Daily Math · 2026-05-06

Question 1

In a room of 23 people, what's the approximate probability that at least two share a birthday? (Assume 365 equally likely birthdays, no twins.)

Accepted Answer

Compute the complement — probability all 23 birthdays are distinct: $$P(	ext{all distinct}) = \frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365} \cdots \frac{343}{365} \approx 0.493$$

So $P(	ext{at least one match}) = 1 - 0.493 \approx 0.507$, just over 50\%.

Answer

$\approx 6\%$

Answer

$\approx 23\%$

Answer

$\approx 50\%$

Answer

$\approx 90\%$

Question 2

A square is inscribed in a circle of radius $r$. What is the area of the square in terms of $r$?

Accepted Answer

The diagonal of the inscribed square equals the circle's diameter, $2r$. For a square with diagonal $d$, the side length is $d/\sqrt{2}$, so the area is $(d/\sqrt{2})^2 = d^2 / 2$.

Here $d = 2r$, so area $= (2r)^2 / 2 = 4r^2 / 2 = 2r^2$.

Answer

$r^2$

Answer

$2r^2$

Answer

$\pi r^2$

Answer

$4r^2$

Question 3

What is the remainder when $7^{100}$ is divided by $5$?

Accepted Answer

Work modulo 5. Note $7 \equiv 2 \pmod 5$, so $7^{100} \equiv 2^{100} \pmod 5$.

Now look at powers of 2 mod 5: $2^1 = 2$, $2^2 = 4$, $2^3 = 8 \equiv 3$, $2^4 = 16 \equiv 1$. The cycle has length 4.

Since $100 = 4 \cdot 25$, $2^{100} = (2^4)^{25} \equiv 1^{25} = 1 \pmod 5$.

Answer

$0$

Answer

$1$

Answer

$2$

Answer

$3$

Problems & worked solutions

Problem 1 — Probability

Problem 2 — Geometry

Problem 3 — Number Theory