Question

Find the sum of last three digits of the number $N=7^{100}-3^{100}$.

Answer 1

Given that,

$7^{100}-3^{100}=?$

$7^{100}=(10-3{)}^{100}$

We know that,

$(x+a{)}^n=\sum _{r=0}^n{}^n{C}_r{x}^{n-r}{a}^r$

$(10+(-3){)}^{100}=\sum _{r=0}^{100}{}^{100}{C}_r\left(10{)}^{100-r}\right(-3{)}^r$

If $100-r=3$ last digit$=000$,

$r=97,r>97$

Now,

$7^{100}-3^{100}={}^{100}{C}_{98}{10}^2(-3{)}^{98}+{}^{100}{C}_{9}9{10}^1(-3{)}^{99}+{}^{100}{C}_{100}\left(10{)}^0\right(-3{)}^{100}-3^{100}$

$=\frac{100\times 99}{2}\times 100\times (-3{)}^{98}+100\times 10\times (-3{)}^{99}+1\times 1\times (3{)}^{100}-3^{100}$

$=5\times 99\times 1000\times (3{)}^{98}+1000(-3{)}^{99}$

Hence, the sum of last three digit number $=0$

This is the answer.