What are last three digits of 2003^2002^2001

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asked Jul 21, 2014 in Quantitative Ability by Jamal
retagged Jul 21, 2014 by Campusgate

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We have to find the last three digits in the expression ${{{2003}^{{{2002}^{2001}}}}}$.  So we have to find the remainder when this expression is divided by 1000.
$\displaystyle\frac{{{{2003}^{{{2002}^{2001}}}}}}{{1000}}$ = $\displaystyle\frac{{{{3}^{{{2002}^{2001}}}}}}{{1000}}$.  
Euler totient theorem says ${\left[ {\displaystyle\frac{{{a^{\phi (N)}}}}{N}} \right]_{{\mathop{\rm Re}\nolimits} m}} = 1$.
Here ${\phi (N)}$ is the number of coprimes less than that of N.
$\phi (N) = N\left( {1 - \frac{1}{a}} \right)\left( {1 - \frac{1}{b}} \right)....$ where N = ${a^p}.{b^q}.{c^r}...$
Now $\phi (1000) = 1000\left( {1 - \displaystyle\frac{1}{2}} \right)\left( {1 - \displaystyle\frac{1}{5}} \right) = 400$
So we need to compute $\displaystyle\frac{{{{2002}^{2001}}}}{{400}} = \frac{{{{2002}^{2001}}}}{{16 \times 25}} = \frac{{{{2002}^{1997}}}}{{25}}$
=$\displaystyle\frac{{{{({2^{10}})}^{199}}{{.2}^7}}}{{25}} = ( - 1).128 =  - 128 = 22$
So remainder when $\displaystyle\frac{{{{2002}^{2001}}}}{{400}}$ is 22 $ \times $16=352.
Now $\displaystyle\frac{{{{2003}^{{{2002}^{2001}}}}}}{{1000}} = \frac{{{3^{{{2002}^{2001}}}}}}{{1000}} = \frac{{{{\left( {{3^{400}}} \right)}^K}{{.3}^{352}}}}{{1000}} = \frac{{{3^{352}}}}{{1000}}$
Now ${3^{352}} = {9^{176}} = {\left( {10 - 1} \right)^{176}} = .....{}^{176}{C_2}{(10)^2} - {}^{176}{C_1}(10) + {1^{176}}$
=0-760+1 = 241
answered Jul 21, 2014 by Campusgate
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