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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 · 2026-05-07

Three problems · Compound Interest, Logarithms, Combinatorics

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

#1 · Compound Interest·Easy
A credit card charges 24\% APR, compounded monthly. What's the effective annual rate (EAR)?
  1. A.Exactly 24\%
  2. B.About 26.8\%
  3. C.About 28.5\%
  4. D.About 124\%
Solution
The monthly rate is 24%/12=2%24\%/12 = 2\%. Compounding 12 times: (1.02)12≈1.2682(1.02)^{12} \approx 1.2682, so EAR ≈26.82%\approx 26.82\%.

The published 'APR' understates what you actually pay because it ignores compounding. EAR is the apples-to-apples comparison.
Why 'APR' and 'APY' differ on every credit card and savings account disclosure. APR is the simple rate; APY (or EAR) accounts for compounding. The gap grows with frequency: continuous compounding at 24\% gives e0.24−1≈27.1%e^{0.24} - 1 \approx 27.1\%.
#2 · Logarithms·Medium
If log⁡2x+log⁡2(x−6)=4\log_2 x + \log_2 (x - 6) = 4, what is xx?
  1. A.x=2x = 2
  2. B.x=8x = 8
  3. C.x=−2x = -2
  4. D.Both 88 and −2-2
Solution
Combine: log⁡2[x(x−6)]=4\log_2 [x(x-6)] = 4, so x(x−6)=24=16x(x-6) = 2^4 = 16, i.e. x2−6x−16=0x^2 - 6x - 16 = 0.

Factor: (x−8)(x+2)=0(x - 8)(x + 2) = 0, giving x=8x = 8 or x=−2x = -2.

Check the domain. log⁡2x\log_2 x requires x>0x > 0, and log⁡2(x−6)\log_2(x-6) requires x>6x > 6. Only x=8x = 8 satisfies both, so x=8x = 8 is the only valid solution.
Log equations love to produce extraneous roots. Always check that every candidate xx keeps every log⁡\log-argument strictly positive. Half of the 'tricky' log problems are exactly this domain-check.
#3 · Combinatorics·Hard
How many ways can you arrange the letters of 'BANANA'?
  1. A.6060
  2. B.120120
  3. C.360360
  4. D.720720
Solution
Total letters: 6. Repeats: 3 A's, 2 N's, 1 B.

Distinct permutations =6!3!⋅2!⋅1!=7206⋅2=72012=60= \dfrac{6!}{3! \cdot 2! \cdot 1!} = \dfrac{720}{6 \cdot 2} = \dfrac{720}{12} = 60.

The denominator divides out the orderings we'd otherwise double-count by treating the three A's as distinguishable.
The multinomial coefficient. If you write the A's as A1A2A3A_1 A_2 A_3, you get 6!=7206! = 720 arrangements — but 3!3! of them produce the same word once you erase the subscripts. Same logic for the two N's. Divide them out.

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