Problem 02: 5/3

Conjugates are the red herring here.

(4√2 + √2)÷(4√2 - √2) = 5√2 ÷ 3√2 =5/3

Problem 03: 2(4√2 + √2)÷(4√2 - √2) = 5√2 ÷ 3√2 =5/3

Draw a vertical line down the middle and the students will quickly see that the two halves are 2x2 and the red portion is 1+1 = 2

Problem 04: 112
Choose three consecutive integers: x-1, x, x+1

(x+1)³ - x³ = 666 + x³ - (x-1)³

x³ + 3x² + 3x +1 - x³ = 666 + x³ - x³ + 3x² - 3x + 1

Stuff cancels all over the place: 6x = 666; x = 111; Answer is 112

Problem 05: 1(x+1)³ - x³ = 666 + x³ - (x-1)³

x³ + 3x² + 3x +1 - x³ = 666 + x³ - x³ + 3x² - 3x + 1

Stuff cancels all over the place: 6x = 666; x = 111; Answer is 112

Problem 06: 17

(x+2)² = x² + 8²

x² + 4x + 4 = 64 + x²

4x = 60; x=15; hypotenuse = 17. Even if you missed the Pythagorean triple.

Problem 07: 15x² + 4x + 4 = 64 + x²

4x = 60; x=15; hypotenuse = 17. Even if you missed the Pythagorean triple.

Problem 08: 36

Problem 09: 117

Problem 10: 59

Problem 11: 2

9

3

complete the square for U = 3

u² + 18u + 81 = 243 + 81 = 324 = 18²

u + 9 = +- 18

3

Problem 12: 13/14^{x}+ 2(3^{x+2}) = 2433

^{2x}+ 18(3^{x}) = 243complete the square for U = 3

^{x}u² + 18u + 81 = 243 + 81 = 324 = 18²

u + 9 = +- 18

3

^{x}= 9, -27 .... 2 is only real answer.Problem 13: 930

Problem 14: 140√2

ab = 28; ac = 20; bc=70

a²b²c² = 28*20*70

abc = √(4*7*4*5*7*2*5) = 140√2

Problem 15: 6/31a²b²c² = 28*20*70

abc = √(4*7*4*5*7*2*5) = 140√2

In these simple dice rolling questions, I like to imagine the full sample space of 36 possibilities. We are told that 6 is not an option and that occurs 5 times, leaving 31 total.

11 12 13 14 15 16

21 22 23 24 25 26

31 32 33 34 35 36

41 42 43 44 45 46

51 52 53 54 55 56

61 62 63 64 65 66

Thus, P(7) = 6/31

Problem 16: 4811 12 13 14 15 16

21 22 23 24 25 26

31 32 33 34 35 36

41 42 43 44 45 46

51 52 53 54 55 56

61 62 63 64 65 66

Thus, P(7) = 6/31

Problem 17: 62

Build a rectangle around the figure and subtract the four triangles.

10*10 - ½(4*7 + 3*7 + 3*3 + 6*3) = 62

Problem 18: 1/510*10 - ½(4*7 + 3*7 + 3*3 + 6*3) = 62

Problem 19: 6/55

Problem 20: 16

Problem 21: 10

Problem 22: 1/13

Problem 23: 8

Problem 24: 4013

Problem 25: 6

log

144

2

b=4c and a=2c; answer 6

Problem 26: a=3, b=2/13_{144}(3^{a}* 2^{b}) = c144

^{c}= 3^{a}* 2^{b}2

^{4c}* 3^{2c}= 2^{b}* 3^{a}b=4c and a=2c; answer 6

Problem 27: 1/6

Problem 28: 49

Problem 29: 5π/8

Problem 30: 300

Problem 31: 64/175

Problem 32: 18√3

ABC is a 30-60-90, with B=90

Problem 33: 3^{o}. AC is 12, split 3,3, and 6. If the altitude is BN, then AN=3 and BN=3√3, making the area = 18√3Problem 34: 339

