Answer: A

Solution:

The sequence repeats every 4 terms.
How many times will the pattern $1,2,3,4$ occur in the first 2005 terms?
Since 2005 divided by 4 gives a quotient of 501 and a remainder of 1, then the first 2005 terms contain the pattern $1, 2, 3, 4$ a total of 501 times (ending at the 2004th term).
Also, the 2005th term is a 1.
Therefore, the sum of the first 2005 terms is $501(1+2+3+4)+1=501(10)+1=5011$.

Answer: C

Solution:

Since the arrow is pointing between 9.6 and 9.8, it is pointing to a rating closest to 9.7.

Answer: D

Solution:

Since $\sqrt{5+n}=7$ and $7=\sqrt{49}$, then $5+n=49$, so $n=44$.

Answer: E

Solution:

Since the area of the square is 144, each side has length $\sqrt{144}=12$.
The length of the string equals the perimeter of the square which is $4 \times 12=48$.
The largest circle that can be formed from this string has a circumference of 48 or $2\pi r=48$.
Solving for the radius $r$, we get $r=\frac{48}{2\pi}\approx7.64$.
The maximum area of a circle that can be formed using the string is $\pi(\frac{48}{2\pi})^2 \approx\pi(7.64)^2\approx 183$.

Answer: D

Solution:

First, we find the coordinates of the points \(P\) and \(Q\) in terms of \(k\) by finding the points of intersection of the graphs with equations \(y=x^2\) and \(y=3kx+4k^2\).
Equating values of \(y\), we obtain \(x^2 = 3kx+4k^2\) or \(x^2 - 3kx-4k^2 = 0\).
We rewrite the left side as \(x^2 - 4kx + kx + (-4k)(k) = 0\) which allows us to factor and obtain \((x-4k)(x+k)=0\) and so \(x=4k\) or \(x=-k\).
Since \(k>0\), \(P\) is in the second quadrant and \(Q\) is in the first quadrant, then \(P\) has \(x\)-coordinate \(-k\) (which is negative).
Since \(P\) lies on \(y=x^2\), then its \(y\)-coordinate is \((-k)^2 = k^2\) and so the coordinates of \(P\) are \((-k,k^2)\).
Since \(Q\) lies on \(y=x^2\) and has \(x\)-coordinate \(4k\), then its \(y\)-coordinate is \((4k)^2 = 16k^2\) and so the coordinates of \(Q\) are \((4k,16k^2)\).
Our next step is to determine the area of \(\triangle OPQ\) in terms of \(k\).
Since the area of \(\triangle OPQ\) is numerically equal to 80, this will give us an equation for \(k\) which will allow us to find the slope of the line.
To find the area of \(\triangle OPQ\) in terms of \(k\), we drop perpendiculars from \(P\) and \(Q\) to \(S\) and \(T\), respectively, on the \(x\)-axis.

The area of \(\triangle OPQ\) is equal to the area of trapezoid \(PSTQ\) minus the areas of \(\triangle PSO\) and \(\triangle QTO\).
Trapezoid \(PSTQ\) has parallel bases \(SP\) and \(TQ\) and perpendicular height \(ST\).
Since the coordinates of \(P\) are \((-k,k^2)\), then \(SP=k^2\).
Since the coordinates of \(Q\) are \((4k,16k^2)\), then \(TQ=16k^2\).
Also, \(ST=4k-(-k)=5k\).
Thus, the area of trapezoid \(PSTQ\) is \(\frac{1}{2}(SP+TQ)(ST) = \frac{1}{2}(k^2+16k^2)(5k) = \frac{85}{2}k^3\).
\(\triangle PSO\) is right-angled at \(S\) and so has area \(\frac{1}{2}(SP)(SO) = \frac{1}{2}(k^2)(0-(-k)) = \frac{1}{2}k^3\).
\(\triangle QTO\) is right-angled at \(T\) and so has area \(\frac{1}{2}(TQ)(TO) = \frac{1}{2}(16k^2)(4k-0) = 32k^3\).
Combining these, the area of \(\triangle POQ\) equals \(\frac{85}{2}k^3 - \frac{1}{2}k^3 - 32k^3 = 10k^3\).
Since this area equals 80, then \(10k^3 = 80\) or \(k^3 = 8\) and so \(k=2\).
This means that the slope of the line is \(3k\) which equals 6.

Answer: A

Solution:

Since 4 balls are chosen from 6 red balls and 3 green balls, then the 4 balls could include:

• 4 red balls, or

• 3 red balls and 1 green ball, or

• 2 red balls and 2 green balls, or

• 1 red ball and 3 green balls.

