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samus

#1 h Jul 24, 2004, 7:51 PM • 2 Y

Y by Adventure10, Mango247

Since people may be wondering about the answers to some problems, why don't we put all of the detailed solutions on this thread. Some rules
-let's try to post in the general order of questions
-try to explain in as much detail as possible
-If anyone has any questions, lets post in another thread.
-Try to quote the question.

Here's questions 1-4:
1. A space diagonal of a polyhedron is a line that connects two vertices of the polyhedron that do not lie on the same face. How many space diagonals does a cube have?

(A) 0 (B) 2 (C) 4 (D) 8 (E) 28

Answer: Let us call the cube ABCDEFGH, ABCD being the face closer to us with A: top left corner, B: top right corner, C: bottom right corner, D:bottom left corner; EFGH being the face farther away from us with E: top left corner, F: top right corner, G: bottom right corner, H:bottom left corner. The space diagnolls have to be on opposite faces, left/right positions, and up/down positions. So the space diagnols are AG,BH, EC, and FD. There are 4, so the answer is: (C) 4.

2. Find the units digit of the product of the first 2004 odd primes.

(A) 1 (B) 3 (C) 5 (D) 7 (E) 9

Answer: The first odd prime numbers are 3,5,7,11,17..... The product of the first 3 prime #'s is 105, first 4 primes is 735, and so on. Therefore, from this pattern, we determine that the units dgit of the first 2004 odd primes will be 5. Another way to look at it: 5 times any number of odd #'s (such a 2003 more) will always have a unit digit of 5. Therefore the answer is: (C) 5.

3. The area of the triangle bounded by the lines y=x, y=-x, and y=6 is

(A) 12 (B) 12 :sqrt:2 (C) 24 (D) 24 :sqrt:2 (E) 36

Answer: The three intersection points of these lines are (0,0),(-6,6), and (6,6). This forms a triangle of height 6 (from a point (x,6) to (x,0)) and length (base) of 12 ((6,6)-(-6,6)=(12,0)). The area of a triangle is bh/2, so the area of this triangle is 12*6/2=36. Therefore the answer is: (E) 36.

4. How many odd positive integers between 1000 and 2000, inclusive, have all digits different?

(A) 224 (B) 280 (C) 504 (D) 512 (E) 729

Answer: This is a combanatorics/counting problem. There are 4 digits:_ _ _ _. We know that the thousands digit can only be 1 (2000 isn't odd). There are 4 possibilities for the units digit: one of the odd numbers 3,5,7,9 since 1 has already been used and we need all the digits different. The tens digit can have only 8 possibities from the 10 digits; 1 can't be used since it is used in the thousands digit, and another digit can't be used since it has already been used in the units place. The hundreds digit has 7 possibilities: 1 can't be used since it is used in the thousands digit, and another digit can't be used since it has already been used in the units place, and yet another digit can't be used since it has been used in the tens digit. We already know there is 1 possibility for the thousands digit: the number 1. Therefore the total number of possibilities for this number are 1 * 7 * 8 * 4 = 224. The answer is therefore: (A) 224.

Again: lets keep questions in order and POST YOUR ANSWERS!

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beta

#2 h Jul 24, 2004, 8:23 PM • 4 Y

Y by Adventure10, Adventure10, Adventure10, Mango247

Don't we already have a discussion thread??
5. For how many integer values of x will there be a non-degenerate triangle with distinct side lengths 11, 15, and x?

(A) 18 (B) 19 (C) 20 (D) 21 (E) 22

Solution: By Triangle Inequality, 11+15>x, or 26>x. By Triangle Inequality again, 11+x>15, x>15-11=4. Therefore 26>x>4. The total number of possiblity for x is 26-4-1=21. However, we included 11 and 15 in the 21 values of x. Thus 21-19=2, or B.

6. How many right triangles with integer side lengths are such that their side lengths are in a nonconstant geometric sequence?

(A) 0 (B) 1 (C) 2 (D) 4 (E) Infinitely many

Solution: Call the sides a b and c. Since they are in geom. progression they can be labeled like this: a, ar, and ar^2 respectively. For this to be a right triangle a^2+(ar)^2=(ar^2)^2 must be true.

