Official Mock AMC C Solutions

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JSRosen3

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JSRosen3

#1 h Aug 3, 2004, 9:25 PM • 2 Y

Y by Adventure10, Mango247

Let's not post solutions until beta says it's OK. Maybe after he's posted the results.

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joml88

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joml88

#2 h Aug 3, 2004, 9:28 PM • 2 Y

Y by Adventure10, Mango247

1. Since the three values that are squared must be $\geq 0$ (since they are squared) the only way the whole thing can equal 0 is if each of those things squared is 0. So we have

$\begin{eqnarray*} a^2-1 &=& 0 \\ b^2-4 &=& 0\\ c^2-9 &=& 0 \end{eqnarray*}$

So $a=\pm 1, b=\pm 2, and \ c=\pm 3$ So there are 2*2*2=8 ordered pairs. D

2. For this one note that $1+3+5+\ldots+(2n-1)=n^2$ So we then have it equaling 21^2=441 C

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ThAzN1

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ThAzN1

#3 h Aug 3, 2004, 9:35 PM • 2 Y

Y by Adventure10, Mango247

3. How many ordered pairs (x,y) satisfy x+4y=2004?

Note that y>0, and x>0 implies y<501. Thus y=1,2,...500 yield all of the solutions and there are 500 pairs

4. A charity sells 140 benefit tickets for a total of 2001 dollars. Some tickets sell for full price (a whole dollar amount),
and the rest for half price. How much money (in dollars) is raised by the full-price tickets?

Set up an equation... let x = number of full price tickets, d = full price of a ticket.

We have,

xd + (140-x)d/2 = 2001
2xd + 140d - xd = 4002
xd + 140d = 4002
d(x+140) = 4002

Since x+140|4002 and 140 < x+140 < 280, we factor 4002=2*3*23*29, try some numbers, and conclude that x=2*3*29=174, so x=34, d=23, xd=782.

This post has been edited 3 times. Last edited by ThAzN1, Aug 3, 2004, 11:09 PM

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dts

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dts

#4 h Aug 3, 2004, 11:05 PM • 2 Y

Y by Adventure10, Mango247

Rep123max wrote:

23. Basically, we have to get $\cos5\theta$ in terms of $\cos\theta$ and $\sin\theta$ .

By DeMoivere's, we have:
$\cos5\theta+i\sin\theta=(\cos\theta+i\sin\theta)^5$ .

So basically, $\cos5\theta$ will be the real parts of $(\cos\theta+i\sin\theta)^5$ .

When we expand it, the real parts are $\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta$ .

We know that $\cos\theta=\frac{1}{4}$ and by the Pythagorean Identity, $\sin\theta=\sqrt{\frac{15}{16}}$ .

Subbing this into what we got above, it turns out to be $\frac{61}{64}$ .

Because calculator use is permitted on the AMC, there is an easier way: evaluate Arccos(1/4), multiply by 5, and take the cosine of that

. Not very elegant, but probably quicker.

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joml88

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joml88

#5 h Aug 3, 2004, 11:09 PM • 2 Y

Y by Adventure10, Mango247

11. $\cot{10^\circ}+\tan{5^\circ}=$

Express cot and tan in terms of sin and cos:

\[\begin{eqnarray*}
&=& \frac{\cos{10}}{\sin{10}}+\frac{\sin{5}}{\cos{5}}\\
&=& \frac{\cos{10} \cos{5}+\sin{10}\sin{5}}{\sin{10}\cos{5}}\\
&=& \frac{\cos{5}}{\sin{10}\cos{5}}\\
&=& \frac{1}{\sin{10}}\\
&=& \csc{10}\\
\end{eqnarray}\]

E is the answer.

14. Using change of base we get:

$\log_{100!}{2}+\log_{100!}{3}+\ldots+\log_{100!}{100}$

which by the property of adding logs equals:

$\log_{100!}{100!}=1$

C

16. I liked this one.

Square both the equations we are given to get:

$\sin^2{x}-\sin^2{y}-2\sin{x}\sin{y}=\frac{1}{9}$

and

$\cos^2{x}-\cos^2{y}-2\cos{x}\cos{y}=\frac{1}{4}$

Adding them and using the fact that $\sin^2{x}+\cos^2{x}=1$ and the cosine of the sum of angles formula we get(and moving a few terms):

$\cos{x-y}=\frac{59}{72}$
A

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interesting_move

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interesting_move

#6 h Aug 3, 2004, 11:10 PM • 2 Y

Y by Adventure10, Mango247

Rep123max wrote:

23. Basically, we have to get $\cos5\theta$ in terms of $\cos\theta$ and $\sin\theta$ .

By DeMoivere's, we have:
$\cos5\theta+i\sin\theta=(\cos\theta+i\sin\theta)^5$ .

