Question

The last two digits of the number $3^{400}$ are

• A. 81
• B. 43
• C. 29
• D. 01

Answer 1

The correct option is D

01

We want to find the last two digits of $3^{400}$. If we can expand in terms of some number which is a multiple of 10 or 100, we can easily find the last two digits. We know $3^4$ = 81.

$\Rightarrow (3{)}^{400}={81}^{100}$

$\Rightarrow (3{)}^{400}=(80+1{)}^{100}$

$\Rightarrow (80+1{)}^{100}={}^{100}{C}_0(80{)}^{100}+{}^{100}{C}_1(80{)}^{499}+{}^{100}{C}_2(80{)}^{498}+..........+{}^{100}{C}_{99}(80{)}^1+{}^{100}{C}_{100}$

$\Rightarrow (80+1{)}^{100}={}^{100}{C}_0(80{)}^{100}+{}^{100}{C}_1(80{)}^{499}+{}^{100}{C}_2(80{)}^{498}+..........+100\times (80{)}^1+1$

The terms other than the last two terms have more than two hundreds.So the only terms which will decide the last two digits are the last two terms.

$(80+1{)}^{100}={}^{100}{C}_0(80{)}^{100}+{}^{100}{C}_1(80{)}^{499}+{}^{100}{C}_2(80{)}^{498}+...........+8000+1$

$\Rightarrow$ The last two digits are 0 and 1