Determine n ∈ ℕ, n ≥ 2, such that binomn2 = 15.

(n(n-1))/(2) = 15 ⇒ n^2 - n - 30 = 0 ⇒ n ∈ {6, -5}. Only n = 6 is admissible.

Daily Math Problems — May 13, 2026: Matrices, Combinatorics, Logs

Problems & worked solutions

RO M1

Problem 1 — Matrices

For $A = (\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array})$ and $B = (\begin{array}{cc} 0 & - 1 \\ 5 & 2 \end{array})$ , $A + B$ equals:

Show answer & worked solution

A. $(\begin{array}{cc} 1 & 1 \\ 8 & 6 \end{array})$ ✓ correct
B. $(\begin{array}{cc} 1 & 3 \\ 8 & 2 \end{array})$
C. $(\begin{array}{cc} 1 & - 1 \\ - 2 & 2 \end{array})$
D. $(\begin{array}{cc} 0 & - 2 \\ 15 & 8 \end{array})$

Add componentwise: $(\begin{array}{cc} 1 & 1 \\ 8 & 6 \end{array})$ .

RO M1

Problem 2 — Permutations & Combinations

Determine $n \in N$ , $n \geq 2$ , such that $(\binom{n}{2}) = 15$ .

Show answer & worked solution

A. $4$
B. $5$
C. $6$ ✓ correct
D. $15$

$\frac{n (n - 1)}{2} = 15 \Rightarrow n^{2} - n - 30 = 0 \Rightarrow n \in {6, - 5}$ . Only $n = 6$ is admissible.

International

Problem 3 — Logs

The solution set of the inequality $\log_{1 / 3} (x^{2} - 4) > 0$ is:

Show answer & worked solution

A. $x \in (- \infty, 0)$
B. $x \in (0, 2)$
C. $x \in (- \sqrt{5}, - 2) \cup (2, \sqrt{5})$ ✓ correct
D. $x \in (\sqrt{5}, \infty)$
E. $x \in (2, \infty)$
F. $x \in (0, \sqrt{5})$

Domain constraint: $x^{2} - 4 > 0 \Rightarrow x \in (- \infty, - 2) \cup (2, + \infty)$ .

Inequality: $\log_{1 / 3} (x^{2} - 4) > 0 = \log_{1 / 3} 1$ . Since the base $\frac{1}{3} < 1$ , $\log_{1 / 3}$ is decreasing, so the inequality flips: $x^{2} - 4 < 1$ , i.e. $x^{2} < 5$ , so $x \in (- \sqrt{5}, \sqrt{5})$ .

Intersection with the domain: $[(- \infty, - 2) \cup (2, + \infty)] \cap (- \sqrt{5}, \sqrt{5}) = (- \sqrt{5}, - 2) \cup (2, \sqrt{5})$ .

Problems & worked solutions

RO M1

Problem 1 — Matrices

For $A = (\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array})$ and $B = (\begin{array}{cc} 0 & - 1 \\ 5 & 2 \end{array})$ , $A + B$ equals:

Show answer & worked solution

A. $(\begin{array}{cc} 1 & 1 \\ 8 & 6 \end{array})$ ✓ correct
B. $(\begin{array}{cc} 1 & 3 \\ 8 & 2 \end{array})$
C. $(\begin{array}{cc} 1 & - 1 \\ - 2 & 2 \end{array})$
D. $(\begin{array}{cc} 0 & - 2 \\ 15 & 8 \end{array})$

Add componentwise: $(\begin{array}{cc} 1 & 1 \\ 8 & 6 \end{array})$ .

RO M1

Problem 2 — Permutations & Combinations

Determine $n \in N$ , $n \geq 2$ , such that $(\binom{n}{2}) = 15$ .

Show answer & worked solution

A. $4$
B. $5$
C. $6$ ✓ correct
D. $15$

$\frac{n (n - 1)}{2} = 15 \Rightarrow n^{2} - n - 30 = 0 \Rightarrow n \in {6, - 5}$ . Only $n = 6$ is admissible.

International

Problem 3 — Logs

The solution set of the inequality $\log_{1 / 3} (x^{2} - 4) > 0$ is:

Show answer & worked solution

A. $x \in (- \infty, 0)$
B. $x \in (0, 2)$
C. $x \in (- \sqrt{5}, - 2) \cup (2, \sqrt{5})$ ✓ correct
D. $x \in (\sqrt{5}, \infty)$
E. $x \in (2, \infty)$
F. $x \in (0, \sqrt{5})$

Domain constraint: $x^{2} - 4 > 0 \Rightarrow x \in (- \infty, - 2) \cup (2, + \infty)$ .

Intersection with the domain: $[(- \infty, - 2) \cup (2, + \infty)] \cap (- \sqrt{5}, \sqrt{5}) = (- \sqrt{5}, - 2) \cup (2, \sqrt{5})$ .