4a+4b+4c=124, so a+b+c=31. Square it

a²+ab+ac+ba+b²+bc+ca+cb+c²=961

2ab + 2ac + 2bc = 622

a² + b² + c² = 339, which is also d²

Problem 35: 250a²+ab+ac+ba+b²+bc+ca+cb+c²=961

2ab + 2ac + 2bc = 622

a² + b² + c² = 339, which is also d²

Assume first that p is odd. The list of integers is symmetric:

x-2, x-1, x, x+1, x+2; 5x = 5

In general, then px = 5

p=5, x=5

p=25, x=5

p=125, x=5

p=625, x=5

Assume p is even.

x and x+1 are the middle two terms.

x-3, x-2, x-1, x, x+1, X+2, x+3, x+4 : 8 terms => 8x + 4

In general, p terms = px + .5p = p(x+1/2).

Since x is integer, we need a 2: q(2x+1) = 5

5(2x+1)=5

25(2x+1)= 5

125(2x+1)=5

625(2x+1)=5

Problem 36: 46x-2, x-1, x, x+1, x+2; 5x = 5

^{7}In general, then px = 5

^{7}, making p a power of 5.p=5, x=5

^{6}. 5 termsp=25, x=5

^{5}. 25 termsp=125, x=5

^{4}= 625 terms.p=625, x=5

^{3}. Impossible because x was the "middle" number and we can't have 312 on either side.Assume p is even.

x and x+1 are the middle two terms.

x-3, x-2, x-1, x, x+1, X+2, x+3, x+4 : 8 terms => 8x + 4

In general, p terms = px + .5p = p(x+1/2).

Since x is integer, we need a 2: q(2x+1) = 5

^{7}.5(2x+1)=5

^{7}: x=5^{6}, means ten terms.25(2x+1)= 5

^{7}: x=1562, 50 terms125(2x+1)=5

^{7}: x=310, 250 terms625(2x+1)=5

^{7}: x=62, impossible for x to be the middle.
log

log

U + 10/U = 7

U² + 10 = 7U

U² - 7U + 10 = 0 ... (U-5)(U-2) = 0 .... log

a

since 45² = 2025, a² = b has 43 candidates in range. 2² = 4 through 44² = 1836

46 candidates.

Problem 37: 19_{a}b + 10log_{b}a = 7log

_{a}b + 10/log_{a}b = 7U + 10/U = 7

U² + 10 = 7U

U² - 7U + 10 = 0 ... (U-5)(U-2) = 0 .... log

_{a}b = 2, 5a

^{5}= b has three candidates in range: 2,32; 3,243; 4;1024since 45² = 2025, a² = b has 43 candidates in range. 2² = 4 through 44² = 1836

46 candidates.

Problem 38: a=9 and b=11

Two equations:

386

146

Starting with the second: 1a² + 4a + 6 = 1b² + 0b + 2

a² + 4a + 4 = b²

(a+2)² = b² gives us a+2 = b

Substitute into equation 1

3a² + 8a + 6 = 2b² + 7b + 2

3a² + 8a + 6 = 2(a+2)² + 7(a+2) + 2

expand

3a² + 8a + 6 = 2a² + 8a + 8 + 7a + 14 + 2

3a² + 8a + 6 = 2a² + 15a + 24

a2 - 7a - 18 = 0

(a+2)(a-9) gives us a = -2, 9

a = -2 is impossible so we're left with a=9 and thus b=11

back to number 38.

Problem 39: 224x³386

_{a}= 272_{b}146

_{a}=102_{b}Starting with the second: 1a² + 4a + 6 = 1b² + 0b + 2

a² + 4a + 4 = b²

(a+2)² = b² gives us a+2 = b

Substitute into equation 1

3a² + 8a + 6 = 2b² + 7b + 2

3a² + 8a + 6 = 2(a+2)² + 7(a+2) + 2

expand

3a² + 8a + 6 = 2a² + 8a + 8 + 7a + 14 + 2

3a² + 8a + 6 = 2a² + 15a + 24

a2 - 7a - 18 = 0

(a+2)(a-9) gives us a = -2, 9

a = -2 is impossible so we're left with a=9 and thus b=11

back to number 38.

