Primes reciprocated

Number Theory Level 4

Find the sum of all primes $P$ such that the decimal expansion of $\frac{1}{P}$ has a fundamental period $5$ .

Details and Assumptions:

As an explicit example, $\frac{1}{3}=0.33333\ldots$ has fundamental period $1$ and $\frac{1}{101}=0.00990099\ldots$ has fundamental period $4$ .

The answer is 312.

2 solutions

Abdeslem Smahi
Aug 7, 2015

Let $P$ be prime so:

If $\frac{1}{P}=0.\overline{a} \implies 9=a.p \implies p=3$

If $\frac{1}{P}=0.\overline{ab} \implies 99=\overline{ab}.p \implies p=11$

If $\frac{1}{P}=0.\overline{abc}\implies 999=\overline{abc}.p \implies p=37$

If $\frac{1}{P}=0.\overline{abcd}\implies 9999=\overline{abcd}.p \implies p=101$

If $\frac{1}{P}=0.\overline{abcde}\implies 99999=\overline{abcde}.p \implies p=41 ,or,p=271$

So the answers are $271,41$

$(a,b,c,d,e,p)=(3,3,3,3,3,3)$ also works. There are no constraints on $a,b,c,d,e$ other than that they're integers in $[0,9]$ , so $3+41+271=315$ is the answer.

mathh mathh - 5 years, 10 months ago

Those who answered 315 have been marked correct. The problem has been edited and the correct answer is now 312.

Calvin Lin Staff - 5 years, 10 months ago

Shouldn't p. abcde = 9999(4 nines (10^5-1))

batman rules - 1 year, 4 months ago

Advaith Kumar
May 12, 2020

the simplest way to solve: notice that 1/p = 0.abcdeabcde.... wher p is a prime 10^5 / p = abcde.abcdeabcde.... => 10^5 - 1/p = abcde So p | 99999 now test p = 3,41,271 ; 41,271 works => 271 + 41 = 312 is the answer

Primes reciprocated

The answer is 312.

2 solutions

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