Daily Math Problems — May 7, 2026: Compound Interest, Logarithms, Combinatorics

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Problems & worked solutions

International

Problem 1 — Compound Interest

A credit card charges $24 %$ APR, compounded monthly. What's the effective annual rate (EAR)?

Show answer & worked solution

A. $Exactly 24 %$
B. $About 26.8 %$ ✓ correct
C. $About 28.5 %$
D. $About 124 %$

The monthly rate is $24 % / 12 = 2 %$ . Compounding 12 times: $(1.02)^{12} \approx 1.2682$ , so EAR $\approx 26.82 %$ .

The published 'APR' understates what you actually pay because it ignores compounding. EAR is the apples-to-apples comparison.

International

Problem 2 — Logarithms

If $\log_{2} x + \log_{2} (x - 6) = 4$ , what is $x$ ?

Show answer & worked solution

A. $x = 2$
B. $x = 8$ ✓ correct
C. $x = - 2$
D. $Both 8 and - 2$

Combine: $\log_{2} [x (x - 6)] = 4$ , so $x (x - 6) = 2^{4} = 16$ , i.e. $x^{2} - 6 x - 16 = 0$ .

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

Check the domain. $\log_{2} x$ requires $x > 0$ , and $\log_{2} (x - 6)$ requires $x > 6$ . Only $x = 8$ satisfies both, so $x = 8$ is the only valid solution.

International

Problem 3 — Combinatorics

How many ways can you arrange the letters of 'BANANA'?

Show answer & worked solution

A. $60$ ✓ correct
B. $120$
C. $360$
D. $720$

Total letters: 6. Repeats: 3 A's, 2 N's, 1 B.

Distinct permutations $= \frac{6!}{3! \cdot 2! \cdot 1!} = \frac{720}{6 \cdot 2} = \frac{720}{12} = 60$ .

The denominator divides out the orderings we'd otherwise double-count by treating the three A's as distinguishable.

Problems & worked solutions

International

Problem 1 — Compound Interest

A credit card charges $24 %$ APR, compounded monthly. What's the effective annual rate (EAR)?

Show answer & worked solution

A. $Exactly 24 %$
B. $About 26.8 %$ ✓ correct
C. $About 28.5 %$
D. $About 124 %$

The monthly rate is $24 % / 12 = 2 %$ . Compounding 12 times: $(1.02)^{12} \approx 1.2682$ , so EAR $\approx 26.82 %$ .

The published 'APR' understates what you actually pay because it ignores compounding. EAR is the apples-to-apples comparison.

International

Problem 2 — Logarithms

If $\log_{2} x + \log_{2} (x - 6) = 4$ , what is $x$ ?

Show answer & worked solution

A. $x = 2$
B. $x = 8$ ✓ correct
C. $x = - 2$
D. $Both 8 and - 2$

Combine: $\log_{2} [x (x - 6)] = 4$ , so $x (x - 6) = 2^{4} = 16$ , i.e. $x^{2} - 6 x - 16 = 0$ .

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

Check the domain. $\log_{2} x$ requires $x > 0$ , and $\log_{2} (x - 6)$ requires $x > 6$ . Only $x = 8$ satisfies both, so $x = 8$ is the only valid solution.

International

Problem 3 — Combinatorics

How many ways can you arrange the letters of 'BANANA'?

Show answer & worked solution

A. $60$ ✓ correct
B. $120$
C. $360$
D. $720$

Total letters: 6. Repeats: 3 A's, 2 N's, 1 B.

Distinct permutations $= \frac{6!}{3! \cdot 2! \cdot 1!} = \frac{720}{6 \cdot 2} = \frac{720}{12} = 60$ .

The denominator divides out the orderings we'd otherwise double-count by treating the three A's as distinguishable.