There is only 1 different-looking way to arrange 4 red balls.
There are 4 different-looking ways to arrange 3 red balls and 1 green ball: the green ball can be in the 1st, 2nd, 3rd, or 4th position.
There are 6 different-looking ways to arrange 2 red balls and 2 green balls: the red balls can be in the 1st/2nd, 1st/3rd, 1st/4th, 2nd/3rd, 2nd/4th, or 3rd/4th positions.
There are 4 different-looking ways to arrange 1 red ball and 3 green balls: the red ball can be in the 1st, 2nd, 3rd, or 4th position.
In total, there are \(1+4+6+4=15\) different-looking arrangements.

4 red balls, or

3 red balls and 1 green ball, or

2 red balls and 2 green balls, or

1 red ball and 3 green balls.

Answer: C

Solution:

For there to be equal numbers of each colour of candy, there must be at most \(3\) red candies and at most \(3\) yellow candies, since there are \(3\) blue candies to start.
Thus, Shuxin ate at least \(7\) red candies and at least \(4\) yellow candies.
This means that Shuxin ate at least \(7 + 4 = 11\) candies.
We note that if Shuxin eats \(7\) red candies, \(4\) yellow candies, and \(0\) blue candies, there will indeed be equal numbers of each colour.

Answer: C

Solution:

The probability that the arrow stops on the largest section is 50% or \(\frac12\).
The probability that it stops on the next largest section is 1 in 3 or \(\frac13\).
Thus, the probability that the arrow stops on the smallest section is \(1-\frac12-\frac13=\frac66-\frac36-\frac26=\frac16\).

Answer: 09

Solution:

We begin by tracing what happens when \(T=337\).
We start with tokens labelled \(1, 2, 3, \ldots, 335, 336, 337\) arranged around a circle.
We remove the first token (1), move 2 tokens along and remove that token (3), move 2 tokens along and remove that token (5), and continue around the circle until we remove tokens 335 and 337.
This leaves tokens \(2, 4, 6, \ldots, 332, 334, 336\). These tokens differ by 2.

On the second pass, we start with tokens labelled \(2, 4, 6, \ldots, 332, 334, 336\), which differ by 2.
Because the last token was removed on the first pass (337), the first token is not removed on the second pass, which means that we remove every other token starting with 4.
This means that the remaining tokens differ by 4, and are \(2, 6, 10, \ldots, 326, 330, 334\).

On the third pass, we start with \(2, 6, 10, \ldots, 326, 330, 334\), which differ by 4.
Because the last token (336) was removed on the previous pass, we remove every other token starting with 6.
The remaining tokens differ by 8 and are \(2, 10, 18, \ldots, 314, 322, 330\).

On the fourth pass, we start with \(2, 10, 18, \ldots, 314, 322, 330\), which differ by 8.
Because the last token (334) was removed on the previous pass, we remove every other token starting with 10.
The remaining tokens differ by 16 and are \(2, 18, 34, \ldots, 290, 306, 322\).

On the fifth pass, we start with \(2, 18, 34, \ldots, 290, 306, 322\), which differ by 16.
Because the last token (330) was removed on the previous pass, we remove every other token starting with 18.
The remaining tokens differ by 32 and are \(2, 34, 66, 98, 130, 162, 194, 226, 258, 290, 322\).

On the sixth pass, we start with \(2, 34, 66, 98, 130, 162, 194, 226, 258, 290, 322\), which differ by 32.
Because the second last token (306) was removed on the previous pass, we remove ever other token starting with 2.
This leaves \(34, 98, 162, 226, 290\), which differ by 64.

On the seventh pass, we remove starting with the second token, which leaves \(34, 162, 290\), which differ by 128.

On the eighth pass, we remove starting with the first token, which leaves \(162\).

This tells us that the smallest possible value of \(T\) is at least 162 and at most 337.

Next, we will show that \(T = 209\) also gives a final token of 162 by working through the various passes. (This is a place where how we discover the answer is more than likely different than how we justify the answer.)
Before first pass: \[1, 2, 3, \ldots, 207, 208, 209\] After first pass (removing every other token starting with 1): \[2, 4, 6, \ldots, 204, 206, 208\] After second pass (removing every other token starting with 4): \[2, 6, 10, \ldots, 198, 202, 206\] After third pass (removing every other token starting with 6): \[2, 10, 18, \ldots, 186, 194, 202\] After fourth pass (removing every other token starting with 10): \[2, 18, 34, \ldots, 162, 178, 194\] After fifth pass (removing every other token starting with 18): \[2, 34, 66, 98, 130, 162, 194\] After sixth pass (removing every other token starting with 2): \[34, 98, 162\] After seventh pass (removing 98): \[34, 162\] This leaves 162 after the eighth pass, since 34 will be removed.
Therefore, when \(T = 209\) and when \(T = 337\), the final token is 162.