Factor an a^2 from the left sides and divide by a^2 on both sides and we are left with 1+r^2=r^4. To find r we use quad form. and take the square root of it(when using quad form we find r^2 so we need to take the sq. root). You can find this and figure out it is irrational. But the sides must be integers. With integer sides and an irrational ratio there are no solutions.

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nr1337

#3 h Jul 24, 2004, 8:24 PM • 2 Y

Y by Adventure10, Mango247

Yay number 5!

5. For how many integer values of x will there be a non-degenerate triangle with distinct side lengths 11, 15, and x?

(A) 18 (B) 19 (C) 20 (D) 21 (E) 22

First we look for x such that 11 + x > 15 and x <= 15. Thus x > 4 from our first inequality, so the minimum x is 5. Next we look for 11+15 > x. So 26 > x, and the maximum x is 25. There are 25-5+1 = 21 numbers between 5 and 25 inclusive BUT WAIT! The problem said DISTINCT side lengths! Thus 11 and 15 don't work for x, so we subtract 2. 21 - 2 = 19, thus our answer is B.

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#4 h Jul 24, 2004, 8:40 PM • 2 Y

Y by Adventure10, Mango247

nr1337 wrote:

BUT WAIT! The problem said DISTINCT side lengths!

<Evil cackle> :twisted:

Only 6 people got this right!!!

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joml88

#5 h Jul 24, 2004, 8:41 PM • 2 Y

Y by Adventure10, Mango247

7. 7. A box contains 11 marbles, numbered 1, 2, 3, ... , 11. If 6 marbles are drawn simultaneously at random, and the probability that the sum of the numbers drawn is odd is D/Q, with D and Q relatively prime positive integers, find D + Q.

(A) 3 (B) 17 (C) 331 (D) 346 (E) 349

First note that:
Even+Even=even, Even+Odd=odd, and Odd+odd=even and that we have 6 odds and 5 evens in our set of numbers.

We are going to try to find the number of times that it makes it odd over the total number of times. You can choose 6 from a group of 11 in ${11\choose 6}=462$

Now to find how many ways it will be even we break it into cases.

1st case- 6 evens... this doesn't create an odd

2nd-5 evens and 1 odd...this creates odd numbers. There is one way to get 5 evens and 6 to get an odd making 6 total ways.

3rd-4 evens 2 odd... none

4th-3 even 3 odds... this works. 10 ways to choose 3 odds and 20 for the evens, 200 ways

5th-2 evens 4 odds...none.

6th-1 even 5 odds... works. 5 ways for the evens and 6 for the odds 30.

So we have $\frac{200+30+6}{462}=\frac{118}{231}$ .

D+Q=349 (E)

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#6 h Jul 24, 2004, 8:44 PM • 2 Y

Y by Adventure10, Mango247

7. A box contains 11 marbles, numbered 1, 2, 3, ... , 11. If 6 marbles are drawn simultaneously at random, and the probability that the sum of the numbers drawn is odd is

\frac{D}{Q}

, with D and Q relatively prime positive integers, find D + Q.

(A) 3 (B) 17 (C) 331 (D) 346 (E) 349

Solution: There are 6 odd numbers and 5 even numbers. If the sum of the six numbers chosen is odd, then we have 3 cases:
1)5 odd and 1 even
2) 3 odd and 3 even
3) 1 odd and 5 even.
In case 1), the total number of ways to get the result is (6C5)(5C1)
In case 2), the total number of ways to get the result is (6C3)(5C3)
In case 3), the total number of ways to get the result is (6C1)(5C5)
Thus, the total number of way of getting the sum of 6 numbers odd is
(6C5)(5C1)+(6C3)(5C3)+(6C1)(5C5)=236
The total number of ways to draw 6 marbles simultaneously is 11C6
The probability is 236/11C6=118/231. 118+231=349, the answer is E.

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#7 h Jul 24, 2004, 9:04 PM • 2 Y

Y by Adventure10, Mango247

8. Find the number of integers

such that

k^3 + k^2 + k + 7

is divisible by

k+1

.

