## Math Series 1

1. Given A e^x + B e^-x = 4 cosh(x) - 5 sinh(x), determine the values of A and B.
1. A = 1/2, B = -4 1/2
2. A = -1/4, B = 4 1/4
3. A = -1/2, B = 4 1/4
4. A = -1/2, B = 4 1/2
Solution:
It is known that,
$$cosh(x) = \frac{e^x + e^{-x}}{2}$$ $$sinh(x) = \frac{e^x - e^{-x}}{2}$$
By substitution,
\begin{align} Ae^x + Be^{-x} & = 4 \left( \frac{e^x + e^{-x}}{2} \right) - 5 \left( \frac{e^x - e^{-x}}{2} \right)\\ Ae^x + Be^{-x} & = \left( \frac{4-5}{2} \right) e^x + \left( \frac{4+5}{2} \right) e^{-x} \\ \end{align} By partial fractions, \begin{align} A & = \left( \frac{4-5}{2} \right)\\ A & = -1/2 \\ B & = \left( \frac{4+5}{2} \right)\\ B & = 9/2\\ \end{align}

2. The average value of function f(x) = sin(x) - x on the interval [0, $\pi$].
1. -0.75419
2. -0.95518
3. -0.73424
4. -0.93418
Solution:
Formula for mean,
$$mean = \frac{\int_a^b f(x)dx}{b-a}$$
By substitution,
\begin{align} mean & = \frac{\int_0^{\pi} sin(x)-x}{\pi-0} \\ & = -0.93418 \\ \end{align}

3. Given $\Gamma(1.5) = 0.8862$, determine $\Gamma(3.5)$.
1. 3.32325
2. 2.45453
3. 2.34560
4. 3.32190
Solution:
Using Gamma Function:
$$\Gamma(n) = (n-1)!, n \ge 1$$ or $$\Gamma(n) = \int_0^\infty x^{n-1} e^{-x} dx$$

Solution 1: Using factorial \begin{align} \Gamma(1.5) & = (1.5-1)! \\ & = (0.5)! \\ & = 0.8862\\ \end{align} \begin{align} \Gamma(3.5) & = (3.5-1)! \\ & = (2.5)! \\ & = (2.5)(2.5-1)(2.5-2)! \\ & = (2.5)(1.5)(0.5)! \\ & = (2.5)(1.5)(0.8862) \\ & = 3.32325\\ \end{align} Solution 2: Using Gamma Function
Use a large number to replace infinity. In this case, let's use 20. \begin{align} \Gamma(3.5) & = \int_0^{20} x^{3.5-1} e^{-x} dx \\ & = \int_0^{20} x^{2.5} e^{-x} dx\\ & = 3.32333\\ \end{align}

4. Given $\Gamma(1.5) = 0.8862$, determine $\Gamma(-0.5)$.
1. -0.2312
2. -2.3533
3. -3.5448
4. -1.3456
Solution:
\begin{align} \Gamma(1.5) & = (1.5-1)! \\ & = (0.5)! \\ & = 0.8862\\ \end{align} \begin{align} \Gamma(-0.5) & = (-0.5-1)! \\ & = (-1.5)! \text{ < Eq. 1>} \\ \\ \Gamma(1.5) & = (0.5)! \\ & = (0.5)(0.5-1)(0.5-2)! \\ 0.8862 & = (0.5)(-0.5)(-1.5)! \\ \end{align} Substituting Eq. 1, \begin{align} 0.8862 & = (0.5)(-0.5)\Gamma(-0.5) \\ \Gamma(-0.5) & = -3.5448\\ \end{align}

5. Find the remainder when 137^153 is divided by 18.
1. 8
2. 2
3. 17
4. 1
Solution:
Factor 137^3 by 18 (the denominator). $$\frac{ (137^3)^{51} }{ 18 } = \frac{(142852 \times 18+17)^{51}}{18}$$ Here, the remainder is somewhere in the 17^51/18. Try to reduce the exponent then factor by 18. \begin{align} remainder \Rightarrow \frac{17^{51}}{18} & = \frac{(17^3)^{17}}{18} \\ & = \frac{(272 \times 18+17)^{17}}{18} \\ \end{align} The remainder is somewhere in the 17^17/18. Repeat the process. \begin{align} remainder \Rightarrow \frac{ 17^{17} }{18} & = \frac{ 17 \times 17^{16} }{18} \\ & = \frac{ 17 \times (17^4)^4 }{18} \\ & = \frac{17}{18} \cdot \frac{(4640 \times 18 + 1)^4}{18} \\ \end{align} \begin{align} remainder & \Rightarrow \frac{17 \cdot (1)^4}{18} = \frac{17}{18}\\ & \Rightarrow 17\\ \end{align}

