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.
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.
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