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I know that the a number with more than 100 prime factors must be larger than

$2 ^ {100}$, so it must have more than 30 digits but i am having trouble with proof.

I was given Hint: every prime number is $≥ 2$. Can someone help me connect the two ideas.

Michael Hardy
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user240650
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1 Answers1

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Call the $100$ prime factors $p_1,p_2,p_3,\ldots,p_{100}$. Then \begin{align} p_1 & \ge 2 \\ p_2 & \ge 2 \\ p_3 & \ge 2 \\ & {}\,\,\, \vdots \\ p_{100} & \ge 2 \end{align} Therefore $$ p_1 \times p_2 \times p_3 \times \cdots \times p_{100} \ge \underbrace{2\times2\times2\times\cdots\times2}_\text{100 factors} $$ and that last number has $31$ digits.

Michael Hardy
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  • How do you call in english, when instead of proving "$A\implies B$" you prove "$B'\implies A'$" (obversion? Con-sth?), cause that is what you do here, isn't it? I think that may be confusing for OP. – quapka May 14 '15 at 22:42
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    Contrapositive. – Rellek May 14 '15 at 22:42
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    @quapka Proof by Contraposition, considers the contrapositive proposition. –  May 14 '15 at 23:39