Largest 9-Digit Prime Number

From ProofWiki
Jump to navigation Jump to search

Theorem

The largest prime number with $9$ digits is $999 \, 999 \, 937$.


Proof

Consider the numbers $\sqbrk {999 \, 999 \, 9ab}$.

Since $999 \, 999 \, 000$ is divisible by $2, 3, 5, 7, 11, 13$,

if $\sqbrk {9ab}$ is divisible by these primes, so is $\sqbrk {999 \, 999 \, 9ab}$.

After this elimination the only $\sqbrk {ab} > 37$ that remains are:

$41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97$

We have:

\(\ds 999 \, 999 \, 941\) \(=\) \(\ds 113 \times 149 \times 59 \, 393\)
\(\ds 999 \, 999 \, 943\) \(=\) \(\ds 5 \, 623 \times 177 \, 841\)
\(\ds 999 \, 999 \, 947\) \(=\) \(\ds 163 \times 6 \, 134 \, 969\)
\(\ds 999 \, 999 \, 953\) \(=\) \(\ds 29 \times 31 \times 773 \times 1 \, 439\)
\(\ds 999 \, 999 \, 961\) \(=\) \(\ds 3 \, 673 \times 272 \, 257\)
\(\ds 999 \, 999 \, 967\) \(=\) \(\ds 3 \, 257 \times 307 \, 031\)
\(\ds 999 \, 999 \, 971\) \(=\) \(\ds 193 \times 5 \, 181 \, 347\)
\(\ds 999 \, 999 \, 977\) \(=\) \(\ds 2 \, 971 \times 336 \, 587\)
\(\ds 999 \, 999 \, 983\) \(=\) \(\ds 337 \times 2 \, 967 \, 359\)
\(\ds 999 \, 999 \, 989\) \(=\) \(\ds 4 \, 327 \times 231 \, 107\)
\(\ds 999 \, 999 \, 991\) \(=\) \(\ds 67 \times 14 \, 925 \, 373\)
\(\ds 999 \, 999 \, 997\) \(=\) \(\ds 71 \times 2 \, 251 \times 6 \, 257\)

And we do have $999 \, 999 \, 937$ is prime.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Largest_9-Digit_Prime_Number&oldid=464823"