1. The first exam has 25−12 = 13 problems not on the second exam, and the second exam has 25−12 = 13 problems not on the first exam, so the total number of problems is 12+13+13 = 38.

2. Let `x` be the unknown side length. If the unknown side is the longest, then `x<8.1+1.4 = 9.5`; if the side of length 8.1 is the longest, then `8.1<x+1.4 => 6.7 < x`. So `6.7<x<9.5` ` ` and the only even integer in this interval is 8.

3. Subtract the first equation from the second to get `3ex+3ey=3e => x+y=1 => x = 1-y`. Substitute `x=1-y` ` ` into the first equation to get:

`pi (1-y) + (pi+e)y = pi+2e => ey = 2e => y = 2 => x=1-2 =-1` ` `

Thus `a=-1` and `b=2` so `b-a = 2-(-1) = 3`.

4. Factor 2014 = 2×19×53, reduce each factor by 1 to get 1, 18, 52, but 18 is not a factor of 52.

Factor 2015 = 5×13×31, reduce each factor by 1 to get 4, 12, 30, but 12 is not a factor of 30.

Factor 2016 = 2×2×2×2×2×3×3×7, reduce each factor by 1 to get 1, 1, 1, 1, 1, 3, 3, 6, and check that 1 is factor of 3 and 3 is factor of 6.

5. The x-intercept of `y = 2x+b` ` ` occurs at `(-b/2,0)`, while the x-intercept of `y = mx-6` ` ` occurs at `(6/m,0)`. So `-b/2 = 6/m => -mb = 12 => mb+12 = 0`.

6. Solve `1/a + 1/b = 1/n` for a to get `a = 1/(1/n-1/b)`. Create a table on the calculator for `y = 1/(1/3 - 1/x)` (because n = 3 is the smallest answer listed) starting at 1 with difference 1 and note that (12,4) and (6,6) are the only integer pairs in the table with a≥b. Repeat for `y = 1/(1/4-1/x)` and note that (20,5), (12,6) and (8,8) all appear in the table.

7. Let b = 5d and note that `c = sqrt(2013-25d^2-a^3)` must be an integer. Create a table for `y = sqrt(2013-25x^2-1^3)` on the calculator, starting at 1 with increment 1; scroll down and note there are no integer values for Y. Now do likewise for `y = sqrt(2013-x^2-2^3` and `y = sqrt(2013-25x^2-3^3)` , noting that there are no integer outputs. Finally, create a table for `y=sqrt(2013-25x^2-4^3)` and note that a=4, b = 5(2) = 10 and c = 43 works, so a+b+c=57.

8. Because A is repeated, A must be 3, 5 or 11 (if A has two prime factors, there won't be enough factors left for the the other letters). If A = 3, one of the other letters must be 11 and another must have 11 as a factor, making it at least 55, which is greater than 27. If A = 5, then one of the other letters must be 11 and another must have 11 as a factor, making it at least 33, which is greater than 27. So A = 11. One of the remaining letters must have two factors, and that letter must be 15, or otherwise an integer will be repeated. So 3+5+7+15 = 30.

9. Let `P(x) = Ax^3+Bx^2+Cx+D`. Thus `P(0) * P(3) = D(27A+9B+3C+D) = 139` so D = 1 because 139 is prime and A, B, C and D are all non-negative, making `27A+9B+3C+D>D`. So:

`27A+9B+3C=138 => 9A+3B+C = 46`

Similarly, `P(1) * P(2) = (A+B+C+1)(8A+4B+2C+1) = 689 = 13 * 53`  so

`A+B+C=12` and `8A+4B+2C=52 => 4A+2B+C=26`

Plugging `C=12-A-B` into the other two equations yields

`{(8A+2B=34),(3A+B=14):} => {(4A+B=17),(3A+B=14):} => A = 3 => B=5 => C =4`

So `P(-1) = -A+B-C+D = -3+5-4+1 = -1`.

10. Graph `y = cos(t) - cos((180/pi)t)` ` ` on the interval [0,2π] using radian mode and and use the calculator's zero feature to find that `t approx 0.108`.

