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There is a sequence of numbers: $7,97,997,9997,99997,999997 ......$ Find the tenth term of this number which is not a prime.

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Comments

Since it is obviously not an infinite sequence of primes, it should be fastest to check the first few terms.

$9997 = 13\times 769$

Edit 1: By Fermat's Little Theorem, if $9997$ is prime, then $2^{9997} \equiv 2\ (\text{mod } 9997).$ But $2^{9997} \text{ mod } 9997 = 8192.$ You can check this with an exponential expansion of $2^{9997},$ or with Euler's theorem; both are tedious. In the words of Fermat himself, I would provide the proof, if I were not afraid to be too long.

Edit 2: the note now asks for the tenth composite in the sequence; I have a Java solution which shows it to be $9999999999997$ (term $13$ , implying there are at least $9$ more consecutive composites after the first) but proving this seems extremely tedious as I am only familiar with FLT and Euler's theorem. Does anyone have an idea? I know there is at least one more prime in the sequence, for example term $17.$

Caleb Townsend - 6 years, 3 months ago

By inspection?

Yogesh Verma - 6 years, 3 months ago

Yes, it is by far the fastest way to solve this problem, but I have added a FLT solution.

Caleb Townsend - 6 years, 3 months ago

Eg if 999999....999997 (n 9's) is the first term which is not a prime factor, then that term is $n+1$ .

Bryan Lee Shi Yang - 6 years, 3 months ago

Note edited, sorry.

Bryan Lee Shi Yang - 6 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$