Mathcounts Marathon - Page 8 - XP Math - Forums

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 10-29-2007 #71 Temperal Guest   Posts: n/a Yeah, a Platonic solid.
 10-30-2007 #72 hunter34 Guest   Posts: n/a My attempt. You know that the surface area of the tetrahedron is $a^2\sqrt{3}$, making the side length $\frac{19}{\sqrt{3}}$. This gives the volume:$\frac{6859\sqrt{3^3}\sqrt{2}}{324}$ You all may laugh at this failing attempt.
 11-01-2007 #73 Temperal Guest   Posts: n/a Next problem?
 11-02-2007 #74 hunter34 Guest   Posts: n/a Wait is that right? If $x+\displaystyle\frac{1}{x}=3$, what is $x^{3}+\displaystyle\frac{1}{x^3}$?
 11-03-2007 #75 Sillysidley   Join Date: Oct 2006 Posts: 822 $3^3=x^3+\frac{1}{x^3}+3(3)$, it's 18. __________________ .
 11-04-2007 #76 Temperal Guest   Posts: n/a Okay, so here's the next one: The 9-digit number abb,aba,ba3 is a multiple of 99 for some pair of digits a and b. What is b - a ?
 11-04-2007 #77 Sillysidley   Join Date: Oct 2006 Posts: 822 Using properties for divisibilities by 9 and 11, we determine a+b is 6 mod 9 b-a is 4 mod 11. Since they are digits, b-a is 4. __________________ .
 11-05-2007 #78 Temperal Guest   Posts: n/a Good. What is the smallest multiple of 24 that is a perfect cube?
 05-13-2008 #79 theredsky Guest   Posts: n/a (x+y)^6 Using Pascal's triangle, we could figure out the problem x^6 + 6*x^5*y + 15*x^4*y*2 + 20*x^3*y*3 + 15*x^2*y^4 + 6*x*y^5 + y^6 Do you notice a pattern?
 05-13-2008 #80 theredsky Guest   Posts: n/a first, you factor 24 = 2^3 * 3 So the next perfect cube would be 2^a multiple of three * 3^3 The answer is 24 * 9, which is equal to 216 next question: WITH FULL SOLUTIONS PLEASE Fractions a/b and c/d are called neighbor fractions if their difference (ad - bc)/(bd) has a numerator of positive or negative 1, that is, ad - bc = positive or negative one. Prove that If a/b and c/d are neighbor fractions, then (a + b)/(c + d) is between them and is a neighbor fraction for both a/b and c/d; moreover, no fraction e/f with positive integer "e" and "f" such that f is less than b + d is between a/b and c/d. [/img]

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