Which member of the family?

Number Theory Level 2

Which member of the sequence is prime?

$101, 10101, 1010101, 101010101,\dots$

The answer is 101.

1 solution

A Former Brilliant Member
Nov 23, 2019

If the number of $1's$ in any member of the sequence be even, then that member will be divisible by $101$ and hence composit. If the number of $1's$ be odd, say $2i+1$ , then that member can be expressed as $\dfrac{(10^{2i+1}-1)(10^{2i+1}+1)}{99}$ . For all $i$ greater than $0$ , both the factors are greater than $99$ . Hence that member is also composit. Therefore the only prime member of the sequence is $101$

Thank you, nice solution

Hana Wehbi - 1 year, 6 months ago

A more explicit way to show that $\frac{(10^{2i+1}-1)(10^{2i+1}+1)}{99}$ is an integer: $(10^{2i+1}-1)\equiv 1^{2i+1}-1 \equiv 0\,$ (mod $9$ ) and $(10^{2i+1}+1)\equiv (-1)^{2i+1}+1 \equiv 0\,$ (mod $11$ ), therefore $(10^{2i+1}-1)(10^{2i+1}+1)\equiv 0\,$ (mod $99$ ).

Ricardo Moritz Cavalcanti - 5 months ago

Which member of the family?

The answer is 101.

1 solution

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