1

How can I approach this problem of comparison between these two numbers. Any hints please.

$A = 1000^{1000}$ or $B = 1\times 3\times 5\times \dots \times 1997$

Thomas Andrews
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mohamez
  • 1,521

3 Answers3

8

For any $a \geq 1$, we have $$(1000+a)(1000-a) < 1000^2$$ Hence, $$\prod_{a=1,3,5}^{999}(1000+a)(1000-a) < \prod_{a=1,3,5}^{999} 1000^2$$ This gives us $$1 \cdot 3 \cdot 5 \cdots 1995 \cdot 1997 \cdot 1999 < 1000^{1000}$$

Adhvaitha
  • 20,259
4

Hint: $(1000-x)\cdot (1000+x)<1000\cdot 1000$.

3

Apply GM $\le$ AM to the $999$ numbers $1, 3, 5, \ldots, 1997$, we have

$$1 \times 3 \times \cdots \times 1997 \le 999^{999} < 1000^{1000}$$

achille hui
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