Exponents Inequality Bound

Question

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

Accepted Answer

D. Left-hand side: $$\prod_{k=1}^{20} (3^k)^k = 3^{\sum_{k=1}^{20} k^2}$$ Sum of squares formula: $\sum_{k=1}^{20} k^2 = \frac{20 imes 21 imes 41}{6} = 2870$. So $ ext{LHS} = 3^{2870}$. Right-hand side: $$\prod_{j=21}^{20+m} 3^j = 3^{\sum_{j=21}^{20+m} j}$$ Sum of AP from $21$ to $20+m$ with $m$ terms: $$ ext{Sum} = \frac{m}{2} [21 + (20+m)] = \frac{m(41+m)}{2}$$ We require $ ext{LHS} < ext{RHS} \implies 2870 < \frac{m(41+m)}{2} \implies m(41+m) > 5740$. Testing options: For $m = 57$: $57 imes 98 = 5586 < 5740$. For $m = 58$: $58 imes 99 = 5742 > 5740$. Thus, the smallest natural number $m$ is $58$.

Answer

57

Answer

56

Answer

59

Answer

58

Exponents Inequality Bound

Explanation

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