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useless


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I rocked test #3
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Coming into this test, given the ungodly hype, I was rather scared. But, as it turns out, I scored an all time high on the test with a 202. 62 attempted and 12 wrong...  I must say that I am rather proud, but still, as always, I have a few questions....


#22) .03999 = ___________


I got 1/25, but that equals .04. Should I assume since .399999... is infinately close to .04 that it is .04? Also, there are no elipses in the written problem, so I can see that some people would write 3999/100,000.


#31) 12^2 + 84^2 = _____________


I approached this with sheer brute arthematic. 144 + 7056 =7200. Apparently, there is some other trick that you are trying to emphasize here, especially since its not in the form we discussed last time.


(I've never seen a scenario when it was absolutly neccesary to use the double and half method before, I thought #34 was very clever)


#36) 5  2/5 * 55  2/5 = _______________ 


I did 5 2/5 * 5  2/5 * 11 = 29 4/25 *11 = wrong answer (b/c 11 * 5.4 is not 55.4 IDIOT)... How do we do this one?  (i also tried (((5 + 55)/2)*2/5) + 55*5 + 4/25 and thats off too.)


#50) 46^3 = __________________


I did 50^3 and then arbitrarily subtracted a lot and got it wrong. Whats the best way to approach this one?


#53) 143342 (base 5)  / 4 has a remainder of ________________


Another guess, another failure. How does this one work?


#56) 1^2 + 2^2 + 3^2 .... + 10^2 = _______________


I know how to do the cubes, but not the squares. Any ideas?


#60) 143342 (base 5) = ______________ (base 10)


I was thinking somethinking like   3,125 + blah. I gave up there and wrote 6800 and lost. Faster and more accurate method I need!


#66) 7^11 MOD 11 = ___________


I tried to look for a pattern but that did not work out too well. I looked at the units digit : 7^1 = 7 (r 4) / 7^2 = 9 (r 5) 7^3 = 3 (r 2) 7^4 = 7 (r 3)  7^5 ( r 10) It would've worked had I been able to do this in my head quickly. Is there any other way?


#67) If (20,-21) = (r, theta) then r = _____________


No clue here. Please explain this. Is this another polar coordinates question?


 #74) 11/15 - 14/17 = ________________


Trick?


#75) The dot product of vectors (2,4) and (4.-5) is ____________


How do dot products work?


#79) 6^3  + 4^3 + 2^3 - 5^3 -3^3 -1^3 = _________


The sum of the even cubes minus the sum of the odd numbers any trick here?


#80) pi^5 + e^5 = __________________


No clue again.


Thats it!


- Zack -



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Anonymous

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Congrats! 202 is a lot better. Everybody's gaining on scores except me.  I just want to know how bad would I be without practice. If you've receive my mail, Zack, I was only 6 points higher and I didn't even get to #60.


#31. n^2 + (7n)^2 = 50n^2.


#36. Foil it. 5 2/5 * 55 2/5 = 5*55 + (5+55) * 2/5 + 4/25 = 275 + 24 + 4/25 = 299 4/25; or 5.4 * 55.4 = 5.4*50 + 5.4^2 = 270+29.16 = 299.16


#50. If you haven't memorized 23^3=12167, I really don't know what to do. I believe Mr. Jones gave an account on this one on the previous post.


#60. The remainder of any number in base m divided by m-1 is that of the sum of the digits of the number divided by m-1. Therefore, find the remainder when (1+4+3+3+4+2)/4.


#66. This question would take you less than a second if you realize that it's a 10-term cycle and the 11th term is the same as the first term 7=7 mod11.


#67. It is polar. Imagine a triangle with legs 20 on the x-axis and -21 on the y-axis, find the length of the hypotenuse, which is 29. The Pythagorean triple is a bit unusual -- (20, 21, 29).


#74. No trick that I know of. However, the multiplication is quite easy.


#75. For vectors u(x1, y1) and v(x2, y2), u dot v = x1x2 + y1y2. Therefore, the answer is 2*4+4*-5 = -12


#79. Mr. Jones also gave the formula on the last post.


#80. pi^2 is about 10, and e^3 is about 20, e^2 is about 7.3 -- these information should be sufficient enough to solve the problem. 


Best Wishes



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fht


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Sorry, I jumped around a bit...


#60's solution should be #53's.


#56. Use the formula: n(n+1)(2n+1)/6, where n=10.


#60. 200000base5 = 6250base10 should be good enough.


Best Wishes



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vvr1590


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haha,


I tried doing that on number 60, except I was stupid and still thinking in base 10 so i got really confused...


my logic was, 1*5^5 - 5*5^4 (because of the 14...,) which made me think, wouldn't that be zero? so I screwed that one up


 



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Vinay Ramasesh
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#36) See here.
 
#66) As long as you have a^b divided by b, the remainder is always a if a and b are relatively prime to each other.  So in the case of 7^11 / 11 we know the remainder is 7.
 
So 12^17 divided by 17 = 12
or 14^53 divided by 53 = 14
so on
 
#50) I didn't think to do 23^3 x 8.  Pretty ingenious way of doing that problem.
 
#74) Cross-multiply and subtract.  That's all there is to this one.

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Anonymous

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22) More simply, 0.999...=1. Exactly, no rounding. This is because the real number line is continuous. There exists a continuum of real numbers between any two distinct points on the number line. Between 0.999... and 1, there is no number (0.999... repeating faster than normal?). So, we say they are equal.

I used to ask what 1/3 as a decimal was. The answer's always 0.333... Now, multiply that by 3. Are you going to multiply 1/3 by three or 0.333... by 3? Still, this leaves an empty feeling that asks "where did that little infinitesimal part go?" Well, Leibniz said infinitesimals can be regarded as 0 in addition. Whatever.

56) Notice that n(n+1)(2n+1)/6=(t_n)(2n+1)/3, where t_n is the nth triangular number.

66) Mr. Jones is trying to refer to Fermat's Little Theorem. a^b divided by b has remainder a if b is PRIME. a and b don't have to be relatively prime. There is a generalization of Fermat's that is extremely good to know, the Euler Totient Theorem. If a and b are relatively prime, a^phi(b)=1(mod b). The totient function, phi(b), counts the numbers relatively prime to b (between 0 and b).

Okay, you can use this to do problems like

Question: 5^32 divided by 12 has a remainder of...

Solution: GCD(5,11)=1. phi(12)=(2^1)(3-1)=4. 5^(32-4*8)=1, remainder 1.

Question: 7^57 divided by 15 has a remainder of...

Solution: GCD(7,15)=1. phi(15)=(5-1)(3-1)=8.
7^57=((7^8)^7)*(7^1)=(1^7)*(7) (mod 15). Remainder 7.

Last Question: 4^11 divided by 13 has a remainder of...

Last Solution: GCD(4,13)=1. phi(13)=13-1=12. 4^(11-12)=4^(-1). Uggh, -1! That sucks! Here's how. 4^11=x(mod 13). 4^12=4x=1(mod 13). Go through until you find a multiple of 4. 13+1,26+1,39+1. x=10.

67) Oh, I just realized. This isn't germane. Oh well. You can skip it.

Become familiar with all those Pythagorean triples whose longest leg is one less than the hypotenuse. Amazingly, you can produce ALL possible triples with this...

a=2nm
b=n^2-m^2
c=n^2+m^2
{integers 0

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Anonymous

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Oops, 11 is supposed to be 12 on the first solution under 66.

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