Simplify by canceling an x³ and then by substituting u = x³. Now for some long division. Yay!

So each column adds one and projecting down the line gets you to a remainder of 224x³

Problem 40: 3/2u^{666}+2u^{663}+3u^{660}+ u^{657}+ ...

-----------------------

u^{3}-1) u^{669}+ u^{666}+ u^{663}+ u^{660}+ ... u^{669}- u^{666}-----------

2u^{666}+ u^{663}2u^{666}-2u^{663}------------

3u^{663}+ u^{660}etc.

So each column adds one and projecting down the line gets you to a remainder of 224x³

Hang on. I messed that up. Must revisit.

Good thing I wasn't taking the test for real.

u^{668}+ u^{667}+ u^{666}+ u^{657}+ ... . ----------------------- u-1) u^{669}+ 0 + 0 + u^{666}+ 0 + 0 + u^{663}+ u^{660}+ ... u^{669}- u^{668}----------- u^{668}+ 0 u^{668}- u^{667}------------ u^{667}+ u^{666}u^{667}- u^{666}------------ 2u^{666}+ 0

When you are finished, you will have 224/u-1 but we can't leave it there. We removed a x³ from numerator and denominator so the last bit is really 224x³ / (x

^{6}-x^{3}) so the remainder is really 224x³.Good thing I wasn't taking the test for real.

d² = x² +(2x)² +(5-x)² + (2x)² + (5-2x)² + (5-x)² + (5-2x)² + x²

Expand all that and cancel/clean up

sum of d² = 20x² - 30x + 50

min when d(sum)=0

40x - 60 = 0

x = 3/2

Kids get in trouble here because they want to differentiate the square root or get off the idea that the sum of the distances must be at a minimum.

Problem 41: 150Expand all that and cancel/clean up

sum of d² = 20x² - 30x + 50

min when d(sum)=0

40x - 60 = 0

x = 3/2

Kids get in trouble here because they want to differentiate the square root or get off the idea that the sum of the distances must be at a minimum.

Cleaning up Jonathan's comment a bit:

The semiperimeter, S = (5+6+7)/2 = 9.

Heron's:

A = √(9*4*3*2) = √216 = 6√6.

To get Altitude (height):

Area = ½*5*h = 6√6 so h = 12/5√6

Use CA = 6 and h to find that CP

Using h and Pythagoras:

q

q

q

q

Sum:

4(12/5*√6)² + (1+16+81+196)/25.

(4*144*6 + 1+16+81+196)/25 = 3750/25

= 150

Eric P said:

First find cos(C) by law of cosines

Cos(C)=[6²-7²-5²]/(-2*7*5)=19/35

q

q

q

q

Lining it up this way and using some quick mental math tricks makes the addition simpler: SUM=4(49)+30-2(7)(10)(19/35)=150

I think I have to side with Eric on this one.

The semiperimeter, S = (5+6+7)/2 = 9.

Heron's:

A = √(9*4*3*2) = √216 = 6√6.

To get Altitude (height):

Area = ½*5*h = 6√6 so h = 12/5√6

Use CA = 6 and h to find that CP

_{h}= 6/5.Using h and Pythagoras:

q

_{1}= (1/5)² + h²q

_{2}= (4/5)² + h²q

_{3}= (9/5)² + h²q

_{4}= (13/5)² + h²Sum:

4(12/5*√6)² + (1+16+81+196)/25.

(4*144*6 + 1+16+81+196)/25 = 3750/25

= 150

Eric P said:

First find cos(C) by law of cosines

Cos(C)=[6²-7²-5²]/(-2*7*5)=19/35

q

_{1}² = 49 +1 -2(7)(1)(19/35)q

_{2}² = 49 +4 -2(7)(2)(19/35)q

_{3}² = 49 +9 -2(7)(3)(19/35)q

_{4}² = 49 +16-2(7)(4)(19/35)Lining it up this way and using some quick mental math tricks makes the addition simpler: SUM=4(49)+30-2(7)(10)(19/35)=150

I think I have to side with Eric on this one.

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