MaxWong

This one is answered earlier: https://web2.0calc.com/questions/exponentials_2

Prob(pen from first box) = \(\dfrac{2}{2 + 2} = \dfrac12\). (It is equally as likely to choose a plain pencil or a pen, because the numbers are the same.)

Prob(crayon from second box) = \(\dfrac{5}{5 + 7} = \dfrac5{12}\).

The events do not impact each other, so the case of both events happening is just the product of the two probabilities. Can you finish it from here?

There are (99 - 10 + 1)^2 = 8100 ways to choose two such numbers. 

99 - 10 + 1 = 90 cases out of all cases has two same numbers chosen,

2(98 - 10 + 1) = 178 cases out of all cases has two numbers with difference = 1.

2(97 - 10 + 1) = 176 cases out of all cases has two numbers with difference = 2.

2(96 - 10 + 1) = 174 cases out of all cases has two numbers with difference = 3.

Therefore, the required probability is \(\dfrac{90 + 178 + 176 + 174}{8100} = \dfrac{103}{1350}\).

Substitute x = 2.

\(f(x) = 3^x\\ f(2) = 3^2\)

Now the answer is obvious.

Note that g(x) is irrelevant in this question. Since the graphs of even functions are symmetric about the line x = 0, if we reflect (2, -2) across x = 0, we get another point on the graph of f(x). Can you do that on your own?

Note that \(\begin{cases}5^{100} \equiv 0 \pmod{125}\\5^{100} = 25^{50} \equiv 1^{50} = 1 \pmod 8\end{cases}\)

Using Chinese Remainder Theorem, the integer x with \(\begin{cases}x \equiv 0 \pmod{125} \\x \equiv 1 \pmod 8\end{cases}\) is unique mod 1000.

Also, note that \(625 = 5(125) \equiv 0 \pmod{125}\) and \(625 = 78 \times 8 + 1 \equiv 1 \pmod{8}\).

That means if \(\begin{cases}x \equiv 0 \pmod{125} \\x \equiv 1 \pmod 8\end{cases}\), then \(x \equiv 625\pmod{1000}\).

Then \(5^{100} \equiv 625 \pmod{1000}\), so the last three digits are 625. The sum of final three digits is 6 + 2 + 5 = 13.

In fact, it is 625 mod 1000 instead.

Assuming the coefficients are always 1, 4, 1, 4, 1, 4 so on:

\(\dfrac1{3^1} + \dfrac4{3^2} + \dfrac1{3^3} + \dfrac4{3^4} + \cdots\\ = \dfrac13 \left(1 + \dfrac19 + \dfrac1{9^2} + \cdots\right) + \dfrac4{3^2} \left(1 + \dfrac19 + \dfrac1{9^2}+\cdots\right)\\ = \left(\dfrac{1}{3} + \dfrac49\right)\left(\dfrac1{1 - \dfrac19}\right)\\ = \dfrac78\)

\(z^2 + 2 + \dfrac1{z^2} = \left(z + \dfrac1z\right)^2 = 1\\ z^2 + \dfrac1{z^2} = 1 - 2 = -1\)

The digit in the ones place is three more than the digit in the tens place.

Judging only by this, the number could be 14, 25, 36, 47, 58, 69. There is only one prime number in this list.