Algebra - Numbers Divisible by 9 (Friendly Version)

How do you know when a number is divisible by 9? If the sum of its digits is divisible by 9, then the original number is divisible by 9! For example, 6525 is divisible by 9 because the sum of its digits (6 + 5 + 2 + 5 = 18) is divisible by 9 (18/9 = 2). But, how and why does this happen?9 times 1 is 9, and 9 is divisible by 9; 9 times 2 is 18, and 18 is divisible by 9; 9 times 3 is 27 and 27 is divisible by 9; so, in general, 9 times any number is divisible by 9. 9x is divisible by 9; 9y is divisible by 9. It doesn't matter; x can be a number or a long sum of numbers; just as long as you multiply by 9, you can always divide a 9 out. Let us remember 9x. We want to get a number in the form of 9x. Then we know it has to be divisible by 9.Take 6525 for example! Now ask yourself if the mathematical sentences are true or not.

Is sentence below true?

6525 = 6(1000) + 5(100) + 2(10) + 5(1)

Yes; look at the metric system: 6m + 5dm + 2cm + 5mm = 6525mm

Is sentence below true?

6525 = 6(999 + 1) + 5(99 + 1) + 2(9 + 1) + 5

Again, yes because 10 = 9 + 1; 100 = 99 + 1, and so on.

Is the sentence below true?
6525 = 6(999) + 5(99) + 2(9) + 6 + 5 + 2 + 5

Yes, because we are allowed to distribute numbers like this. For example, x(x + 2) = x^2 + 2x, or 25 = 5(5) = 5(2 + 3) = 5(2) + 5(3) = 10 + 15 = 25. Now notice that the numbers on the right are the sum of the number 6525, which is 6 + 5 + 2 + 5 = 18, and 18 is divisible by 9 because 9 times 2 =18, and again 9 times any number is divisible by 9.

Is this sentence below true?
6525 = 9(111)(6) + 9(11)(5) + 9(2) + 9(2)

Yes, because all we did was factor out a 9 from each term. The last term was the sum of the digits 6525.This sentence below is true because we can factor a 9 from each term, we couldnt have done this if the sum of the digits in 6525 was NOT divisible by 9, so we need the sum of the digits to equal a number that's divisible by 9 in order to factor it out at this step!

6525 = 9[6(111) + 5(11) + 2(1) + 2]

Now remember, we said 9 times ANY NUMBER is divisible by 9, well let's make that number [6(111) + 5(11) + 2(1) + 2], now we have shown not only how but why if the sum of digits are equal to a number divisible by 9, that number MUST be divisible by 9, and vis versa. Here are all the steps in order:

6525 = 6(1000) + 5(100) + 2(10) + 5(1)6525 = 6(999 + 1) + 5(99 + 1) + 2(9 + 1) + 5
6525 = 6(999) + 5(99) + 2(9) + 6 + 5 + 2 + 5
6525 = 9(111)(6) + 9(11)(5) + 9(2) + 9(2)
6525 = 9[6(111) + 5(11) + 2(1) + 2]

6525/9 = 6(111) + 5(11) + 2(1) + 2725 = 666 + 55 + 2 + 2 = 725

Grammar - The Grammar of 'Either'

What is the grammatical function of the word 'either'? If you are confused, you're not alone!

Let's look at 'either' when it is used as a pronoun: 'Marry either of the two women.' sounds grammatically correct and it is. 'Marry any of the three women.' not only sounds better than 'Marry either of the three women.' but is also the grammatically correct choice. Just as 'between' is used to talk of 2 and only 2 entities and 'among' for 3 or more entities, 'either' is used to talk of 2 and only 2 entities while 'any' for 3 or more entities. In other words, use 'either' when talking about two things and use 'any' when talking about more than two things.

Nothing seems straightforward in grammar though. During the Old English (6th cen to 10th cen) and early Middle English (10th cen to 13th cen) periods, 'either' was taken to mean 'each of two' or 'both'. It wasn't until the late Middle English period where 'either' took on the disjunctive sense of meaning 'one or the other (but not both)'. The disjunctive sense was also covered by the word 'outher'; but, 'outher' became obsolete around the 16th cen, and so 'either' once again denoted the two prior meanings. Over the next 5 centuries the two prevailing meanings competed with each other. Eventually, the disjunctive meaning became dominate in Modern English.

Nonetheless, the original meaning has left a remaining residue of confusion in current language speakers. The Oxford English Dictionary actually recommends that the original meaning "must often be avoided on account of [its] ambiguity".

However, using 'either' is preferred over 'any' for whence a sentence talking of 3 or more entities yields an ungrammatical sentence. Thus, the sentence, 'President Nixon was either a good president, a bad president, or the best president' is a grammatical exception and therefore correct since using 'any' in this situation would yield an ungrammatical sentence.

Abstract Algebra - Numbers Divisible by 9 (Technical Proof)

This is a technical proof of how if you sum the didgets of any number and that sum is divisble by 9, then the original number is divisible by 9.

Let ai denote a digit in mod 10 and let a1a2…an denote a positive integer with an n number of digits. Then, 9a1a2…an Î Z+ if and only if 9Si = 1, n ai Î Z+.

Proof. Factoring out a 10n from the nth digit gives

a1a2…an – 1an
= 10(n – 1)a1 + 10(n – 2)a2 + … + 10(n – (n – 1))an – 1 + 10(n – n)an
= 10(n – 1)a1 + 10(n – 2) + … + 10an – 1 + an

Since [(10n – 1) + 1] = 10n,

= [(10n – 1 – 1) + 1]a1 + [(10n – 2 – 1) + 1]a2 + … + [(10 – 1) + 1]an – 1 + an.

Distributing ai to each term gives

= (10n – 1 – 1)a1 + a1 + (10n – 1 – 1)a2 + a2 + … + (10 – 1)an – 1 + an – 1 + an

Because 9(10n – 1) "n Î Z+, 9 can be factored out from each (10n – 1) to leave (S i = 0, n 10i).

= 9a1(S i = 0, n – 1 10i) + 9a2(S i = 0, n – 2 10i) + … + 9an – 1 + a1 + a2 + … + an (5)

The number a1a2…an – 1an contains the sum Si = 1, n ai. Let 9Si = 1, n ai = R. Factoring out 9 from each term gives

a1a2…an – 1an = 9a1(S i = 0, n – 1 10i) + 9a2(S i = 0, n – 2 10i) + … + 9an – 1 + 9R
= 9[a1(S i = 0, n – 1 10i) + a2(S i = 0, n – 2 10i) + … + an – 1 + R]

Because a1a2…an – 1an can factor out 9, 9a1a2…an – 1an. This could not have occurred if R did not factor out nine as well. Further, if 9a1a2…an – 1an, then a1a2…an – 1an gives a 9R term by equation (6). 9R = Si = 1, n ai and 9Si = 1, n ai. QED