Hard Math Problems

Test Preparation SAT Preparation
21

<p>For the first one I cleared the denominator to get 68x^8-18x^6-11x^4+1=0, then graphed this in my calculator to see that there are 2 roots (since neither of these are 0 they are roots of the original as well).</p>

<p>If you need a more rigorous solution let me know. Otherwise I have very lttle interest in this problem.</p>

22

<p>For the second one, the worst case scenario would be if the remaining 30 balls were one color, say black. Then there are 90, 90, 90, and 30 balls. So the answer is 4*29+1 = 117, choice (D)</p>

23

<p>Then never mind. The thing is I got this question from my Math Olympiad test and I was not able to use any calculator during the test ugh!</p>

24

<p>Well you can use calculus here - take the derivative (of the expression after clearing the denominator), and factor out x^3. You are left with an equation that is reducible to a quadratic with the substitution u=x^2. After solving this quadratic equation you should find that there are no critical numbers. Using the first derivative test you will find that the function is increasing up to 0 and decreasing after 0, yielding 2 roots.</p>

<p>Hopefully I have this right, since I haven’t actually attempted to do any of this.</p>

25

<p>Ok your help is much appreciated. For the second question I got (D) too; however, I used another method: because black or white balls are 1/9 of green, blue, and red balls, then for every 9 balls, there has to be ( assuming but not necessarily) one black or white ball. The only number that is divisible by 9 is 117. Is this right approach?</p>

26

<p>@DrSteve i aint no math genius but shouldnt Q2 of post #19 be A? Assuming 15 white and 15 black… if we draw 29 red, 29 blue, 29 green, 15 white, 15 black… we have 117 but not 30 of one color… so shouldnt the worst case scenario be 118?? i might be wrong but just asking…</p>

27

<p>“Didn’t quite get it, but here is another problem. Any help will be appreciated.
Let a and b be positive integers such that 7 divides a^2 + b^2. When ab - 1 is divided by7, the remainder is
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6”</p>

<p>I came up with another short and easy answer! Let’s a^2+b^2= 7k then (a+b)^2 = 7k + 2ab. Since (a+b) is divisible by 7, ab must be divisible by 7. Therefore, ab-1 remainder= 6.
Source: my 16 year old brain.</p>

28

<p>"@DrSteve i aint no math genius but shouldnt Q2 of post #19 be A? Assuming 15 white and 15 black… if we draw 29 red, 29 blue, 29 green, 15 white, 15 black… we have 117 but not 30 of one color… so shouldnt the worst case scenario be 118?? i might be wrong but just asking…"
You are correct. you just lack 1 ball since you are supposed to draw one what ever ball to have a 30th ball of the same color.
Source: my 16 year old brain.</p>

29

<p>@JRswish and hamburger</p>

<p>Upon rereading the question you are probably correct. My answer is based on the possibility that there can be 0 white marbles. But the first line does say that there are 5 DIFFERENT colors, so I think you are right.</p>

30

<p>@hamburger</p>

<p>That’s a nice solution to the remainder problem.</p>

31

<p>" Let’s a^2+b^2= 7k then (a+b)^2 = 7k + 2ab"</p>

<p>Why? How did u come up with that @hamburger??</p>

32

<p>Oh wow! I don’t even know why I posted this easy question! Thank you again DrSteve. These are another questions: The number of real roots of the equation
-8x^4 - 18x^2 - 11 + 1/x^4 = 0 is
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4</p>

<p>I do it step by step and it eventually comes out as this:
(x^2+1)^2(-(2x^2+1)(4x^2-1)) DIDNT USE A CALCULATOR! So the answer is 2 since (x^2+1)^2 and (2x^2+1) cannot be 0. 4x^2 -1 = 0 <=> x= 0.5 or x = -0.5</p>

33

<p>A^2 + b^2 +2ab= (a+b)^2 = 7k +2ab. Got it?</p>

34

<p>Anymore problems guys? If not, Im turning in!</p>

35

<p>Yeah, I got 118 for the marbles one too.</p>

36

<p>Here’s the link guys:
<a href=“http://i1327.■■■■■■■■■■■■■■■/albums/u678/hamburger110/706652659efa999c2a524ffab115c1a5_zpsb817a117.jpg[/url]”>http://i1327.■■■■■■■■■■■■■■■/albums/u678/hamburger110/706652659efa999c2a524ffab115c1a5_zpsb817a117.jpg&lt;/a&gt;&lt;/p&gt;

<p>For this </p>

<p>Oh wow! I don’t even know why I posted this easy question! Thank you again DrSteve. These are another questions: The number of real roots of the equation
-8x^4 - 18x^2 - 11 + 1/x^4 = 0 is
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4</p>

<p>I do it step by step and it eventually comes out as this:
(x^2+1)^2(-(2x^2+1)(4x^2-1)) DIDNT USE A CALCULATOR! So the answer is 2 since (x^2+1)^2 and (2x^2+1) cannot be 0. 4x^2 -1 = 0 <=> x= 0.5 or x = -0.5</p>

<p>Source: my 16 year old brain.</p>

37

<p>Haizzz. No one saying thanks?</p>

38

<p>Great! Thank you all, you’ve been very helpful.</p>

39

<p>@ Afaloo :
It’s not an SAT question, but it’s hard though.
There are three kinds of food, and you want to choose 5 dishes from the three kinds of food, so how many choices you have? ( you can choose 5 dishes from one kind of food if you want)</p>

<p>If ABCAB is different than ABABC then the answer is 3^5= 243</p>

<p>Actually it’s not. If ABCAB is not different than AABBC, then it’s way easier: just list them out!</p>

<p>0A, 0B, 5C.
0A, 1B, 4C
0A, …
B runs from 0 to 5.
1A, 0B, 4C.
1A, 1B, 3C.
1A, 2B, 2C.
1A, 3B, 1C.
1A, 4B, 0C.</p>

<p>B runs from 0 to 4.
2A, 0B, 3C
2A, 1B, 2C.
2A, 2B, 1C.
…</p>

<p>Contine doing so!</p>

<p>Then you have the sum =6(since B runs from 0 to 5) 5+ 4 + 3 +2 +1= 21</p>

<p>Source: my 16 year old brain.</p>