(A) 2 (B) 4 (C) 6 (D) 8 (E) Infinitely many

Solution: $k^3+k^2+k+7$ must be divisible by $k+1$ , so divide it to get $\displaystyle k^2+1+\frac{6}{k+1}$ . If it's divisible, that will have to be an integer, meaning $k+1|6$ . From this, we see $\pm(k+1)\in \{1,2,3,6\}$ So there are 8 possible values, 1, 2, 5, 0, -2, -3, -4, -7 for k, and the answer is d.

9. Circle A is inscribed in an isosceles right triangle and circle B is circumscribed about the same triangle. The ratio of the area of circle A to the area of circle B is

(A)

\frac{1}{2}

(B)

3 - \sqrt{2}

(C)

3 + \sqrt{2}

(D)

3 - 2\sqrt{2}

(E)

3 + 2\sqrt{2}

Solution: Without Loss of generality let the length of the leg of right triangle be 1. Thus the length of the hypotenuse is $\sqrt2$ .Since the center of the circumscribed circle of a right triangle is the midpoint of the hypotenuse of the right triangle. The length of the circumradius is $\frac{\sqrt2}{2}$ . Now note that the area of the triangle is $1(1)/2=\frac{1}{2}=rs$ , where r is the inradius, and s is the semiperimeter. The semiperimeter= $\frac{1+1+\sqrt2}{2}$ . Thus the inradius $=\frac{1}{2}/ \frac{2+\sqrt2}{2}=\frac{1}{2+\sqrt2}$ . Inradius: Circumradius= $\frac{1}{2+\sqrt2}:\frac{\sqrt2}{2}=\frac{\sqrt2}{2+\sqrt2}=\sqrt 2-1$ . Ratio of area=ratio of radius^2
= $(\sqrt 2-1)^2=3-2\sqrt2$ , or D

This post has been edited 1 time. Last edited by beta, Jul 25, 2004, 2:33 AM

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#8 h Jul 24, 2004, 9:11 PM • 2 Y

Y by Adventure10, Mango247

Thx to all solution posters.

This is what a getting started problem solver needs.

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joml88

#9 h Jul 25, 2004, 2:47 AM • 2 Y

Y by Adventure10, Mango247

10. A box contains 2 pennies, 4 nickels, and 6 dimes. Six coins are drawn randomly and without replacement. Given that the probability that the value of the coins drawn is at least 50 cents is m/n, with m and n relatively prime positive integers, find m + n.

(A) 961 (B) 1015 (C) 1051 (D) 1056 (E) None of these

The total number of ways to choose 6 is . We are left with finding how many ways to get 50 or over.

We can have
dddddd
dddddn
dddddp or
ddddnn
There is one way to get the first case.

For the second case you can chose the 5 d's in ways and the n in 4 ways(since order doesn't matter). So we have 6*4=24 ways for this case.

For case 3, the d's can be chosen in 6 again. The p can be chosen in 2 which means there are 12 for this case.

For the last case, the d's can be picked in ways and the n's in ${4\choose 2}=6$ ways. This makes 90 ways.

So the answer is 127/924. m+n= 1051 (C)

11. 11. Given that $i^2=-1$ , for how many integers n is $(n+i)^4$ an integer?

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

Expanding the binomial we get
$n^4+4n^3i+6n^2i^2+4ni^3+i^4=n^4+4n^3i-6n^2-4ni+1$

Taking away the parts that we know are integers we are left with:
$4n^3i-4ni=4ni(n+1)(n-1)$
This is only an integer when it is 0. so $n\in \{-1,0,1\}$ . (D)

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#10 h Jul 25, 2004, 4:21 AM • 2 Y

Y by Adventure10, Mango247

Pork_Chop8 wrote:

5. I missed the distinct part. Great trick, evil problem writer.

Hey, I resent that...

Only 6 people out of 29 got that one right.

Also, I think this thread was intended have the solutions in order of the questions...so you may want to edit your post, or move it to the other discussion thread where solutions are not being discussed in order.

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beta

#11 h Jul 25, 2004, 2:32 PM • 2 Y

Y by Adventure10, Mango247

12. A quadrilateral circumscribed about a circle has three of its side lengths 132, 162, and 201. What is the positive difference between the minimum and maximum lengths for the fourth side?