So basically, $\cos5\theta$ will be the real parts of $(\cos\theta+i\sin\theta)^5$ .

When we expand it, the real parts are $\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta$ .

We know that $\cos\theta=\frac{1}{4}$ and by the Pythagorean Identity, $\sin\theta=\sqrt{\frac{15}{16}}$ .

Subbing this into what we got above, it turns out to be $\frac{61}{64}$ .

Why is $\cos5\theta+i\sin\theta=(\cos\theta+i\sin\theta)^5$ .?

-interesting_move

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dts

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dts

#7 h Aug 3, 2004, 11:12 PM • 2 Y

Y by Adventure10, Mango247

#10: Because $\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\sqrt{a}-\sqrt{b}}{a-b}$ , the expression becomes $\sqrt9-\sqrt8+\sqrt8-\sqrt7+\sqrt7-\sqrt6+\sqrt6-\sqrt5+\sqrt5-\sqrt4+\sqrt4-\sqrt3=3-\sqrt3$ , and we are done.

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ThAzN1

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ThAzN1

#8 h Aug 3, 2004, 11:12 PM • 2 Y

Y by Adventure10, Mango247

Shouldn't it be

\cos 5\theta + i\sin 5\theta = (\cos\theta + i\sin\theta)^5

?

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joml88

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joml88

#9 h Aug 3, 2004, 11:17 PM • 2 Y

Y by Adventure10, Mango247

I fixed it. I was having a lot of trouble getting the eqnarray to work. I still have to get used to using that

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white_horse_king88

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white_horse_king88

#10 h Aug 3, 2004, 11:34 PM • 2 Y

Y by Adventure10, Mango247

Number 7 - The Tricky One.

Realize that you form a 30 degree angle when you turn 150 degrees left, so you have a triangle with sides 3 and rt3, but that is it. The other side can be rt3, or it can be 2rt3 depending on whether or not the triangle is a 30-60-90 or a 30-120-30. So (E). Undetermined.

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white_horse_king88

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white_horse_king88

#11 h Aug 3, 2004, 11:40 PM • 2 Y

Y by Adventure10, Mango247

Number 9) Find the unit digit of the sum

1 + 2^5 + 3^5 + ... 2005^5

.

Notice, that the continuation pattern for the units digit is at the most in groups of 4, so the 5th will always be the original digit. So, the last digit can be determined by 1+2+3+...2005 = 2005(2006)/2 = ... 5. The last digit is 5, or (C).

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beta

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beta

#12 h Aug 4, 2004, 12:13 AM • 2 Y

Y by Adventure10, Mango247

15)
Let the tetrahedron be ABCD, with BC and AD opposite, and let their midpoints be M and N, respectively. Note that $\Delta ABC \Delta BCD$ are equilateral triangles. Therefore length of $AM=\frac{\sqrt3}{2} =BD$ . Point M must lie on the perpendicular bisector of AD, so $MN \perp AD$ , ANM is a right triangle, $AN=\frac{1}{2}$ . By Pythagorean Theorem, $MN=\frac{\sqrt2}{2}$

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white_horse_king88

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white_horse_king88

#13 h Aug 4, 2004, 1:37 AM • 2 Y

Y by Adventure10, Mango247

Aren't you the Mock AMC C creator? Why are you posting the solutions - just wondering.

For Number 13) To find the number of digits in 5^89, you take log_5^89. You know that log_2 = 0.3010, and you know that log_2 + log_5 = log_10 = 1, so log_5 = 0.699. log_5^89 = 89(log_5) = 89(0.699). This is if you don't have a scientific calculator on you. You get 62.something, and since it is 10^62.some, you have 63 digits. If you have a scientific calc, you can just plug in the numbers.

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beta

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beta

#14 h Aug 4, 2004, 1:49 AM • 2 Y

Y by Adventure10, Mango247

white_horse_king88 wrote:

Aren't you the Mock AMC C creator? Why are you posting the solutions - just wondering.

Why can't I post the solutions

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