11. Sketch the quadrilateral and angle bisector:

Because BCD is a right triangle with sides 6 and 8, BD = 10. Because AD = 10, triangle ADB is an isoceles triangle, so angles BAD and ABD are equal. Let the measure of ABD be 2θ so that the measure of BAE is θ and the measure of BEA is π−3θ. Using the Law of Cosines on triangle ADB:

`10^2=10^2+6^2-2*10*6 cos(2 theta) => cos(2 theta) = 3/10 => theta = 1/2 arccos(3/10)`

Using the Law of Sines on triangle BAE:

`(BE)/sin(theta) = (BD)/sin(pi-3 theta) => BE = (6sin(1/2 arccos(3/10)))/sin(pi-3/2arccos(3/10)) = 3.75`

12. The first possibility is that the intercepts for the first line are (0,2) and (4,0), so that `y = 2-1/2 x`;and that the intercepts for the second line are (0,4) and (6,0), so that `y = 4-2/3 x`. Solving these equations simultaneously yields x = 12 and y = -4, so 3a+b = 36−4 = 32.

The next possibility is that the intercepts for the first line are (0,2) and (4,0), and that the intercepts for the second line are (0,6) and (4,0), so that the lines intersect at x = 4 and y = 0, making 3a+b = 12+0 = 12.

The third possibility is that the intercepts for the first line are at (0,4) and (2,0), and that the intercepts for the second line are at (0,4) and (6,0), so that the lines intersect at x = 0 and y = 4,making 3a+b = 4.

The final possibility is that the intercepts for the first line are at (0,4) and (2,0), so that `y = 4-2x`; and that the intercepts for the second line are at (0,6) and (4,0), so that `y = 6 - 3/2 x`. Solving these equations simultaneously yields x = -4 and y = 12, so 3a+b = -12+12 = 0.

13. Let d be the duration of her first trip. So the duration of her second trip was d+2, the duration of her third trip was d+4, etc. If she made n trips, then

`d+(d+2)+ * * * +(d+2(n-1)) = nd + 2(1+2+ * * * +(n-1)) = nd + 2 ((n-1)(n))/2 = nd + n(n-1) = n(d+(n-1)) = 366 = 2 * 3 * 61`

If n ≥ 61, then d < 0. If n = 2, then d = 182, and if n = 3, then  d = 120; in neither case would the first trip be less than 3 months. So n = 6, and thus d+5 = 61, so d = 56. The trip durations must have been 56, 58, 60, 62, 64 and 66 days in length.

14. There are 26 = 64 possible 6-digits strings: 32 start with a 0 and 32 start with a 1. Each string beginning with a 0 (for example, 011010) has a counterpart beginning with a 1 (for example, 100101) such that the sum of these two strings is 111111. Omit all strings beginning with a 1, leaving the 32 strings beginning with a 0. In order for one of these remaining string to have a reveral, it must begin and end with 0. List these 16 strings:

000000
000010  000100  001000  010000
000110  001010  001100  010010  010100  011000
001110  010110  011010  011100
011110

There are six reversal pairs, so omit the strings highlighted in green above, leaving 32−6 = 26 strings.

However, we can now re-introduce the counterparts to these six green strings that begin with 1:

111011  110111
110101  101011
101001  100101

for a total of 26+6 = 32 strings.

Because the measures of angles QPR and RPS are equal, each measure must be 22.5° and the quadrilateral must consist of two congruent triangles (via AAS congruence). The measure of angle PRS must be 180°−45°−22.5° = 112.5° and using the Law of Sines on triangle SPR:

`(RS)/sin(22.5^o) = 8sqrt(2)/sin(45^o) = 16 => RS = 16sin(22.5^o)`

so the area of the quadrilateral is given by:

`2 * 1/2 * RS * PR * sin(PRS) = 16 * 8 sqrt(2) *sin(22.5^o) * sin(112.5^o) = 64`

16. Note that `a^2 = 2+2b^2 => a = sqrt(2+2b^2)`, so create a table for `y = sqrt(2+2x^2)` starting at 1 with increment 1. Scroll down, looking for integers in the Y column. the first pair is (7,10), but here `a =10`. Continue scrolling until you see (41,58), so `a-b=58-41=17`.

17. There are 5×4×3×2×1 = 120 possible arrangements of the 5 digits. List the arrangements for which no two adjacent digits are consecutive integers:

13524    14253
24135    24153    25314
31425    31524    35142    35241
41352    42513    42531
52413    53142

so the probability is `14/120 = 7/60` .

18. When the original triangle (shown in red below) rotates 90° counterclockwise, it sweeps out a quarter-circle of radius 4 in the first quadrant, and a quarter-circle of radius 3 in the second quadrant overlapping the resulting triangle (show in blue below).

The area swept out in the first quadrant is `1/2 * 4^2(pi/2) = 4pi`.