(A) 60 (B) 78 (C) 93 (D) 138 (E) 231

Solution:
Lemma: If a circle can be inscribed in a quadrilateral ABCD, then AB+CD=AD+BC, so AD=AB+CD-BC.
Proof: Let the circle be tangent at AB, BC, CD, AD at W, X, Y, Z respectively.
Thus AW=AZ, BW=BX, CY=CX, DY=DZ. Add all the equations up yields
AB+CD=BC+AD as desired.
Therefore, let AD be the fourth side, AD is at minimum when AB+CD=132+162=294, and BC=201. AD=294-201=93.
AD is at maximum when AB+CD=201+162=363, BC=132. AD=363-132=231.
Thus 231-93=138, or D.

13. For how many primes

is

p^4 + p^5

the square of an integer?

(A) 0 (B) 1 (C) 2 (D) 3 (E) Infinitely many

Solution: First we factor $p^4(p+1)$ .
So we need the square root of this to be an integer. $p^4$ will come out of the square so we are left with finding were $\sqrt{p+1}$ is an integer. In other words we want $p+1=a^2$ where a is an integer.

Subtract by one on both sides and factor to get $p=(a+1)(a-1)$ Wait a minute a prime is only divisible by one and itself. So this could only work when a-1=1. this makes p=3 and 3 is prime so we only have one solution, or B

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#12 h Jul 25, 2004, 2:42 PM • 2 Y

Y by Adventure10, Mango247

14. If

,

, and

are positive numbers satisfying

x + \frac{1}{y} = 4

,

y + \frac{1}{z} = 1

, and

z + \frac{1}{x} = \frac{7}{3}

, then find

xyz

.

(A)

\frac{2}{3}

(B)

(C)

\frac{4}{3}

(D)

(E)

\frac{7}{3}

Solution: $(x+\frac{1}{y})(y+\frac{1}{z}) (z+\frac{1}{x})$
$=xyz+x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xyz}$
$=\frac{28}{3}$

Since
$x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$
$=(x+\frac{1}{y})+ (y+\frac{1}{z})+(z+\frac{1}{x})$
$=\frac{22}{3}$
Sub that in, and let $xyz=k$
$k+\frac{1}{k}+\frac{22}{3}=\frac{28}{3}$
$3k^2-6k+3=0$
$(k-1)^2=0$
$k=1$ , or B.

15. If

\frac{\pi}{2} < \alpha < \frac{3\pi}{4}

,

\cos{\alpha} + \sin{\alpha} = \frac{1}{3}

, and

\left|\cos{2\alpha} + \sin{2\alpha}\right|= \frac{n + \sqrt{p}}{q}

, where

,

, and

are relatively prime positive integers and

is not divisible by the square of any prime, find

n+p+q

.

(A) 34 (B) 35 (C) 36 (D) 37 (E) 38

Solution: $(\sin{\alpha}+\cos{\alpha})^2$
$=(\sin^2{\alpha}+\cos^2{\alpha})+2\sin{\alpha}\cos{\alpha}$
$=1+\sin{2\alpha}$
$=\frac{1}{9}$
Solve for $\sin{2\alpha}$
to get $\sin{2\alpha}=-\frac{8}{9}$
$\cos{2\alpha}=\pm \sqrt{1-(\frac{8}{9})^2}$
$=\pm \frac{\sqrt{17}}{9}$
Because of our restrictions, $\cos{2\alpha}$ is negative
$=-\frac{\sqrt{17}}{9}$
absolute value of $\cos{2\alpha}+\sin{2\alpha}=\frac{\sqrt{17}+8}{9}$
17+8+9=34, or A.

16. If

r_1

,

r_2

, and

r_3

are the solutions to the equation

x^3 - 4x^2 - 4x + 17 = 0

, find

r_1^3 + r_2^3 + r_3^3

.

