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

Step 1: Let \(E(m) = \frac{m(m+41)}{2}\) be the RHS exponent; we want the smallest \(m\) with \(E(m) > 2870\). Note the increment \(E(m) - E(m-1) = m + 20\), an easy number to add each time.
Step 2: Start from a convenient anchor, \(E(50) = \frac{50 \times 91}{2} = 2275\), and add the increments \(m+20\) one at a time: \(E(51)=2275+71=2346\), \(E(52)=2346+72=2418\), \(E(53)=2418+73=2491\), \(E(54)=2491+74=2565\), \(E(55)=2565+75=2640\), \(E(56)=2640+76=2716\), \(E(57)=2716+77=2793\), \(E(58)=2793+78=2871\).
Step 3: \(E(57)=2793\) is still below \(2870\), but \(E(58)=2871\) crosses it. So the smallest qualifying \(m\) is 58.
\[ \boxed{m=58} \]