Finally, we show that if the final token starting with \(T\) tokens is 162, then \(T \geq 209\), which will tell us that the smallest value of \(T\) is 209.
Suppose that \(T \leq 209\) and that the token remaining after the final pass is 162.
Before each pass, the remaining tokens differ by a power of 2, since we start by removing every other token from a list that differs by 1, then every other token from a list that differs by 2, and so on.
The smallest powers of 2 are 2, 4, 8, 16, 32, 64, 128, 256.

Since 162 is left after the last pass (this will turn out to be the eighth pass), the remaining tokens must have differed by 128 before the eighth pass, and thus were 34, 162. (Since \(T \leq 209\), then there could not be a token numbered \(162 + 128 = 290\).) Also, if the tokens differed by 64 before the eighth pass, there would have been tokens labelled 34 and 98 that were both removed.
Thus, before the eighth pass, the tokens were 34, 162 and 34 was removed.

Before the seventh pass, the tokens differed by 64.
Thus, these were 34, 98, 162. We note that the last token was not removed on this pass, and so the first token is removed on the eighth pass, as expected.
Also, there cannot be a token numbered \(162 + 64 = 226\), since \(T \leq 209\).

Before the sixth pass, the tokens differed by 32.
Thus, these were 2, 34, 66, 98, 130, 162, 194. The last token cannot have been 162 since the last token must be removed on this pass so that the second token (98) is removed on the seventh pass. Thus, \(162 + 32 = 194\) must be the last token here. (Note that 226 was already rejected earlier.)

Before the fifth pass, the tokens differed by 16.
Since the first token (2) is removed on the sixth pass, the last token is not removed on the fifth pass.
This means that the tokens before this pass were \[2, 18, 34, 50, 66, 82, 98, 114, 130, 146, 162, 178, 194\] On this pass, 178 is the last token removed. (Note that \(194 + 16 = 210\) is too large.)

Before the fourth pass, the tokens differed by 8.
Since the second token (18) is removed on the fifth pass, the last token is removed on the fourth pass.
This means that the tokens before this pass were \[2, 10, 18, \ldots, 162, 170, 178, 186, 194, 202\] The token 202 must be included since 194 remains for the fifth pass. (Note that \(202 + 8 = 210\) is too large.)

Before the third pass, the tokens differed by 4.
Since the second token (10) is removed on the fourth pass, the last token is removed on the third pass.
This means that the tokens before this pass were \[2, 6, 10, 14, 18, \ldots, 186, 190, 194, 198, 202, 206\] The token 206 must be included since 202 remains for the fourth pass. (Note that \(206 + 4 = 210\) is too large.)

Before the second pass, the tokens differed by 2.
Since the second token (6) is removed on the third pass, the last token is removed on the second pass.
This means that the tokens before this pass were \[2, 4, 6, 8, \ldots, 198, 200, 202, 204, 206, 208\] The token 208 must be included since 206 remains for the third pass. (Note that \(208 + 2 = 210\) is too large.)

Before the first pass, the tokens differed by 1.
Since the second token (4) is removed on the second pass, the last token is removed on the first pass.
This means that the tokens before this pass were \[1, 2, 3, 4, \ldots, 204, 205, 206, 207, 208, 209\] The token 209 must be included since 208 remains for the second pass. (Note that \(209 + 1 = 210\) is too large.)

Therefore, we must have at least 209 tokens for the final token to be 162, and so the smallest possible value of \(T\) is 209, whose rightmost two digits are 09.

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Answer: A

Solution:

\(H\) is a perfect square and is \(1\) more than \(D\).
The perfect squares from \(1\) to \(8\) inclusive are \(1\) and \(4\). Since \(H\) is one more than \(D\), \(H\) cannot be equal to \(1\) (since \(D\neq0\)), and so \(H=4\) and \(D=3\).
\(B\) is the largest prime number in the set, and so \(B=7\).
\(C\) is a multiple of both \(G\) and \(D\). Since \(D=3\), then \(C=6\) and \(G\) is equal to either \(1\) or \(2\) (since \(6\) is a multiple of each of these).
The value of \(B+G\) is even, and since \(B=7\), then \(G=1\).
The letters which have not been assigned values are \(A\), \(E\) and \(F\), and the integers which have not been assigned are \(2\), \(5\) and \(8\).
Since \(5\) and \(8\) are in the same row and \(A\) and \(E\) are the two remaining unassigned letters that are in the same row, then \(A\) and \(E\) are \(5\) and \(8\) in some order, which leaves \(F=2\).