(A) -68 (B) -4 (C) 0 (D) 61 (E) 68

Solution:First off, rename $r_1=a; r_2=b; r_3=c$ (so that you don't get all the r's confused). Since they are solutions to the equation, the equation $x^3-4x^2-4x+17=(x-a)(x-b)(x-c)$ . Expanding the right side gives $x^3-(a+b+c)x^2+(ab+bc+ac)x-abc=0$ . Therefore, $a+b+c=4; abc=-17$ ; $ab+bc+ac=-4$ . Now take $a+b+c=4$ and multiply both sides by (a+b+c), giving $a^2+b^2+c^2+ab+ac+bc=16$ . Replace $ab+ac+bc$ with -8, swing the -8 to the other side, then multiply both sides by $(a+b+c)$ again. Now you have (with a little bit of simplification):
$a^3+b^3+c^3+a(ab+ac)+b(ab+bc)+c(ac+bc)=96$
Change the equation we already had to ab+ac=-4-bc and put that into the equation we have, then repeat for the other two with the other variables. It now looks like this:
$a^3+b^3+c^3+a(-4-bc)+b(-4-ac)+c(-4-ab)=96$ which is equivalent to this:
$a^3+b^3+c^3-4(a+b+c)-3abc=96$ Substituting from our equations for a, b, and c, gives us $a^3+b^3+c^3-16+51=96$ ; therefore, $a^3+b^3+c^3=61$ and the answer is (D).

17. Find

\sum_{k=0}^{49} (-1)^k  {99 \choose 2k}

.

(A)

(B)

-2^{49}

(C)

2^{49}

(D)

-2^{50}

(E)

2^{50}

Solution:Note that
$(1-i)^{99}=1-{99 \choose 1}i-{99 \choose 2}+ {99 \choose 3}i+{99 \choose 4}.....$
$=\displaystyle\sum_{k=0}^{49} (-1)^k {99 \choose 2k}+i\sum_{k=0}^{49} (-1)^k {99 \choose {2k+1}}$
And $1=-(\sqrt2)(\cos\frac{3\pi}{4}), -1=-(\sqrt2)(\sin\frac{3\pi}{4})$
By DeMorvie's Theorem
$(1-i)^{99}=-((\sqrt2)(\cos\frac{3\pi}{4})+i(\sqrt2)(\sin\frac{3\pi}{4}))^{99} =-(\sqrt2)^{99}(\cos\frac{(99)(3)\pi}{4}+i(\sin\frac{(99)(3)\pi}{4}))$
Thus
$\displaystyle\sum_{k=0}^{49} (-1)^k {99 \choose 2k}$
$=-(\sqrt2)^{99}(\cos\frac{297\pi}{4})$
$=-2^{49}(\sqrt2)(\cos\frac{\pi}{4})$
$=-2^{49}$ , or B.

That took A LOT of Copy and Paste.

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joml88

#13 h Jul 26, 2004, 1:58 AM • 2 Y

Y by Adventure10, Mango247

18. A rising number is a positive integer, each digit of which is larger than each of the digits to its left, and no digit of which is 0. For example, 12689 is a rising number. When all five-digit rising numbers are listed from smallest to largest, the 97th number in the list does not contain the digit

(A) 4 (B) 5 (C) 6 (D) 7 (E) 8

The smallest is 12345. With 4 as the second number we have 5 that work(i.e.12345,12346,12347,12348,12349).
If we keep the first 3 the same and keep increasing the second number by 1 we will get 4, 3, 2 ,and 1 that work for each one. Therefore there are 5+4+3+2+1=15 rising numbers of the form 123_ _ . Continuing in this way we find that there are 4+3+2+1=10 rising numbers of the form 124_ _. 6 with a 5 in the third, 3 with a 6 and 1 with a 7. This makes 35 of the form 12_ _ _. For 13_ _ _ there are 4+3+2+1 + 3+2+1 + 2+1 + 1= 20. For 14_ _ _ we have 10. For 15_ _ _ there are 4. For 16_ _ _ there is 1. This makes a grand total of 70.
For 2 as the first number we do the same thing. We find that 24589 is the 96 so 24678 is the 97. This doesn't contain a 5 so (B)

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#14 h Jul 26, 2004, 3:23 AM • 2 Y

Y by Adventure10, Mango247

Joml, I understood your solution. Thx.

I got one question though. How do you know that 24589 is the 96th number in that pattern?

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#15 h Jul 26, 2004, 5:16 PM • 2 Y

Y by Adventure10, Mango247

kool_dudy wrote:

Joml, I understood your solution. Thx.

I got one question though. How do you know that 24589 is the 96th number in that pattern?

