Question

${53}^{53}-{33}^{13}$ is divisible by

• A. $3$
• B. $10$
• C. $33$
• D. $53$

Answer 1

Answer: (B)

$10$

The above expression can be written as,
$(50+3{)}^{53}-(30+3{)}^{13}$

$={(}_0^{53}\text{C}\left(50{)}^{53}\right(3{)}^0{+}_1^{53}\text{C}(50{)}^{52}$

$=(3{)}^1+......{+}_{52}^{53}\text{C}(50{)}^1\left(3{)}^{52}{+}_{53}^{53}\text{C}\right(50{)}^0\left(3{)}^{53}\right)-{(}_0^{53}\text{C}\left(30{)}^{53}\right(3{)}^0{+}_1^{53}\text{C}(30{)}^{52}$

$=(3{)}^1+......{+}_{52}^{53}\text{C}(30{)}^1\left(3{)}^{52}{+}_{53}^{53}\text{C}\right(30{)}^0\left(3{)}^{53}\right)$

Here we can see that all the terms in both the expression except the nth term has either $50$ or $30$,
Now lets look at the $n^{th}$ term,
$={(}_{53}^{53}\text{C}\left(50{)}^0\right(3{)}^{53}{-}_{53}^{53}\text{C}\left(30{)}^0\right(3{)}^{53})$

$=3^{53}-3^{13}$

$=3^{13}{3}^{40}-3^{13}$

$=3^{13}(3^{40}-1)$
We know, $3^{40}$ can be written as, $\left(\right(3^2{)}^2{)}^{10}$
$=\left(\right(9{)}^2{)}^{10}$
$=(81{)}^{10}$
The last digit of $81$ is $1$ so whatever power it is raised by, its last digit will always remain $1$,
Therefore our expression, $(3^{40}-1)$,
will always have $0$ as its last digit,
Now we can see that all the terms are having $0$ at the end, so they will be divisible by $10$.