Now you take the remainder of that, 2, and raise it to the 7th power, which is 128. Take the remainder when 7 is divided into 128, and you get 2. That is the answer, right?
2) This one is pretty neat. Your base here is 12^2. Let a^2 = 12^2, so that we can derive this
60^2 = 25a^2, and 48^2 = 16a^2, (because 12^2 * 5^2 = 60^2, and 12^2 * 4^2)
So now... we get 41a^2 - a^2 which equals 40a^2.
So lets apply this formula!
40*12*12 = 5760.
Those were two very awesome questions. Keep em coming! This site is making me awesome at NS. Just gotta get faster, and I'll be competetive.
3^14 /7: because (3,7)=1, the remainders as the power goes up form cycles of 6 - all possible remainder except that of 0, that is. so 3^14 = 3^2 %7 = 2
Yeah, we're gonna be NS champs by the time the school year starts.
For the summer I'm going from 1989-1990 tests all the way up to 2006. I ordered Dr. Numsen 2004_2005 tests on CD (sent check on may 16th.. still don't have them ) so I will go through those after I'm done with the UIL ones. Not sure what I will do after that.
3^14 /7: because (3,7)=1, the remainders as the power goes up form cycles of 6 - all possible remainder except that of 0, that is. so 3^14 = 3^2 %7 = 2
can you go over this thoroughly? i was never clearly explained how get remainders when a numberis raised to a large exponent. (3,7) = 1, means GCF i assume, what if it isn't? any other special cases?
3^7 %7 = (3^6 *3)%7 = (3^6 %7)*3%7 = 1*3 = 3 -- the beginning of the second cycle.
There are several points I intend to demonstrate here:
1. (a+or*b)%k = (a%k +or* b%k)%k, thus you can break down large number operations into smaller remainder operations
2. (a^b)%k = [(a%k)^b]%k, this is most useful in the second kind of remainder question you get in UIL.
3. look at the remainder as the power goes up: it covers all of %7's possible remainder except 0 because 7 can never divide 3's powers because 3 itself is not divisible in 7. Note that it's a 6-element cycle that repeats at the 7th term and so on, so the original question, 3^14, has 2 full cycle in it already, so get rid of them 3^(14-6-6)=3^2, then %7=2.
4. in order for the cycle to repeat, the remainder at the end of the cycle has to be 1, as in 3^6 %7, so you can conclude: for (a, k)=1 and integer m, a^(mk-1) %k = 1, where mk-1 is the term number for every end of the cycle.
5. if (a, k)>1, the cycle will be smaller for sure. Now that you've mention it, I will find some generalization when I have time, but make up some problem, try them yourself, and see what you can figure out.
thanks for explanation. if you want to use this on the test, it would have to be memorized ahead of time the number of terms for each cycle from 2-9 huh?
No, it would be definitely too much to memorize. Memorizations only help you to SIMPLIFY thing. Most of the times they can't and won't get you the answer.
Since for a number, n, has n-1 nonzero remainders, n-1 is how long each cycle is. And after your problem is simplified, you just do'em.
You can also find the pattern on the spot, and then put the remainder to whatever power you need (which, after reducing, should not be big); that's how I've always done'm. I'm sure that's one of the reason why I'm still not UP THERE.
is this what u mean by breaking it down on the spot? this will take forever and i probably won't be able to keep track of the numbers. for any number n, it has n-1 nonzero remainders. is n the base, exponent, divisor? dooesn't the whole problem change if one is chagned?
Who says you have to follow EVERY step that I wrote?! I was trying to demonstrate my points by adding all those extra steps in. Well, I seem to have done a bad job clarifying and explaining. Sorry.
Now, let's see the 4^14 %7 problem. I think the rules work just as well.
4^1 %7=4
4^2 %7=2
4^3 %7=1
4^4 %7=4
And there the cycle starts again. It's a cycle of 3, now the cycle of 6 still works because 6 is 2 cycles of 3. Thus you can still reduce 4^14 to 4^2 by substracting 2 full cycles of 6, then %7=2.
Now if I see this problem, this would be all I do in my head
4%7=4 //find the 1st remainder of the cycle
16%7=2 //find the 2nd remainder of the cycle
(2*4)%7=1 //find the 3rd remainder of the cycle, note that 1 is likely the end of it
1*4=4 //the remainder repeats, confirm a 3-term cycle(this step is not even needed)
4^14 %7=4^2 %7=2
If you have any questions, please ask and notify the exact place where you got lost. I'll do my best.
Recall one of the rules that basically states: 2 #s adding and multiplying is like their remainders following the same operations.
So e.g, you have a natural #, n, and n to some power, p, divided by another natural #, k, has a remainder of 1. We know that a cycle starts with n^1 %k = n, and if n^p %k = 1, and you do n^(p+1), its remainder, by the rule above, should be (n^p %k)* (n^1 %k) = (1)*(n) = n, which marks the beginning of another cycle. The key is any # times 1 is itself, and itself is coincidentally the start of a cycle. So if you see 1 as a remainder, you'll know it's the end of a cycle.
after reading it over 10+ times, i couldn't get it. but it's ok, i probably won't be able to understand, hopefully later i will. thanks for the explanation anyways.
Don't you give up! I haven't given up yet! I want to make it understood, and I don't want to feel incapable of effective explanations.
Now, You must have started to get confused somewhere, is it because of my complex sentence structure and all those commas? or you have trouble dealing with generalizations and the usage of n, k, p, etc? Or is it because you still didn't get how I derived it.
Do you prefer symbols with pointers and arrows to words?
you're great man, i really appreciate it, but i don't think i have to knowledge to really understand it. i havn't taken any number properties classes nor anything like that. and yes the n r p q thing kinda throw me off, hard to follow. i think the //comments allowed me to understand better.
) so I will go through those after I'm done with the UIL ones. Not sure what I will do after that.
