# Remainders again

• Mar 26th 2010, 07:33 PM
banku12
Remainders again
find the remainder when $1992^{1998} - 1955^{1998} - 1938^{1998} + 1901^{1998}$ is divided by $1998$

i tried using eulers theorem and got only this much

$1992^{1980}\equiv 1mod 1998$
similarly for oders...but cudnt get after that

so i need a kind help.

thanku
• Mar 26th 2010, 08:44 PM
banku12
first of all can anyone tell me if this q uses any modular arithmetic..

bec its getting very untidy with congruency...
• Mar 26th 2010, 10:37 PM
Drexel28
Quote:

Originally Posted by banku12
find the remainder when $1992^{1998} - 1955^{1998} - 1938^{1998} + 1901^{1998}$ is divided by $1998$

i tried using eulers theorem and got only this much

$1992^{1980}\equiv 1mod 1998$
similarly for oders...but cudnt get after that

so i need a kind help.

thanku

If you got that they are all congruent to one then wouldnt the expression be congruent to -2=1996?
• Mar 27th 2010, 08:09 AM
chiph588@
Quote:

Originally Posted by banku12
find the remainder when $1992^{1998} - 1955^{1998} - 1938^{1998} + 1901^{1998}$ is divided by $1998$

i tried using eulers theorem and got only this much

$1992^{1980}\equiv 1mod 1998$
similarly for oders...but cudnt get after that

so i need a kind help.

thanku

$1992^{1980}\equiv 1296\mod{1998}$

Here's a hint: $\phi(1998) = 648$ and $1998 = 3\cdot648+54$.
• Mar 27th 2010, 08:18 AM
banku12
seriously not getting(Thinking)

can any one of u plz post a complete soln
• Mar 27th 2010, 09:35 AM
chiph588@
I should have mentioned in my last hint that for Euler's Theorem to work i.e. $a^{\phi(n)}\equiv1\mod{n}$, we need $(a,n)=1$. So Euler's Theorem doesn't apply to every term here.

Try this:

$1992\equiv-6\mod{1998}$

So $1992^{1998}\equiv6^{1998}\mod{1998}$.

Now do something similar to this: http://www.mathhelpforum.com/math-he...remainder.html