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

Given \(a_k = 3^k\). Step 1: Simplify the LHS. \[ (a_k)^k = (3^k)^k = 3^{k^2}. \] So, \[ \text{LHS} = (a_1)^1 (a_2)^2 \dots (a_{20})^{20} = 3^{1^2} \cdot 3^{2^2} \cdots 3^{20^2} = 3^{1^2 + 2^2 + \dots + 20^2}. \] Use the formula \[ 1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}. \] For \(n=20\): \[ \sum_{k=1}^{20} k^2 = \frac{20 \cdot 21 \cdot 41}{6} = 10 \cdot 7 \cdot 41 = 2870. \] Thus \[ \text{LHS} = 3^{2870}. \]
Step 2: Simplify the RHS. \[ \text{RHS} = a_{21} a_{22} \cdots a_{20+m} = 3^{21} \cdot 3^{22} \cdots 3^{20+m} = 3^{21 + 22 + \dots + (20+m)}. \] The exponent is the sum of an arithmetic progression: First term \(= 21\), last term \(= 20 + m\), number of terms \(= m\). Sum: \[ 21 + 22 + \dots + (20+m) = \frac{m}{2}\bigl(21 + (20+m)\bigr) = \frac{m}{2}(41 + m). \] So \[ \text{RHS} = 3^{\frac{m(m+41)}{2}}. \] 
Step 3: Set up the inequality. We need \[ 3^{2870}<3^{\frac{m(m+41)}{2}}. \] Since the base \(3>1\), compare exponents: \[ 2870<\frac{m(m+41)}{2}, \] \[ 5740<m^2 + 41m, \] \[ m^2 + 41m - 5740>0. \] Solve the quadratic equation \[ m^2 + 41m - 5740 = 0. \] \[ m = \frac{-41 \pm \sqrt{41^2 + 4\cdot 5740}}{2} = \frac{-41 \pm \sqrt{1681 + 22960}}{2} = \frac{-41 \pm \sqrt{24641}}{2}. \] Now \[ 157^2 = 24649,\quad 156^2 = 24336, \] so \[ \sqrt{24641} \approx 156.97. \] Positive root: \[ m \approx \frac{-41 + 156.97}{2} \approx \frac{115.97}{2} \approx 57.98. \] Thus the quadratic is positive for \(m>57.98\), so the smallest natural number satisfying the inequality is \[ m = 58. \] 
Step 4: Quick verification around the boundary. Check \(m=57\): \[ \frac{m(m+41)}{2} = \frac{57\cdot 98}{2} = \frac{5586}{2} = 2793<2870 \quad\Rightarrow\quad \text{RHS exponent}<\text{LHS exponent}. \] Check \(m=58\): \[ \frac{58\cdot 99}{2} = \frac{5742}{2} = 2871>2870 \quad\Rightarrow\quad \text{RHS exponent}>\text{LHS exponent}. \] So \(m=58\) is indeed the smallest valid natural number. Therefore, the required smallest \(m\) is \[ \boxed{58}. \]