Assorted questions
OK, these are just some assorted questions that I have, rather than putting many posts, I am putting them all in one as they aren't very difficult, but please post HOW you did it, I want more methods to solve them :)
1) p(x) is an expression. and it satisfies, 2p(x) + p(1/x) = x+1. find p(2)
2) Find the Sum of digits, and number of digits, of the number: (1000^20) -20
3) Find the last integer of the number (((2)^2)^n) + 1 and prove it. only for n>1(not using trials :P )
4) x = (-1 (+/-) rt(-3))/2 find the value of (x^1999 + x^2000) = ?
5)a+b+c = 0 ; (b^2 + a^2 + c^2) / (b^2 - ac) = ?
All of these are 10th grade problems, so should be easy.
Labels: logic, mathemagic
posted by Vago at
9:09 PM
30 Comments:
Hi Vago. I see that (in "Badly made wagons" I have just been preaching to the converted). I won't modify the comment as I was writing in full awareness that others would see it too. I want to encourage the publishing of the method (especially if it is elegant), not simply an uninteresting result.
Has anyone got a 10th grader going spare?
problem 1: 2p(x) + p(1/x) = x+1
Let x -> 1/x => 2p(1/x) + p(x) = 1/x + 1
=> p(1/x) = 1/(2x) + 1/2 -p(x))/2
Substitute into the original equation and doubling
=> 4p(x) + 1/x + 1 - p(x) = 2x + 2
=> p(x) = (2x + 1 -1/x)/3
Tring x = 1 in this and original equation => p(1) = 2/3 (good)
So p(2) = 3/2
I don't recall ever having done a problem quite like that before.
Thank you Vago.
Problem 2: 1000^20 - 20 = 10^60 -20. This has 60 digits.
10^60-20 = 999...99980. So the sum of the digits is 58*9 +8 = 530
I should have done problem 1 a little differently; I should have stuck with 2p(1/x) and made the written form easier on the eye.
Problem 5: a+b+c = 0
=> (a +b +c)² = (a²+b²+c²) +2(ab +bc +ca) = 0
(a²+b²+c²) +2[b(a+c) +ca] = (a²+b²+c²) + 2[b(a+b+c) +ca -b²] = 0
=> (a²+b²+c²) + 2(ca -b²) = 0
=> (a²+b²+c²) = 2(b² -ac)
=> (a²+b²+c²)/(b² -ac) = 2
The other two problems will have to wait. I've got to go to bed - it's 6:42 am here.
Thank you Vago. I'm enjoying these.
Sorry Vago, I should have written those with the more symmetrical forms e.g. (b²+a²+c²)/(b² -ac) = 2
1. can be displayed as a function: p(x)= (2x^2 + x - 1)/ 3x, if p(2) then (8+2-1)/6, 9/3, 3.
2. 1000^20, (10^3)^20, 10^60, and that means a one with 60 zeroes behind it (61 digits), thus -20 it would have 60 digits. sum of digits = (9 * 58) + 8 + 0, 530. If 60 digits and last two are 8 and 0 then 58 are 9.
3. unsure of question: if (2)^2^n thus 2^2n or 2^(2^n. If the first it is impossible all you know is that it is odd. If the latter, it is 7 if n > 1... any even number to an exponent is even and if you add one it is odd. Proof of it is 7: nvm too lazy, jut sumbld the page and was bored
Couldn't resist
Problem 3: I assume you meant 2^(2^n) + 1
For n > 1, the last digit is 7. Sadly I did that by trial and error. I already new (after reading about Graham's number) that the last digit(s) would eventually become fixed as n increased. I'll try better tomorrow, if no-one has done it.
2^(2^n) represents the largest number representable in 2^n bits. When n is 2, 2^n is 4; four bits represent one hex digit. Every higher value of n is an even number of hex digits. Therefore every value of 2^(2^n) where n is greater than 1 is a power of 16.
16 mod 10 is 6. (16 to any power) mod 10 is still 6. Hence any value of 2^(2^n) where n>1 ends in 6.
One more than that, therefore, ends in 7.
Rephrasing that first paragraph in a slightly different, perhaps better way:
When n=2, 2^n = 4 and 2^(2^n) = 16.
When n>2, say n=2+m, 2^n = 2^(2+m) = 2^2 * 2^m = 16*2^m, which is a multiple of 16; hence 2^(2^n) is a power of 16.
ok, these are answers, and comments on comments according to numbers:
1) Yup chris, i normally dont care abt the answer, i prefer finding methods. But i thought ur wagons questions was easy, and I thought some1 else would post their answer (i am really lazy... i agree :))
2) Yep, im spare, in 10th, 14yo, so i can only give u 10th grade ones :)
3) Absolutely correct, but rather than getting a general formula, i substituted x = 2, making it a 1 line, more easier to write, and to see.
