PROVE THAT nnn-n is divisible by 6?

Answer 1

A number is divisible by 6 if it is divisible by 2 and 3.

Look at 333-3 which is 330 The sum of the digits is 6 and it is even so it is divisible by 6 Now consider 222-2 which I picked because unlike 333, 222 has even digits. 222-2=220, one again even number so divisible by 2 but NOT divisible by 3 so NOT divisible by 6

So it look like this is not true for all n

For any odd n, we have the following

1. nnn-n ends in 0 so it is even if we can show it is divisible by 3 we are done. but 777-7 is 770 which is NOT divisible by 3 so it is NOT true.

For some n it is true, but not for all n... Now when will nnn-n be divisible by 3.

only when n+n is a multiple of 3, ie n=33,66, 99 an that is it!

So we could easily prove that nnn-n is divisible by 6 if and only if n=3,6,or 9

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If by nnn, you mean n3, a proof is as follows:

n=0,1,2,3,4, or 5 (mod 6)

If n=0 (mod 6), we have (0 (mod 6))((0(mod 6))2-1)=0 (mod 6). [Since the first term is zero]

If n=1 (mod 6), we have (1 (mod 6))((1(mod 6))2-1)=0 (mod 6) [Since 1-1=0].

If n=2 (mod 6), we have (2 (mod 6))((2(mod 6))2-1)=(2*3) (mod 6) = 6 (mod 6)=0 (mod 6).

If n=3 (mod 6), we have (3 (mod 6))((3(mod 6))2-1)=(3*8) (mod 6) = 24 (mod 6) = 0 (mod 6).

If n=4 (mod 6), we have (4 (mod 6))((4(mod 6))2-1)=(4*15) (mod 6) = 60 (mod 6) = 0 (mod 6).

If n=5 (mod 6), we have (5 (mod 6))((5(mod 6))2-1)=(5*24) (mod 6) = 120 (mod 6) = 0 (mod 6).

If you're not comfortable with the modular arethmetic, you can substitue 6m+_, where the blank is each of the numbers 0 through 5 (since every number can be expressed either as a multiple of six, or as a multiple of six plus some number between 1 and 5 --the remainder when the number is divided by six). Taking our example with 5, you would get:

(n)(n2-1) can be written as (6m+5)((6m+5)2-1), where m is an integer.

Simplifying this, you get:

(6m+5)((6m+5)2-1)

(6m+5)((6m2+60m+25-1)

6m*6m2+6m*60m+6m*25-6m+5*6m2+5*60m+5*25+5(-1)

6m*6m2+6m*60m+6m*25-6m+5*6m2+5*60m+5*24

Since m is an integer and each term is divisible by 6, (n)(n2-1) is divisible by 6 for integers that can be expressed as 6m+5. You would then repeat the process for each of 0 through 4 to complete the proof. Clearly, if you are comfortable with it, modular arithmetic is the less cumbersome way to proceed.