Question

Let $n>1$ be an integer. Which of the following sets of numbers necessarily contains a multiple of $3$ ?

• A. $n^{19}-1,n^{19}+1$
• B. $n^{19},n^{38}-1$
• C. $n^{38},n^{38}+1$
• D. $n^{38},n^{19}-1$

Answer 1

The correct option is B $n^{19},n^{38}-1$

$n$ can be either of the form $3k$ or $3k+1$ or $3k+2$ for some $k\in N$

Case $\left(1\right):$
Let $n=3k$
$n^{19}=3^{19}{k}^{19}$
and $n^{38}=3^{38}{k}^{38}$
Then $n^{19}-1,n^{19}+1,n^{38}-1,n^{38}-1$ are not multiples of $3.$

Case $\left(2\right):$
Let $n=3k+1$
$n^{19}=(1+3k{)}^{19}$
$=1+19\cdot 3k+\cdots$
$=1+3p_1,p_1\in N$

$n^{38}=(1+3k{)}^{38}$
$=1+38\cdot 3k+\cdots$
$=1+3p_2,p_2\in N$
Then $n^{19}-1$ and $n^{38}-1$ are multiples of $3$
and $n^{19}+1$ and $n^{38}+1$ are not multiples of $3.$

Case $\left(3\right):$
Let $n=3k+2$
$n^{19}=(2+3k{)}^{19}$
$=2^{19}+19\cdot 2^{18}\cdot 3k+\cdots$
$=2^{19}+3q_1,q_1\in N$
$=(3-1{)}^{19}+3q_1$

$n^{38}=(2+3k{)}^{38}$
$=2^{38}+38\cdot 2^{37}\cdot 3k+\cdots$
$=2^{38}+3q_2,q_2\in N$
$=(3-1{)}^{38}+3q_2$

Now, $(3-1{)}^n=3^n+n{3}^{n-1}(-1{)}^1+\cdots +(-1{)}^n$
Therefore, when $n$ is odd, the remainder when $(3-1{)}^n$ is divided by $3$ is $2.$
When $n$ is even, the remainder when $(3-1{)}^n$ is divided by $3$ is $1.$

$\therefore n^{19}$ leaves a remainder of $2$ when divided by $3.$
$\therefore n^{19}+1$ is a multiple of $3$ but $n^{19}-1$ is not.
Also, $n^{38}$ leaves a remainder of $1$ when divided by $3.$
$\therefore n^{38}-1$ is a multiple of $3$ but $n^{38}+1$ is not.

Taking different pairs in combination, we conclude that $\{n^{19},n^{38}-1\}$ always contains a multiple of $3$ for any $n\in N$