4) It was too easy.. i just wanted to make it 5 :)
5) nice, that is easier than my proof.. mine goes like 7-8 steps.
6) I did tht, cuz i typed b by mistake b4 a, didn't wanna erase
7) heya, Anon (phew Chris, 6 comments back to back!) sorry, first is wrong,2nd right :) NOTE: about the first, u made a mistake, u wrote (8+2-1)/6 = 9/3
u hanged the 6 to 3, hence the whole typo. But answer is correct, just a typo
8) and onwards...IN MY NEXT COMMENT
8) Ross beat u :)
9) Great Ross, I couldn't have phrased as good as this!
10) refer above :)
11) Looooong comment :P
I am happy to see that Chris got interested enough to spend 1 hour or so on this :)
Also, happy to see that nobody has solved the 4th one. it is the toughest of the lot, my friend used 11th and 12th grade math for it, i cant understand any of it... but later, I got a 3 line proof, which even a person who studied algebra for 1 week wud understand...
PS: I forgot about n>1 and i have edited that question.
Prob 3. Of course (doh! to me), any number ending in 6 multiplied by any number ending in 6 also ends in 6. You don't need all those ()s, 2^(2^n) +1, disambiguates completely.
Prob 1. Anonymous just plucked that equation out of nowhere - I don't call that a solution.
Prob 4. Although I've found a neat expression or two, I still don't know how to finish it. That the result is an extremely small number is all I know. Please don't publish the solution yet. I hope your friend did it nicely.
I'm going to look at it now.
Again, thanks for these problems.
Problem 4: x = (-1 + (+/-)√3)/2
=> x² = (1 - (+/-)2√3 +3)/4 = (4 - (+/-)2√3)/4
=> x² = -(-2 + (+/-)√3)/2 = -(-1 + (+/-)√3)/2 +1/2
=> x² = 1/2 -x
=> x(1+x) = 1/2
Let n = x^1999 + x^2000
Then n = x^1999 (1+x) = x^1998 x(1+x) = (x^1998)/2
Aaargh, I can't think of a sweet next move.
So I'll stick the above into the Windows calculator, hey presto!
n ≈ 3.9301067798788005632078545956483 * 10^(-873)
If had used a lesser calculator I'd have done:
log(n) = 1998 log(x) - log2 ≈ -872.4056 ≈ -873 + 0.59440
So n ≈ 10^log(n) ≈ 3.9301 * 10^(-873)
Three liner eh. I shall burn some more grey matter up.
...ooops, I forgot the other root =>
n ≈ 2.2161764 * 10^270. I hope.
I've not given up on a nice solution, yet!
I'm beginning to think that I've done it after all. Except for the skipped obvious details of the other value.
I'll chew this over subconsciously. I don't think that I'm going to do any better, though. I'm happy for you to post your solution now.
umm.. Chris, u have copied it wrong, the question is root of -3 u have put it as +3.
HINT: dont think of it as what i put up, change it to a quadratic equation, and then it will be easier.
Also, how r ur answers "extremely small numbers"??
but yes, HINT2: the answer is smaller than 10 :)
oh, about the answer which is not small.. i meant the 10^270 one.
Also, chris, plz tell me how u use root and square symbol? will b really useful for me.
And, thirdly, is my answer for ur earlier question (in days) correct???
Hi Vago. When there's a new face here, I'm not sure if they have a "nice" solution when they simply publish the result.
I sometimes simply publish a result in order to give others a chance of strutting their stuff and having a target (trouble with that is I'm sometimes slipshod, and publish an incorrect value, even though I used a good method). Perhaps that's arrogance on my part. It's difficult to get the balance right. If I regard a problem as being difficult, then I'll post my reasoning pretty quickly.
I've definitely got some mental blocks on some of the problems. Then, I'm very disappointed when someone publishes a verifiably correct solution, but doesn't show their method.
I also give full credit to someone who publishes an incorrect value, if the method was nice. That wouldn't do if they were really designing a bridge though ;)
In short, I don't (usually) give a monkey's about the final values, I only care about the methods used. That's why I'm not too keen on brute-force methods - I think of it as cheating. Even so, I do like seeing the code, especially if it is nice.
ok, im sorry if I irritated you in anyway Chris, but i think of final values as confirmation that u have done it right.
but just a confirmation, the method for the fourth one is still on, Also, please do post various methods (if u have an easier one) for the other problems.
PS: I shall not b able check for round 10-12 hours.
PPS: again Chris, sorry.
