Daily Math · 2026-07-03

Question 1

Solve $\begin{vmatrix}1&2&3\2&x+1&2x+1\3&2x+1&3x+1\end{vmatrix}=-1$ and find the sum of all solutions.

Accepted Answer

$\bullet$ Expand the determinant by row reduction or cofactors:

$$\begin{vmatrix}1&2&3\2&x+1&2x+1\3&2x+1&3x+1\end{vmatrix}=-x^2+3x-1$$

$\bullet$ Set it equal to $-1$:

$$-x^2+3x-1=-1\ \Longleftrightarrow\ x(x-3)=0$$

$\bullet$ The two roots are $0$ and $3$, so their sum is:

$$0+3=3$$

Answer

$0$

Answer

$3$

Answer

$-3$

Answer

$1$

Answer

$-1$

Answer

$6$

Question 2

On Monday at 7:00 a.m. a monk begins climbing a winding mountain trail, arriving at the summit at 5:00 p.m. The next morning at 7:00 a.m. she begins descending the same trail and reaches the base at 5:00 p.m. There must exist a point on the trail and a clock time at which the monk was at the same place on both days. **Which classical theorem most directly justifies this?**

Accepted Answer

Let $L$ be the trail length. Define

- $u(t)$: the monk's distance from the base on Monday at time $t \in [7, 17]$
- $d(t)$: her distance from the base on Tuesday at the same clock time

Both functions are continuous on $[7, 17]$ (a hiker doesn't teleport).

Now consider $f(t) = u(t) - d(t)$. By the problem statement:

- $u(7) = 0$ and $d(7) = L$, so $f(7) = -L < 0$.
- $u(17) = L$ and $d(17) = 0$, so $f(17) = +L > 0$.

Since $f$ is continuous and changes sign on $[7, 17]$, the **Intermediate Value Theorem** guarantees some $t^* \in (7, 17)$ with $f(t^*) = 0$, i.e. $u(t^*) = d(t^*)$. At that clock time, the monk stands at the same point on the trail on both days. $\blacksquare$

The physical intuition ("imagine two monks: one ascending Monday, one descending Tuesday at the same time — they must meet") collapses into a one-line IVT argument once you let $f$ do the work.

Answer

Mean Value Theorem

Answer

Intermediate Value Theorem

Answer

Rolle's Theorem

Answer

Brouwer Fixed-Point Theorem

Answer

Pigeonhole Principle

Answer

Bolzano–Weierstrass Theorem

Question 3

For $A = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix}$, the entry $(A^n)_{12}$ equals (for $n \ge 1$):

Accepted Answer

By induction, $A^n = \begin{pmatrix} 1 & n \ 0 & 1 \end{pmatrix}$, so $(A^n)_{12} = n$.

Answer

$1$

Answer

$n - 1$

Answer

$n$

Answer

$n^2$

Problems & worked solutions

Problem 1 — Matrices

Problem 2 — Continuity

Problem 3 — Matrices