Some math problems (not SAT's, but really hard)

Test Preparation SAT Preparation
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<p>These aren’t SAT’s math problems, nor are they homework problems. I started preparing for math team few days ago so I found these really hard math problems I didn’t get, maybe someone here can help. Thanks.</p>

<li><p>What is the maximum integer n such that 3^n is a factor of the product of all the odd inegers between 1 and 200?</p></li>
<li><p>When the binomial (x^2 - 1/x)^6 expanded, one of the terms is a constant. What is the value of that constant?</p></li>
<li><p>Given ABC, with points X, Y, and Z lying on sides BC, AC, and AB respectively. Segments AX, BY, and CZ are concurrent at point O. If BX = 4, XC = 5, CY = 3, YA = 6, and AB = 7. Find the length of AZ.</p></li>
</ol>

<p>By the way, these are from AMTNJ math contest.</p>

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<p>2 isnt there like a way u can expand the binomial? no matter the degree? im too lazy to do that
3. idk wat concurrent indicates lol, but my guess AZ=2? is that a regular triangle =S?</p>

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<p>These questions don't really belong in this forum, CollegeConfidential. In any case, I'll help you with 1.</p>

<ol>
<li>Basically, the question is, if I have 1x3x5x...x199 on the numerator, how many 3's can I have on the denominator so that you get a whole number when the numerator is divided by the denominator? Perhaps there's a way that we can get rid of extraneous numbers. Let's start with a simpler problem: Look at 1x3x5x7x9. What would the maximum integer n be? What if I eliminated 5, so you get 1x3x7x9? Would n be the same? Why? If a number z is not a multiple of 3, and you multiply it by another number p that is not a multiple of 3, would the product be a multiple of 3? Take the number 27. That is 3x3x3. Multiply it by 29 and 31. Does the product of 27x29x31 have a greater factor that is a power of 3? Why? If it has no effect, then is it possible to eliminate any extraneous numbers?</li>
</ol>

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<p>I was just going to say that :P</p>

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<p>so basically then you multiply 3<em>6</em>9...*195 (largest odd multiple of 3 under 200), take the prime factors, and see how many 3's you have there</p>

<p>EDIT: actually 6, 12, etc wouldn't be there since it's not odd :/ sry about that</p>

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<h1>2 is 15</h1>

<p>(x^2 - 1/x)^6 expanded is
x^12-6x^9+15x^6-20x^3-((6)/(x^3))+((1)/(x^6))+15</p>

<p>for #3, if you sketch it out AZ looks like ~5. assuming the answer is in integers.</p>

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<p>Bleh I give up, this is what I have so far.</p>

<p>angle A: 67.1146
angle B: 67.1146
Angle C: 45.7708</p>

<p>AZ has to be larger than 3.5 because the other segments are uneven, thus CZ cannot be perpendicular to side ab</p>

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<ol>
<li>When the binomial (x^2 - 1/x)^6 expanded, one of the terms is a constant. What is the value of that constant?</li>
</ol>

<p>you can also solve this using binomial theorem (HAHA this was just on my trial which I did last week) using T(n+1) = 6Cn*(x^2)^(6-n)(-1/x)^n and then finding where the power of x=0 then you sub that value of n in and you get both the term and the value of the number... uch hard to type but message me if you want me to explain it!</p>

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<p>basically, what you're left with is 3<em>9</em>15*21, </p>

<p>which is (3<em>1)</em>(3<em>3)</em>(3<em>5)</em>(3<em>7)... (3</em>65).</p>

<p>So you're trying to get as many 3's on there </p>

<p>as you can. First let's not worry about how </p>

<p>many 3's would be there for the second factor </p>

<p>within each parenthesis (i.e. don't worry </p>

<p>about (3<em>3, 3</em>9, 3*15, etc), just count how </p>

<p>many 3's parenthesis there are. You'd realize </p>

<p>that there's (65-1)/2+1, or 33 parenthesis. </p>

<p>So now the problem becomes 3^33*</p>

<p>(1<em>3</em>5<em>7</em>9<em>...</em>65). (Now if this is a </p>

<p>multiple choice eliminate anything equal to or </p>

<p>less than 33, haha (; ). Now let's try to </p>

<p>figure out how many 3'd we can pull off inside </p>

<p>of this parenthesis. By using the same </p>

<p>technique, we see that 63= 21*3. So the </p>

<p>parenthesis part becomes (3<em>1)</em>(3<em>3)</em>(3*5)</p>

<p><em>...</em>(3*21). (21-1)/2+1, or 11 total </p>

<p>parenthesis. So it's 3^11<em>(1</em>3<em>5</em>...*21).</p>

<p>See the pattern?</p>

<p>At this step I'd just celebrate by counting </p>

<p>the number of 3's with my fingers xD. It's </p>

<p>(3<em>1)</em>(3<em>3)</em>(3<em>5)</em>(3*7). That's five 3's in </p>

<p>there. </p>

<p>So now add! 33+11+5, or 49. Done!</p>

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<p>ahh notepad screwed up my formatting. </p>

<p>Btw this post has the highest reply:views ratio! Now it's 8:9, but it should have been 8:8, since I refreshed this page once :/</p>

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<p>uh no, i think the views doesnt count anymore =S
if u see other threads, the views are always 1 higher than the posts</p>

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<p>anyone found the answer to #3?
o_o
its kinda irritating not knowing the answer/ how to do it.</p>

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<p>^ #3 when he says ABC is he saying a triangle ABC or just 3 points A,B,C?</p>

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<p>I get AZ as ~5. Rough estimation :D</p>