Proof into finding

Number Theory Level 5

Find last digit of positive integer $m$ if

$\large\ 1991 | { 12 }^{ m } + { 9 }^{ m } + { 8 }^{ m } + { 6 }^{ m }$

1 solution

Kushal Bose
Sep 23, 2016

$2^{2 m}+9^m+8^m+6^m$

Let $2^m=p ,3^m=q$ .Then the expression becomes $p^2 q+q^2+p^3+pq \\ =(p^2+q)(p+q) \\ =(4^m+3^m)(2^m+3^m)$

Factrorizing $1991=11 \times 181$

Here $11,181$ both are prime numbers so applying Euler's Theorem :

$2^{10} \equiv 1 \pmod{11}$ and $3^{10} \equiv 1 \pmod{11}$

$2^5 \equiv 10 \pmod{11}$ and $3^5 \equiv 1 \pmod{11}$

$2^5+3^5 \equiv 0 \pmod{11}$

Putting $m=5$ in another part becomes divisible by $181$ .

If $m!=5$ is not a solution then we have to take \(m=10+5\=15) and continue so on.

So, minimum value will be $5$

This is a good start. However, your explanation has slight gaps in it. The issues are:

You have to show that there exists an integer that satisfies the condition
You have to show that all integers which satisfy the condition has a last digit of 5

Keep writing more solutions and you will get the hang of this!

Calvin Lin Staff - 4 years, 8 months ago