Question

Find the last two digits of $3^{1997}$.

• A. $67$
• B. $63$
• C. $80$
• D. $56$

Answer 1

The correct option is B $63$

This is same as asking what is remainder when $3^{1997}÷100$
$3^4\equiv 81mod100$
$3^8\equiv 61mod100$
$3^{12}\equiv 41mod100$
$3^{16}\equiv 21mod100$
$3^{20}\equiv 1mod100$

Now, $3^{40},3^{60},3^{80},3^{100},....,3^{1980}$ all are $\equiv 1mod100$

We know $3^{16}\equiv 21mod100$

$3^{17}\equiv 21\times 3mod100$

$3^{17}\equiv 63mod100$

$\therefore 3^{1997}\equiv 3^{1980}\times 3^{17}$

since, $3^{1980}\equiv 1mod100$

and $3^{17}\equiv 63mod100$

$\therefore 3^{1997}\equiv 63mod100$

$\therefore$ Last two digit is 63