Get as close as you can with the method he used up to 70, and then list the rest.

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Ultimatemathgeek

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Ultimatemathgeek

#16 h Jul 28, 2004, 4:54 PM • 2 Y

Y by Adventure10, Mango247

Thanks every one
More solutions please

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#17 h Jul 28, 2004, 6:29 PM • 2 Y

Y by Adventure10, Mango247

19.

\bigtriangleup ABC

has all side lengths distinct. Two altitudes of this triangle have lengths 4 and 12. If the length of the third altitude is also an integer, find its maximum value.

(A) 4 (B) 5 (C) 6 (D) 7 (E) None of these

Solution: Let the length of AB be a, BC be 1, AC be c. Let the altitude of length 4 and 12 be opposite to AB and BC respectively. Let the length of the third altitude be h. The area of ABC=base*altitude/2 $=\frac{4a}{2}=\frac{12}{2}=\frac{hc}{2}$ . Multiply by 2 yields 4a=12=hc, thus a=3
Since AB, BC, AC must satisfy triangle inequality, c+1>3, c>2. Since hc=12,
h<6. The maximum integral value of h is 5, or (B).

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#18 h Jul 28, 2004, 7:00 PM • 2 Y

Y by Adventure10, Mango247

20. It is given that

(\log_z{3})(\log_x{3})+ (\log_y{3})(\log_x{3}) + (\log_y{3})(\log_z{3}) = 6(\log_z{3})(\log_x{3})(\log_y{3})

and

x<y<z

. If

,

, and

are in a geometric sequence with common ratio 2, and

x + y + z = \frac{m}{n}

, with m and n relatively prime positive integers, find

m + n

.

(A) 61 (B) 62 (C) 63 (D) 64 (E) 65

Solution: Note that $(\log_m{n})(\log_n{m})=1, \log_m{n}=\frac{1}{\log_n{m}}$ . Divide both side of the original equation by $(\log_z{3})(\log_x{3})(\log_y{3})$ we have $\frac{1}{\log_y{3}}+\frac{1}{\log_z{3}}+\frac{1}{\log_x{3}}=6$ , or $\log_3{y}+\log_3{z}+\log_3{x}=6$ ,
$\log_3{xyz}=6$ , which yields $xyz=3^6=729$ . Since x, y, z are in geometric squence of common ratio 2, y=2x, z=2y=4x. xyz=8x^3=729, solve for x yields $x=\frac{9}{2}$ . $x+2x+4x=7x=\frac{63}{2}$ . m+n=63+2=65, or (E).

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Ultimatemathgeek

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Ultimatemathgeek

#19 h Jul 29, 2004, 6:13 AM • 2 Y

Y by Adventure10, Mango247

Where can I find the solutions for the rest of the test? Please, I need this.
I appreciate it

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#20 h Mar 20, 2005, 10:48 PM • 2 Y

Y by Adventure10, Mango247

22. Find the number of distinct pairs of integers (x, y) such that 0<x<y and sqrt{1984} =sqrt{x} +sqrt{y}

(A) 0 (B) 1 (C) 3 (D) 4 (E) 7

you have to factor out 1984 which is 2^6*31
root(1984) simplifies to 8root(31)

x has to be 31 and that multiplied by a perfect square for root(X) and root(y) to have a common factor of root 31

when x is 31*1, the other number is (8-root(1))squared*31 which is 49*31 and is 1519
(31,1519) is another solution

then you find the next perfect square

when x is 31*4, the other number is (8-root(4))squared*31 which is 36*31 and is 1116
(124,1116) is another solution

when x is 31*9, the other number is (8-root(9))squared*31 which is 25*31 and is 775
(279,775) is another solution

you can do this for the next perfect square until you realize that x and y will be the same in that

therefore:
(279,775),(124,1116), and (31,1519) are the 3 ordered pairs that work

the answer is (C). or 3

p.s. sorry for first response, i felt i had to correct myself

This post has been edited 4 times. Last edited by kaycubs, Mar 21, 2005, 3:14 AM

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#21 h Mar 21, 2005, 2:24 AM • 2 Y

Y by Adventure10, Mango247

sorry, i didnt know calculators werent allowed, look back up for the non- calculator solution

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