The region in the second quadrant can be split into a sector of a circle of radius 3 and a triangle with a side of length 4 along the y-axis. The quarter-circle (with equation `x^2+y^2=9`) and the hypotenuse of the triangle (with equation`y = 4/3 x + 4`) intersect at:

`x^2+(4/3 x + 4)^2 = 9 => 25/9 x^2 + 32/3 x +7 = 0 => x = 25x^2+96x+63 = 0 => (x+3)(25x+21) = 0 => x = - 21/25 =>y = (4/3)(- 21/25)+4 = 72/25`

So the area of the second-quadrant sector is `1/2 * 3^2 arctan(72/21)` and the area of the triangle is `1/2 (21/25)(4) = 42/25`, so the total area is `4pi + 9/2 arctan(24/7) + 42/25 approx 20.04`.

19. Suppose `x^2+mx+n` factors as `(x+p)(x+q) = x^2 + (p+q)x+pq => m = p+q, \ n = pq`, and that `x^2+mx-n` factors as `(x+r)(x-s) = x^2 + (r-s)x-rs => m = r-s, \ n = rs`. Thus `p+q = r-s => r = p+q+s` and `pq=rs => pq = (p+q+s)s => n = (m+s)s` for some integer s.

We need `m^2-4n ge 0 => m^2 - 4(m+s)s ge 0 => m^2 -4ms -4s^2 ge 0` for some positive integer s, and furthermore we need `sqrt(m^2-4ms-4s^2)`  to be an integer.

Create a table for the function `y = sqrt(x^2-4x-4*1^2` starting a 1 with increment 1. The only resulting integer output for x≤100 occurs when x = 5, yielding m = 5 and n = (5+1)(1) = 6. Note that `x^2+5x+6 = (x+2)(x+3)` and `x^2+5x-6 = (x+6)(x-1)`, so we have 1 such pair corresponding to s = 1.

Now create a table for `y = sqrt(x^2-4*2x-4*2^2)`, which yields m = 10 and n = 24, as well as m = 13 and n = 30. Note that `x^2+10x+24 = (x+6)(x+4)` and `x^2+10x-24 = (x+12)(x-2)`, while `x^2+13x+30 = (x+3)(x+10)` and `x^2+13x-30 = (x+15)(x-2)`, so we have 2 such pairs corresponding to s = 2.

Next create a table for `y = sqrt(x^2-4*3x-4*3^2)`, which yields m = 15 and n = 54, m = 17 and n = 60, and m = 25 and n = 84. Note that `x^2+15x+54 = (x+6)(x+9)` and `x^2+15x-54 = (x+18)(x+3)`, `x^2+17x+60 = (x+5)(x+12)` and `x^2+17x-60 = (x+20)(x-3)`, and `x^2+25x+84 = (x+4)(x+21)` and `x^2+25x-84 = (x+28)(x-3)`, so we have 3 such pairs corresponding to s = 3.

Then create a table for `y = sqrt(x^2-4*4x-4*4^2`, which yields m = 20 and n = 96, m = 26 and n = 120, and m = 41 and n = 180; only the first pair has both integers below 100, and we can verify that `x^2+20x+96 = (x+8)(x+12)` and `x^2+20x-96 = (x+24)(x-4)` , so there is 1 such pair corresponding to s = 4.

For s= 5, the smallest pair is m = 25 and n = 150, so the total number of pairs satisying all conditions is 1+2+3+1 = 7.

20. Both ACD and BCD must be right triangles, since they are inscribed in a semicircle, and both have side lengths that are Pythagorean triples: ACD has sides of length 14, 48 and 50, and BCD has sides of length 30, 40 and 50. The area of ACD is thus `1/2 (14)(48) = 336` and the area of BCD is thus `1/2 (30)(40) = 600`. There are two possible configurations of these triangles inside the semicircle; here is one:

In the other configuration, A and B are both in the same quadrant and we have:

so only in the first configuration will AB > 25 hold. The total area will be `336+600-I=936-I`, where I is the area of the intersection of the triangles.

`I = 1/2 (50)(CE) sin(DCE) = 25(CE)(14/50) = 7CE`

To find CE, use the Law of Sines:

`(CE)/sin(CDE) = 50/sin(CED) => CE = (50(30/50))/sin(CED) = 30/sin(CED)`

but

`sin(CED) = sin(pi - ACD-BDC) = sin(ACD+BCD) = sin(ACD)cos(BDC)+cos(ACD)sin(BDC) = (14/50)(40/50)+(30/50)(48/50) = 4/5`

so `CE = 30/(4/5) = 75/2 => I = 7(75/2) = 262.5` and the total area is 936−262.5 = 673.5.