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A daily ritual for math-brained humans. One fresh quiz drops every morning at 6am — three problems, sized for a coffee break.

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Daily · 2026-05-05

Three problems · Combinatorics, Compound Interest, Exponential Functions

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

#1 · Combinatorics·Medium
An airplane row has 3 seats: window, middle, aisle. You're seating 3 people: Alice prefers the window, Bob prefers the aisle, Carol has no preference. In how many of the 3!=63! = 6 possible orderings is everyone happy (i.e., neither Alice nor Bob ends up away from their preferred seat)?
  1. A.11
  2. B.22
  3. C.33
  4. D.66
Solution
Alice must take the window seat (1 way). Bob must take the aisle seat (1 way). Carol takes the middle (1 way). Only 1×1×1=11 \times 1 \times 1 = 1 ordering satisfies both preferences.

More generally — when each person's preference is for a distinct seat, only the permutation that matches preferences exactly is 'fully happy'.
A toy version of a real airline-software problem: assigning seats to optimize stated preferences. With more seats and more passengers it becomes a classic bipartite matching problem.
#2 · Compound Interest·Medium
How long, in years, does it take for an investment to double at 7% annual interest, compounded annually? (Use the closest answer.)
  1. A.About 7 years
  2. B.About 10 years
  3. C.About 14 years
  4. D.About 70 years
Solution
We need tt such that (1.07)t=2(1.07)^t = 2, i.e. t=ln⁡2ln⁡1.07≈0.6930.0677≈10.24t = \dfrac{\ln 2}{\ln 1.07} \approx \dfrac{0.693}{0.0677} \approx 10.24 years.

The banker's shortcut: the Rule of 72 — divide 72 by the interest rate. 72/7≈10.372 / 7 \approx 10.3. Quick, accurate enough.
The Rule of 72 is one of the most useful pieces of finance math you can carry around. It's a logarithm in disguise — ln⁡2≈0.693\ln 2 \approx 0.693, and dividing by a small rr behaves close enough to dividing by ln⁡(1+r)≈r\ln(1+r) \approx r when rr is small.
#3 · Exponential Functions·Hard
Solve for xx: 2x+1=3x\quad 2^{x+1} = 3^{x}.
  1. A.x=log⁡32x = \log_3 2
  2. B.x=ln⁡2ln⁡3−ln⁡2x = \dfrac{\ln 2}{\ln 3 - \ln 2}
  3. C.x=log⁡23x = \log_2 3
  4. D.x=ln⁡2ln⁡3x = \dfrac{\ln 2}{\ln 3}
Solution
Take ln⁡\ln of both sides: (x+1)ln⁡2=xln⁡3(x+1) \ln 2 = x \ln 3.

Expand: xln⁡2+ln⁡2=xln⁡3x \ln 2 + \ln 2 = x \ln 3.

Collect xx: ln⁡2=x(ln⁡3−ln⁡2)\ln 2 = x(\ln 3 - \ln 2).

So x=ln⁡2ln⁡3−ln⁡2≈0.6930.405≈1.71x = \dfrac{\ln 2}{\ln 3 - \ln 2} \approx \dfrac{0.693}{0.405} \approx 1.71.
Whenever the unknown is in two different exponential bases, ln⁡\ln both sides and turn it into a linear equation in xx. The answer is dimensionally a logarithm, but it's a single real number — about 1.71 here.

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