6. Find the remainder when 37^100 is divided by 29.
1. 23
2. 25
3. 17
4. 2
Solution:
Using Fermat's Little Theorem:
$$\frac{t^{p-1}}{p}, remainder \Rightarrow 1$$ Given that, \begin{align} t & = \text{integer} \\ p & = \text{prime number}\\ \end{align}
Since the denominator is a prime number, \begin{align} \frac{37^{100}}{29} & = \frac{ \left( 37^{29-1} \right)^3 \times 37^{16}}{29} \\ remainder \Rightarrow & \frac{(1)^3 \times 37^{16}}{29} \\ & = \frac{(37^4)^4}{29} \\ & = \frac{(64626 \times 29 + 7)^4}{29} \\ \end{align} The remainder is somewhere in the 7^4/29. \begin{align} remainder & \Rightarrow \frac{ 7^4 }{29} = 82 \frac{23}{29} \\ & \Rightarrow 23. \end{align}

7. Find the remainder when 45^1000 is divided by 31.
1. 17
2. 23
3. 25
4. 1
Solution:
Since the denominator is a prime number, use Fermat's Little Theorem. \begin{align} \frac{45^{1000}}{31} & = \frac{ \left( 45^{31-1} \right)^{33} \times 45^{10}}{31} \\ remainder \Rightarrow & \frac{(1)^{33} \times 45 \times (45^3)^3}{31} \\ & = \frac{(31 \times 1 + 14) \times (2939 \times 31 + 16)^3}{31} \\ \end{align} \begin{align} remainder & \Rightarrow \frac{ 14 \times 16^3 }{31} = 1849 \frac{25}{31} \\ & \Rightarrow 25. \end{align}

8. Today is Monday, 4 July 2022. What day of the week will be 2^9833 days from now?
1. Tuesday
2. Thursday
3. Wednesday
4. Friday
Solution:
$$\frac{2^{9833}}{7}$$ Since the denominator is a prime number, use Fermat's Little Theorem. \begin{align} \frac{2^{9833}}{7} & = \frac{ \left( 2^{7-1} \right)^{1638} \times 2^{5}}{7} \\ remainder & \Rightarrow \frac{(1)^{1638} \times 2^5}{7} \\ & = 4 \frac{4}{7} \\ remainder & \Rightarrow 4 \\ \end{align} The day is 4th day from Monday. Therefore, the answer is Friday.

9. A stack of bricks has 61 bricks at the bottom layer, 58 bricks in the second layer, 55 bricks in the third layer and so on until there are 10 bricks in the topmost layer. Find the total number of bricks.
1. 525
2. 639
3. 789
4. 144
Solution:
$$\text{Sum} = 61 + 58 + 55 + ... + 10$$ The difference between the terms is 3 and since the difference is constant, this is an Arithmetic Progression.
First, we need to get how many stacks are there.
Arithmetic Progression
$$a_n = a_1 \pm (n-1)d$$ \begin{align} a_n & - \text{nth term} \\ d & - \text{difference} \\ n & - \text{number of terms}\\ \end{align}
Solving, \begin{align} a_n & = a_1 \pm (n-1)d \\ 10 & = 61 - ( n - 1 )(3) \\ n & = 18 \\ \end{align} To get the sum,
$$\text{Sum of all terms} = \frac{n}{2} \left( a_1 + a_n \right)$$
\begin{align} \text{Sum of all terms} & = \frac{18}{2} \left( 61 + 10 \right) \\ & = 639 \\ \end{align}

10. Allen press carrots by hand to make carrot juice. One-eighth of the juice is extracted from the first pressing. Each subsequent pressing extracts one-eighth of the remaining juice from the carrots. How many times will a bunch of carrots need to press to extract at least three-fourths of the juice?
1. 10
2. 9
3. 11
4. 7
Solution:
It can be seen that the difference for each extracted juice is a common ratio. This is a Geometric Progression.
Geometric Progression
$$a_n = a_1 r^{n-1}$$ \begin{align} a_n & - \text{nth term} \\ n & - \text{number of terms}\\ r & - \text{common ratio} = \frac{a_2}{a_1} = \frac {a_3}{a_4} \\ \end{align}

Solving for the remaining juice after the first press, \begin{align} a_1 & = 1 \\ a_2 & = 1 - \frac{1}{8} \\ & = \frac{7}{8} \\ r & = \frac{a_2}{a_1} \\ & = \frac{7}{8}\\ \end{align} Solving for n, \begin{align} a_n & = a_1 r^{n-1} \\ 1 - \frac{3}{4} & = 1 \cdot \left( \frac{7}{8} \right)^{n-1} \\ 0.25 & = (0.125)^{n-1} \\ n & = 11.3818 \\ n & \approx 11\\ \end{align}

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