Question

For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is:

• A. \(58\)
• B. \(59\)
• C. \(57\)
• D. \(56\)

Hint:

When both sides of an inequality are powers of the same base greater than 1, you can drop the base and compare the exponents directly. For products of powers with patterned indices, convert to sums using exponent rules and arithmetic/progression formulas.

Answer 1

The Correct Option is A

Approach: Everything is a power of 3, so take logs base 3 and the whole problem collapses to comparing two exponents — a sum of squares on the left and an arithmetic-progression sum on the right.

Step 1: LHS exponent. Since \(a_k = 3^k\), \((a_k)^k = 3^{k^2}\), so \[ \text{LHS} = 3^{1^2 + 2^2 + \cdots + 20^2} = 3^{\sum_{k=1}^{20} k^2}. \] Using \(\sum k^2 = \tfrac{n(n+1)(2n+1)}{6}\) with \(n=20\): \[ \frac{20 \cdot 21 \cdot 41}{6} = 2870. \] So LHS \(= 3^{2870}\).

Step 2: RHS exponent. \[ \text{RHS} = 3^{21 + 22 + \cdots + (20+m)}. \] This is an AP of \(m\) terms, first \(21\), last \(20+m\): \[ \text{sum} = \frac{m}{2}\big(21 + 20 + m\big) = \frac{m(m+41)}{2}. \]

Step 3: Compare exponents. Base \(3 > 1\), so the inequality on the powers is the inequality on the exponents: \[ 2870 < \frac{m(m+41)}{2} \implies m(m+41) > 5740. \]

Step 4: Test integers near the boundary. No need for the quadratic formula — just probe: \[ m=57:\ 57 \cdot 98 = 5586 < 5740 \ (\text{fails}), \] \[ m=58:\ 58 \cdot 99 = 5742 > 5740 \ (\text{works}). \] So the smallest \(m\) is \(58\).

\[ \boxed{m = 58} \] (option 1).

Answer 2

Approach: Reduce to comparing exponents of 3, then instead of testing integers near a guessed boundary, solve the resulting quadratic inequality exactly using the quadratic formula.

Step 1: Exponents. \(\text{LHS} = 3^{\sum_{k=1}^{20}k^2} = 3^{2870}\) (using \(\sum k^2 = \frac{20\cdot21\cdot41}{6}=2870\)). \(\text{RHS} = 3^{21+22+\cdots+(20+m)} = 3^{\frac{m(m+41)}{2}}\).

Step 2: Set up the inequality. We need \(\frac{m(m+41)}{2} > 2870\), i.e. \(m^2 + 41m - 5740 > 0\).

Step 3: Solve exactly. The boundary root is
\[ m = \frac{-41 + \sqrt{41^2 + 4(5740)}}{2} = \frac{-41+\sqrt{24641}}{2}. \]
Since \(157^2 = 24649\), \(\sqrt{24641}\) is just under \(157\) (about \(156.97\)), so
\[ m > \frac{-41+156.97}{2} \approx 57.99. \]
The smallest integer above this boundary is \(m = 58\).

\[ \boxed{m = 58} \] option (1).