Factor the repunit ## R_{6}=111111 ## into a product of primes

[Math100]

Homework Statement

Factor the repunit $R_{6}=111111$ into a product of primes.

Relevant Equations

None.

Consider the repunit $R_{6}=111111$.
Then $R_{6}=111111=1\cdot 10^{5}+1\cdot 10^{4}+1\cdot 10^{3}+1\cdot 10^{2}+1\cdot 10^{1}+1\cdot 10^{0}$.
Note that a positive integer $N=a_{m}10^{m}+\dotsb +a_{2}10^{2}+a_{1}10+a_{0}$ where $0\leq a_{k}\leq 9$ is
divisible by $7, 11$, and $13$ if and only if $7, 11$, and $13$ divide the
integer $M=(100a_{2}+10a_{1}+a_{0})-(100a_{5}+10a_{4}+a_{3})+(100a_{8}+10a_{7}+a_{6})-\dotsb$.
This means $a_{0}=a_{1}=a_{2}=a_{3}=a_{4}=a_{5}=1$.
Thus $M=(100+10+1)-(100+10+1)=0$.
Since $7, 11$, and $13$ divide $0$, it follows that $7, 11$, and $13$ divide the repunit $R_{6}$.
Observe that the sum of digits in $R_{6}$ is $1+1+1+1+1+1=6$.
This means $3\mid R_{6}$.
Thus $R_{6}=111111=3\cdot 7\cdot 11\cdot 13\cdot 37$.
Therefore, a product of primes in $R_{6}$ is $3\cdot 7\cdot 11\cdot 13\cdot 37$.

 

[fresh_42]

Homework Statement:: Factor the repunit $R_{6}=111111$ into a product of primes.
Relevant Equations:: None.

Consider the repunit $R_{6}=111111$.
Then $R_{6}=111111=1\cdot 10^{5}+1\cdot 10^{4}+1\cdot 10^{3}+1\cdot 10^{2}+1\cdot 10^{1}+1\cdot 10^{0}$.
Note that a positive integer $N=a_{m}10^{m}+\dotsb +a_{2}10^{2}+a_{1}10+a_{0}$ where $0\leq a_{k}\leq 9$ is
divisible by $7, 11$, and $13$ if and only if $7, 11$, and $13$ divide the
integer $M=(100a_{2}+10a_{1}+a_{0})-(100a_{5}+10a_{4}+a_{3})+(100a_{8}+10a_{7}+a_{6})-\dotsb$.
This means $a_{0}=a_{1}=a_{2}=a_{3}=a_{4}=a_{5}=1$.
Thus $M=(100+10+1)-(100+10+1)=0$.
Since $7, 11$, and $13$ divide $0$, it follows that $7, 11$, and $13$ divide the repunit $R_{6}$.
Observe that the sum of digits in $R_{6}$ is $1+1+1+1+1+1=6$.
This means $3\mid R_{6}$.
Thus $R_{6}=111111=3\cdot 7\cdot 11\cdot 13\cdot 37$.
Therefore, a product of primes in $R_{6}$ is $3\cdot 7\cdot 11\cdot 13\cdot 37$.

Correct.

Or $111111=111\cdot 1001=(3\cdot 37)\cdot (7\cdot 11 \cdot 13)$ also from former results.

 

[fresh_42]

The test with the 3-digit nonalternating sum works for $111$, according to a Wikipedia page about divisibility rules.

 

[pbuk]

That's a very clever solution, mine is much simpler:

• $3 | 111,111$ by the sum of digits rule: $\frac{111,111}{3} = 37,037$
• $\frac{37,037}{37} = 1,001$ by inspection
• It is well known that $11| 10^{(2k + 1)} + 1; \frac{1,001}{11} = 91$
• $91 = 7 \cdot 13$ by inspection