I keep these in a file so I can copy/paste:
¹²³αβγδΔεζηθξρσφλμπωΣΩΓ»«º☺♠♣♥♦♪♫ћ√∞≈≠≡≤≥‖│ ¼ ½ ¾ ⅓ ⅔ ⅛ ⅜ ⅝ ⅞ ₁ ₂ ₃ ₄
The last 4 subscripts don't behave very nicely.
I got most of them from the Windows character map appplet.
The x values are cube root of unity. Using the facts that
1+x+x^2 = 0-----1
x^4 =1 -----2
x^1999 + x^2000 =
x^1999(1+x)=
x^1999(-x^2)(from-1)=
-x^2001=-x^2000 * x=-1*x(from-2)=-x=
1 (-/+) rt(-3)/2
This post has been removed by the author.
This post has been removed by the author.
Deletes due to typos and merging two posts.
Thank you Anonymous, I didn't see the "-" sign under the rt.
No wonder I wasn't making the expected progress.
I quite liked what I had done though.
x = (-1 + (+/-)i√3)/2.
Letting ω = (-1 + i√3)/2 and ω² = (-1 - i√3)/2
ω³ = 1 {i.e. ω is a cube root of unity}
We do get 1 + ω + ω² = 0 => ω + ω² = -1
ω^1999 + ω^2000 = ω^1998 (ω + ω²) = ω^1998(-1) {from a few lines back}
Also ω^1998 = (ω³)^666 = 1^666 = 1
=> ω^1999 + ω^2000 = -1
But ω² is simply the conjugate of ω => same result for ω²
=> x^1999 + x^2000 = -1
Hi last Anonymous. Sorry about stealing you thunder (if I did).
Hi Vago. I've been re-reading the comments. You haven't done anything to irritate me. I think that we may be slightly misunderstanding each other. I'm sure we're going to get along fine together. You presented a good variety of nice problems and it gives me some idea of what you might like me to post.
Hi Vago. Been re-reading again. Re your comment 3). I nearly always substitute late. On this occassion I was curious about what px) looked like. If I substituted early, I'd have lost that.
6) I prefer the b a c ordering.
My long comment Feb 3 7:56 AM, wasn't a response to anything that you said. I was just stating my views so that you knew me better (and to give a hint to other posters).
Of course, getting the right answer is desirable; and it can let you speed read a long explanation.
I can't imagine how I missed your comment about missing the "-" sign in problem 4. Fortunately the last Anonymous made me notice my error before you posted your solution.
your answer is absolutely right, but i like my method, cuz i don't know the cube roots of unit at all, an the aren't 10th, they are 12th
MY method (the 3 line thing)
ok, so the actual question is if x²+x+1 = 0 then... i changed it cuz i thought it would be more trickier :)
so, x²+x+1=0 --------1
multiplying with(x-1)
=> (x-1)(x²+x+1) = 0(x-1)
=> (x-1)(x²+x+1) = 0
=> x³ - 1³ = 0
[there is an identity as such]
=> x³ = 1 ---------2
ok, now,
x²ººº + x^1999
= x^1998 ( x² + x)
= (x^3)^666 ( x² + x)
= (1) (-1).............[from 1 and 2]
=> the answer is -1
ok, this may not seem as 3 steps here, but I have extremely extended it here, so that people may understand it more better.
The normal method I do it is mental, for me it was 3 steps.
1)multiply by (x-1)
2)take x^1998 common
3)substitute, and Voila! the answer!
ok, i thought that i wud try to see the omega stuff u wrote, and try to understand, and now i see that the final steps are exactly the same, like u wrote omega instead of x.
But i still prefer my method, as i dont know cube thing, but according to the earlier values of the omegas, it seems like they are all the values of x³ - 1. but again, the only thing I know about imaginary numbers is:
A) it is imaginary.....
B) they are all like root(-something)
C) they occur in conjugate pairs in polynomial equations.
Here's a picture od the three roots of unity:
http://en.wikipedia.org/wiki/File:3rd_roots_of_unity.svg Note W is being used rather than ω.
If you've done vectors, you should immediately realise that adding the three roots => 0.
The nth roots f unity have a similar pattern: n equally spaced radial lines of length 1. If w is the principal root, then w^0, w^1, w^2, w^3,... w^(n-1) is all the roots. The sum w^0 +w^1 + w^2 +...+w^(n-1) = 0 and w^0 = w^n = (w^m)^n = 1.
I typically try to avoid using quantities that are multi-valued, hence turned the problem into two parts, one for each root. x is multi-valued, ω is single-valued.
I have a degree in electronics (I'm now 57) and I take that sort of stuff for granted.
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