Fractions at 2015

Number Theory Level 2

$\large \frac {2}{2015} \times \frac {6}{2013} \times \frac {10}{2011} \times ... \times \frac {4022}{5} \times \frac {4026}{3} \times \frac {4030}{1}$

What is the remainder if the expression above is divided by 7?

The answer is 1.

1 solution

Tijmen Veltman
May 31, 2015

We can rearrange to find:

$\frac2{2015}\times\frac6{2013}\times\frac{10}{2011}\times\ldots\times\frac{4022}5\times\frac{4026}3\times\frac{4030}1$

$=\frac21\times\frac63\times\frac{10}5\times\ldots \times\frac{4022}{2011}\times\frac{4026}{2013}\times\frac{4030}{2015}$

$=2^{\frac{2016}2}=2^{1008}$ .

Since $2^3=8\equiv 1 (\text{mod }7)$ , we get $2^{1008}=(2^3)^{336}\equiv \boxed{1} (\text{mod }7)$ .

Moderator note:

Yes, that's one way to solve without resorting to Fermat's little theorem. For the sake of variety, can you solve this via Fermat's little theorem?

Same method :)

Nihar Mahajan - 6 years ago

If we use Fermat's little theorem, we get (a weaker result) that $2^6\equiv 1 (\text{mod }7)$ and hence $2^{1008}=(2^6)^{168}\equiv\boxed{1}(\text{mod }7)$ .

Tijmen Veltman - 6 years ago

hey guys, how can i learn mod? i never know it until i'm a brilliant member. thx

Ardianto Kurniawan - 6 years ago

Fractions at 2015

The answer is 1.

1 solution

Moderator note:

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