Ever notice that the digits of the result of any number multiplied by 9 always equals 9. Example: 364 x 9 = 3276. 3+2+7+6=18 and 1+8=9.
Of course. Simple math theory.
Let x represent any number.
x * 9 is the same as (10 * x) - (x);
which, digit-wise, is (x - 1) in the tens place and (10 - x) in the ones place.
Add those two together and you have:
(x - 1) + (10 - x) or
x + (-1) + 10 + (-x) or
x + (-x) + 10 + (-1) which is
0 + 9 which is
9.
Ergo, all numbers multiplied by 9 will have their digits add up to a multiple of 9.
JMCJ
Give to Radiskull!
Along the same lines, the difference between two numbers with transposed digits is always a multiple of nine. Eg: 83-38=45. Handy if you ever have to balance a column of numbers manually.
And all even length palindromes (like 145541) are divisible by 11.
And the “add up the digits” trick works for 3 as well - if a number is divisible by three, the digits add up to something divisible by 3, ultimately 3, 6, or 9.
As **John Corrado]/i] implies, the “digits of a multiple of nine always add up to nine” trick is because we typically represent numbers in base 10.
Generally, if numbers are represented in base ‘n’, then the digits of multiples of (n-1) will always add up to (n-1) (repeating the addition until a single-digit result is gooten).
Also, if (n-1) has non-trivial factors f[1]…f*, then multiples of any factor f[k] will always add up to a multiple of f[k].
“I don’t just want you to feel envy. I want you to suffer, I want you to bleed, I want you to die a little bit each day. And I want you to thank me for it.” – What “Let’s just be friends” really means
Here’s a closely related fact.
Add all the digits of a base ten number together. Add all the digits of that result. Continue on each successive result until you get 1 digit. That final digit is the remainder after you divide the original number by 9. If you get 9, that indicates a remainder of 0.
It follows that all of the intermediate results you got also have that same remainder.
You can shorten the process by eliminating sets of digits in each result that sum to 9 or a multiple of 9. This process is sometimes called “casting out nines.”
Accountants once used this to check their addition. They would cast out nines on the sum of a column, giving a result N. Then they would cast out nines for each entry in the column, then add those results together, casting out more nines as necessary, giving a result M. If M was not equal to N, they knew there was an error. However, if M was equal to N, that did not eliminate the possibility of an error.
The above is also true for any number in base n after dividing by (n-1).
…this is another Moebius sig…b!s sn!qaoW jay+oue s! s!y+…
(adaptation of a WallyM7Sig™ a la quadell)
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Long ago in school this was one of the tricks they taught us to use to determine if one number could be divided by another.
If the number is even, it is divisible by 2.
If the sum of the digits of the number is divisible by 3, the number is divisible by 3.
If the last two digits of the number are divisible by 4, the entire number is divisible by 4.
If the number ends in 0 or 5, it is divisible by 5.
If the number is divisible by both 2 and 3, it is divisible by 6.
If the sum of the digits of the number is divisible by 9, the number is divisible by 9.
If the number ends in 0 it is divisible by 10.
I can’t remember if there was a similar trick for 7 and 8. I think there was one for 8 but not for 7, but cannot remember for sure. Anyone know?
“Sometimes I think the web is just a big plot to keep people like me away from normal society.” — Dilbert
tanstaafl - for 8, last 3 digits. Just like 2 is last digit, 4 last 2 digits. 10 has a factor of 2, so 100 has 2 factors of 2, etc.
The “9’s” situation works in other bases, too.
In base X, any integral multiple of (X-1) has digits that–when added–also equal a multiple of (X-1).
I’ll leave it to the number theorists to explain why.
A lot have already been mentioned, but here’s one that works for 7, 11, and 13:
Starting from the right, take the digits in groups of 3’s and alternate adding and subtracting. Example:
78,592,590,287
287 - 590 + 592 - 78 = 211
211 will have the same remainder as 78,592,590,287 when divided by 7, by 11, and by 13. Actually, you can break the digits into groups of one each for the case with 11. This all comes from the fact that 71113 = 1001.
…ebius sig. This is a moebius sig. This is a mo…
(sig line courtesy of WallyM7)
the process of “casting out nines” to check the correctness of any mathematical operation also works in any other base than decimal. In base n you cast out the (n-1)