Ask Me Help Desk - Divisibility rules 11

I am completing an assignment and had to look for patterns with the number 11.
I now need to prove 2 of my findings but can't can you help please.
If I have the number 81752 I know that 8+7+2 -1+5 = 11 so its divisible by 11.
How do I prove this using albebra
Also if I take the same number now I know its divisible by 11 and
8175-2=
8173
817-3=
814
81-4=
77 = 7x11 so 7 and the numbers I subtracted being 4,3,2, means 7432x11=81752
How do I prove this?

Wow, cool problem. Mostly it's something like

"help me solve 1+x=2. How can I find x?".

Are you familiar with some number theory?

Also, a number is divisible by 11 only if the alternating sum of its digits sums to 11 or of we obtain 0. That is considered as being divisible by 11 as well. Say, we have 385. We know it is divisible by 11. Alternate the digits and we get 3-8+5=0.

As for yours:

8-1+7-5+2=11... check.

Start with some modular congruencies.

$10\equiv -1(mod \;\ 11)$

We can alternate the sign by using odd and even powers of -1:

$10^{k}\equiv (-1)^{k}(mod \;\ 11), \;\ k=1,2,3,....$

$x\equiv a_{0}+a_{1}(-1)+a_{2}(-1)^{2}+a_{3}(-1)^{3}+a^{4}(-1)^{4}+a^{5}(-1)^{5}+\cdot\cdot\cdot\cdot +a_{n}(-1)^{n}$

$\equiv a_{0}-a_{1}+a_{2}-a_{3}+a_{4}-a_{5}\cdot\cdot\cdot\cdot a_{n}(-1)^{n}(mod \;\ 11)$

Therefore,
x is divisible by 11 iff the alternating sum of its digits is divisible by 11.

Galactus thanks for helping but my maths isn't so advanced. Can you explain the 3 lines =plus another line. Does that mean 'which means that'. I understand numbers elevated mean to the power of but what do the numbers below a0-a1+a2 mean. And does mod mean multiple of and then I can try translating! Would you be able to help me with the second part of my question. Seeing I only did basic gcse I feel pleased I found what I did but obviously the proving is the difficult part. But thank you any way so far. And also as a first time user I got a buzz that I even got a reply. Are you in the states? Just curious really.

This symbol: $\equiv$ means equivalent.

As for [math]a_0, a_1, a_2[math] only represent some constant named 'a'. It cold have been any other letter. And the lowered number is there to differentiate between each constant.

As for the other things, I would be glad to help you, but I don't know about modular congruencies :o

I can't give a complete lesson on elementary number theory, but I can explain what the mod business is about.

We have $10\equiv -1(mod \;\ 11)$

Add the 1 to the 10 and divide by 11.

$10^{k}\equiv (-1)^{k}(mod \;\ 11)$

Let k=2:

$100\equiv 1(mod \;\ 11)$

$\frac{100-1}{11}=9$

Say we have $10^{3}\equiv (-1)^{3}(mod \;\ 11)$

$\frac{1000+1}{11}=91$

and so on.

It's about remainders. It is sometimes called clock arithmetic and is very useful when dealing with huge numbers.

Arbitrary example:

Solve $42x\equiv 12(mod \;\ 90)$

We want to find all values of x that satisfy the congruence.

We have gcd(42,90)=6. Since 6 is a factor of 12, there is a solution.

Solving the congruence is the same as solving

$42x=12+90q$ for integers x and q.

This reduces to $7x=2+15q$ or $7x\equiv 2(mod \;\ 15)$

The numbers congruent to 1 modulo 15 are 16,31,46,61,.
and also -14,-29,-44,.

See, they're all off by 1 from being a multiple of 15.

7 is a factor of -14, so we multiply both sides by -2 since $-14\equiv 1(mod \;\ 15)$

Thus, we have $-14x\equiv -4(mod \;\ 15)\rightarrow x\equiv 11(mod \;\ 15)$

The solution is $x\equiv11,26,41,56,71,86(mod \;\ 90)$

Check: $42*11-12=450$ . 90 divides into 450.

I know this is a lot to digest, but a lot of proofs come from this area of number theory. It is a fascinating field and can be used to solve many problems.

Uh huh.. I could understand some things, but not everything :(

I guess I'll have to wait university or next year during my 'spare' time.

Quote:

Originally Posted by Unknown008

Uh huh.. I could understand some things, but not everything :(

I guess I'll have to wait university or next year during my 'spare' time.

Yes, when you get to school you will no doubt take number theory.
Then, you will lern about this stuff. You can always get a book or Google. There is lots out there. Just Google "modular arithmetic".
Or the Chinese Remainder Theorem. That is a goodie.