MaxWong

Let that constant be k.

\((120 + k)^2 = (60 + k)(140 + k)\\ k^2 + 240k + 14400 = k^2 + 200k + 8400\\ 40k + 6000 = 0\\ k = -150\)

Common ratio = \(\dfrac{120 + k}{60 + k} = \dfrac13\)

\(\quad \text{Required sum}\\ = (3 + 13 + 23 + \cdots + 993) + (4 + 14 + 24 + \cdots + 994) + (6 + 16 + 26 + \cdots + 996) + (9 + 19 + 29 + \cdots + 999)\\ = 40(1 + 2 + \cdots + 99) + 100(3 + 4 + 6 + 9)\\ = 200200\)

Let \(\langle \cdot, \cdot \rangle\) denote the inner product. Then 

\(\dfrac{\langle \mathbf u, \mathbf v \rangle}{\|\mathbf u\|\|\mathbf v\|} = \cos 60^\circ = \dfrac12\\ \langle \mathbf u ,\mathbf v \rangle = \dfrac{3\cdot 2}2 = 3\)

\(A = \begin{pmatrix}\mathbf u^T\\\mathbf v^T\end{pmatrix}\\ A \mathbf u = \begin{pmatrix}\mathbf u^T\\\mathbf v^T\end{pmatrix} \mathbf u = \begin{pmatrix}\langle \mathbf u,\mathbf u \rangle\\\langle \mathbf v, \mathbf u\rangle\end{pmatrix} = \begin{pmatrix}\|\mathbf u\|^2\\\langle \mathbf u, \mathbf v\rangle\end{pmatrix} = \begin{pmatrix}9\\3\end{pmatrix}\\\)

Similarly we have

\(A \mathbf v = \begin{pmatrix}\langle \mathbf u, \mathbf v\rangle \\\|\mathbf v\|^2\end{pmatrix} = \begin{pmatrix}3\\4\end{pmatrix}\)

If this is not the notation you are used to, \(\langle \mathbf u,\mathbf v \rangle \) is just \(\mathbf u \cdot \mathbf v\).

Simplifying gives x^2 - mx + 14 = 0. By Vieta's formula, x_1 x_2 = 14 and m = x_1 + x_2. Which integers multiply to 14? (-7, -2), (-14, -1), (7, 2), and (14, 1), assuming |x_1| > |x_2|. So the possible values of m are -9, 9, -15, and 15.

All numbers divisible by 8 has the property that the last 3 digits are also divisible by 8. But note that since the 4-digit number is a palindrome, we are actually counting the number of 3-digit multiples of 8 that has the same first two digits, and does not end with 0. We can then make put the last digit in front to get back the 4-digit palindrome we wanted. The other case is that the number ends with 00(digit), so we need to take that into account as well, but this case is simple because the only possibility is 8008. 

For the first case, we first count the number of 3-digit multiples of 8 that has the same first two digits. They are 112, 224, 336, 440, 448, 552, 664, 776, 888, and 992. So in total 10, but 440 ends with 0 so we exclude that case. Hence, we have only 9 cases here, which gives the palindromes 2112, 4224, 6336, 8448, 2552, 4664, 6776, 8888, 2992 respectively.

Including 8008, the answer is 10.

Let \(P = (x, -x+6)\). We have

Note that \(m_{AO} = \dfrac{8 - (-12)}{2 - 10} = -\dfrac 52\). So the slope of perpendicular bisector of AO is \(\dfrac25\).

Midpoint of AO is \(\left(\dfrac{10+2}2, \dfrac{-12+8}2\right) = (6,-2)\). Since PA = PO, P lies on the perpendicular bisector of AO. Then,

\(\dfrac{(-x+6) - (-2)}{x - 6} = \dfrac{2}5\\ 5(8 - x) = 2(x - 6)\\ 40 - 5x = 2x - 12\\ 52 = 7x\\ x = \dfrac{52}7\)

The point is \(\left(\dfrac{52}7, -\dfrac{52}7 + 6\right) = \left(\dfrac{52}7, -\dfrac{10}7\right)\).

Let \(d = a_2 -a_1\). Note that \(S_5 = 5a_1 + (d + 2d + 3d + 4d) = 5a_1 + 10d\), and \(S_{10} = 10a_1 + (d + 2d + \cdots + 9d) = 10a_1 + 45d\), and \(S_{15} = 15a_1 + (d + 2d + \cdots + 14d) = 15a_1 + 105d\).

So, the given conditions are \(\begin{cases} 5a_1 + 10d = \dfrac15\\ 10a_1 + 45d = \dfrac1{10} \end{cases}\). Solving gives \(d = -\dfrac3{250}\) and \(a_1 = \dfrac8{125}\).

So \(S_{15} = 15a_1 + 105d = -\dfrac3{10}\).

Suppose \(d = a_2 - a_1\). Note that the given conditions are equivalent to:

\(\begin{cases} a_1 + d = -1\\ a_1 + 3d = \dfrac13 \end{cases}\)

Solving gives \(d = \dfrac23\) and \(a_1 = -\dfrac53\).

Multiply by (z + 1)^2 on both sides. Then,

\((3 - z)(z + 1) \geq 2(z + 4)(z + 1)^2\text{ and }z + 1 \neq 0\\ (z + 1)(2(z + 4)(z + 1) -(3-z)) \leq 0\text{ and }z + 1 \neq 0\\ (z + 1)(2z^2 +11z+5)\leq 0 \text{ and }z + 1 \neq 0\\ (z+1)(2z+1)(z+5)\leq 0 \text{ and }z + 1 \neq 0\\ z \leq -5 \text{ or }-1 < z \leq -\dfrac12\)

In interval notation, \(z \in (-\infty, -5] \cup \left(-1, -\dfrac12\right]\)

Similar to the one you posted before, so I will only give partial work:

Comparison between before and after drinking coffee gives

\((t - 3)(4p + 5) = 2tp\\ 4tp - 12p + 5t - 15 = 2tp\\ 2tp - 12p + 5t - 15 =0 \\ 2p(t - 6) + 5(t - 6) = -15\\ (2p + 5)(6 - t) = 15\)

Note that since p is a positive integer, \(p \geq 1\). That means \(2p + 5 \geq 7\), and 2p + 5 is also a divisor of 15. 

What is the only divisor of 15 that is greater than 7?

Please continue on your own from this point. If you are stuck, look at my answer to your previous similar problem. I gave full solution there.

Comparison between before and after drinking coffee gives:

\((t - 11)(4p + 2) = 2tp\\ 4tp - 44p + 2t - 22 = 2tp\\ 2tp + 2t - 44p - 22 = 0\\ 2t(p + 1) - 44(p + 1) = -22\\ (22 - t)(p + 1) = 11\)

How is that possible? The only possibilities are \(\begin{cases} 22-t = 1\\ p+1=11 \end{cases} \text{ or } \begin{cases} 22-t=11\\ p+1=1 \end{cases}\). But p + 1 can't be 1 since p would be 0, which is not positive.

So, the solution is (t, p) = (21, 10), which means the mathematician solved 4p + 2 = 42 problems that day.

The second equation gives r = 1 so the third equation says p + 1 = 9 + p, which has no solution for p.

Hence, the system has no solutions.