The last Prime.

Number Theory Level 4

$\large {\underbrace{77777\ldots77 }_{7919 \text{ times}} }^2$

Find the remainder when the above number is divided by 7919.

The answer is 49.

1 solution

Akshat Sharda
Mar 22, 2016

$\underbrace{77777\ldots 77^2}_{7919 \text{ times}} \pmod{7919}$

The above number can be written as,

$\left(7\cdot \frac{10^{7919}-1}{9}\right)^2 \pmod{7919}$

We know,

$10^{\phi(7919)}=10^{7918}\equiv 1\pmod {7919} \\ \therefore \left( 7 \cdot \frac{10-1}{9} \right)^2\equiv 7^2\equiv \boxed{49} \pmod{7919}$

$• 7919 \text{ is a prime number.}$

Nice written solution.

Although you have to justify the modulo with fractions, which is not always true. $\frac{18} {2} \equiv 9 \equiv 3 \pmod{6}$ $\frac{18} {2} \equiv \frac{0} {2} \equiv 0 \pmod{6}$

Abdelhamid Saadi - 5 years, 2 months ago

Aah! Can you please explain it? Maybe we should have co-prime numbers?

Akshat Sharda - 5 years, 2 months ago

Well written solution +1

A Former Brilliant Member - 5 years, 2 months ago

Thanks .... :-) Nice problem

Akshat Sharda - 5 years, 2 months ago

Can you please explain hw have you re-written $77777.....77^2$ ?

Chirayu Bhardwaj - 5 years, 2 months ago

@Chirayu Bhardwaj – see , the number 10^7919 -1 gives 999.... and after dividing by 9 you get 1111..... and hence multiplying it by 7 we simply get 777....

A Former Brilliant Member - 5 years, 2 months ago

@A Former Brilliant Member – Oh ! did'nt see that. Ty

Chirayu Bhardwaj - 5 years, 2 months ago

The last Prime.

The answer is 49.

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