. 13 - 12 + 3 = 4, so 1453 is 4 (mod 7) which is correct, given that 1449 is divisible by 7.

Another example:

Take the number 187607. Counting from the end, we have the powers of two 1, 2, 4, 8, 16 for five digits. It makes no sense to try divide the first digit, 1, by 32. Instead we divide 18 by 16 and get a quotient of 1 and a remainder of 2. We divide 27 by 8 and get a quotient of 3 and a remainder of 3, We are left with 36 and then a zero. We divide 36 by a and get a quotient of 9 and no remainder. We are left with 2 decimal digits and two descending powers of 2 (2 and 1). As we should always leave the last decimal digit unaltered in conversion from decimal to vigesimal we complete the vigesimal number with 0 as a multiple of 2 and 7 as a multiple of 1. Therefore, collecting the quotients, the vigesimal number we found is 01 03 09 00 07. 7 - 0 + 9 - 3 + 1 = 14, so the decimal 187607 is divisible by 7. 7 + 0 + 9 + 3 + 1 = 20, therefore 187607 leaves a remainder of 1 when divided by 19.

Base 90

One would think that, since 90 is removed 10 units from 100, i.e. a significant number for the decimal system, there would be some formula to convert from base 100 to base 90. We can derive such a formula via the binomial theorem and we just present it here up to the second power of 90 for the sake of simplicity.

We separate the number into groups of two digits starting from the end as usual and we add the second group of digits to the tens of the first group. We then add twice the third group to the tens of the second group, then the third group to the units of the second group and then the third group to the tens of the first group. Now, 91 is similar 11 in decimal. If we subtract and add alternately groups of two digits starting from the end we can establish divisibility by 91 or 7 or 13 or the remainder of the division.

Example: Let the number be 93 92 base 100, 9392 decimal. We just have to add 93 to the tens of the first group. 93 + 9 = 102, so we have 93 102 2, or 93 1022. The greatest multiple of 90 in 1022 is 990, so we take that away. 11 90's are added to 93, so we have 104 32. We convert 104 to base 90, and we find that 9392 decimal = 1 14 32 base 90. 32 - 14 + 1 = 19, so this number leaves a remainder of 19 when divided by 91, 6 when divided by 13, and 5 when divided by 7. Furthermore, 32 + 14 + 1 = 47, so the remainder of the division of 9392 by 89 is 47.

Another example: Let the number be 16807 decimal (=7^5), 1 68 07 base 100. Adding 68 and 1 to the tens of the first group, we get 1 68 697. Adding 1 to the tens and units of the second group (remember to add twice to the tens), we get 1 89 697. Casting out 90's from 697 we get 1 96 67. Converting 1 96 to base 90 we get 2 06 67. 67 - 6 + 2 = 63, so 16807 leaves a remainder of 63 when divided by 91, 11 when divided by 13, and 0 when divided by 7. Furthermore, 67 + 6 + 2 = 75, so the remainder of the division of 16807 by 89 is 75.

Another way of checking divisibility by 23

If only for the purpose of the exercise, we can now examine divisibility by 23, actually divisibility by 92, in the same way we examined divisibility by 91 (see “A divisibility rule for both 7 and 13” below). We convert base 100 to base 101 so we can work with 92 instead of 91. 

Example:

We have seen from the previous post 5 30 57 base 100 to be 5 20 32 base 101. 5 20 32 = 50 + 20 - 5  32 = 65 32 = 65 * 10 + 3  2 - 65. But remember our base of operations is 101, so 65 * 10 = 644.  644 + 30 + 2 - 65 = 611 = 71 - 6 = 65 = 19 (mod 23).

Another example:

234871 (100) = 23  48 - 2*23  71 - 23  =  23  02  48 =  23  02  48 - 2 = 23 02 46 (101)

23 02 46 = 23*10 + 2 - 23  46  =  228 (101) + 2 - 23  46 = 207 (101)  46  =  25 46  =  25*10  +  46  - 25  =  248 (101)  +  46  - 25  =  269  =  87

87 leaves a remainder of 18 when divided by 23 so 234871 (mod 23) = 18

More changes of base

If we change base 100 to base 101, we can readily verify divisibility by 17 (17*6=102) just as if we were using 11 on base 10 in decimal.

Moreover, if we change base 10 to base 11 and then to base 12, divisibility by 13 can be examined by the same token.

Examples: 5 30 57 (100) = 5 30 - 2*5 57 - 5 = 5 20 52 = 5 20 32 (101)

We just reverse-engineered the operation of adding groups of digits using the coefficients of the Pascal triangle (here 2 and 1, 1)

32 - 20 + 5 = 17, so the initial decimal number is divisible by 17.

289 (100) = 287 (101) = 2 87 (101). 87 - 2 = 85 = 5 * 17 so 289 is divisible by 17 (in fact, it's 17's square).

1001 (10) = A * A * A + 1 (11). A * A (11) = 91 (11). 91 * A (11) = 82A (11). 1001 (10) = 82A + 1 (11) = 830 (11). We see that the number is a mulltiple of the base, i.e. 11. 830 (11) = 8 * A1 (12) + 3*B (12) = 688 + 29 (12) = 6B5 (12). B5 + 6 = BB (12), divisible by 11 in base 12, which is 13 decimal.