Daily · 2026-05-06

Three problems · Probability, Geometry, Number Theory

One bite-sized math problem set for the day. Read the statement, think it through, then expand the solution to check your reasoning.

#1 · Probability·Medium

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.)

A. $\approx 6\%$
B. $\approx 23\%$
C. $\approx 50\%$
D. $\approx 90\%$

Solution

Compute the complement — probability all 23 birthdays are distinct:

P(\text{all distinct}) = \frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365} \cdots \frac{343}{365} \approx 0.493

P(\text{at least one match}) = 1 - 0.493 \approx 0.507

, just over 50\%.

The 'birthday paradox' isn't a paradox — it's a counting result that violates intuition. With 23 people there are

\binom{23}{2} = 253

pairs, and each pair has a

1/365

chance of matching. Pairs scale quadratically; people scale linearly.

#2 · Geometry·Easy

A square is inscribed in a circle of radius

r

. What is the area of the square in terms of

r

A. $r^2$
B. $2r^2$
C. $\pi r^2$
D. $4r^2$

Solution

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

A two-line problem if you remember that a square's diagonal is

\sqrt{2}

times its side. The trick is recognizing that the inscribed-square's diagonal *is* the circle's diameter — which is true precisely because all four vertices lie on the circle.

#3 · Number Theory·Medium

What is the remainder when

7^{100}

is divided by

5

A. $0$
B. $1$
C. $2$
D. $3$

Solution

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

Modular arithmetic turns 'compute a 100-digit number, then divide' into 'find a small cycle, then index into it'. This is exactly how RSA-style cryptography handles huge exponents — find Euler's totient, reduce modulo it, done.

Get tomorrow's drop at 6am — and start tracking your streak.

Create an account