# Congruent numbers

• Aug 8th 2006, 08:45 PM
xXxSANJIxXx
Congruent numbers
what is the last 2 digits of 3 raised to 1994 and 7 raised to 1994??

Plz help me this is urgent :)
• Aug 9th 2006, 06:34 AM
Soroban
Hello, xXxSANJIxXx!

I found two different methods for these problems . . .

Quote:

What are the last 2 digits of $\displaystyle 3^{1994}$ ?

Since $\displaystyle 3^{1994} \:=\:(3^2)^{997} \:=\:9^{997} \:=\:(10 - 1)^{997}$, consider the binomial expansion:

. . $\displaystyle 10^{997} - \binom{997}{996}10^{996} + \binom{997}{995}10^{995} - \hdots +$$\displaystyle \binom{997}{3}10^3 - \binom{997}{2}10^2 + \binom{997}{1}10 - 1$
. . . . $\displaystyle \underbrace{\qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad}$
. . . . . . . . . These do not affect the last two digits

Therefore, the last two digits are: .$\displaystyle \binom{997}{1}10 - 1\:=\:997\cdot10 - 1\:=\:99\boxed{69}$

Quote:

What are the last 2 digits of $\displaystyle 7^{1994}$ ?

We note that: $\displaystyle 7^4 = 2401$
. . Hence: $\displaystyle (7^4)(7^4)(7^4)\cdots(7^4)$ will end in $\displaystyle 01$.

Since $\displaystyle 1994 \,=\,4\cdot498 + 2$, we have: .$\displaystyle 7^{1994} \:=\:7^{(4\cdot498 + 2)} \;= \;7^{4\cdot498}\cdot7^2$

. . . . . $\displaystyle =\;\underbrace{(7^4)^{498}}\cdot\,49\quad \Longrightarrow\quad \text{a number ending in }49$
. . (a number ending in $\displaystyle 01$)

Therefore, the last two digits of $\displaystyle 7^{1994}$ are: $\displaystyle \boxed{49}$

• Aug 9th 2006, 07:38 AM
ThePerfectHacker
Quote:

Originally Posted by xXxSANJIxXx
what is the last 2 digits of 3 raised to 1994 and 7 raised to 1994??

Plz help me this is urgent :)

I have a different approach. It involves the phi function.
Since,
$\displaystyle \gcd (3,100)=1$ you have,
$\displaystyle 3^{\phi(100)}\equiv 1 (\mbox{mod } 10)$
To find, $\displaystyle \phi(100)$ prime factorize 100,
$\displaystyle 100=2^4\cdot 5^2$
Thus,
$\displaystyle \phi(100)=100\left(1-\frac{1}{2}\right)\left(1-\frac{1}{5} \right)=\frac{100\cdot 1\cdot 4}{2\cdot 5}=40$
Thus,
$\displaystyle 3^{40}\equiv 1 (\mbox{mod }100)$
Raise both sides to the 50th power,
$\displaystyle 3^{2000}\equiv 1 (\mbox{mod }100)$
Express as,
$\displaystyle 3^{1994}\cdot 3^6 \equiv 1 (\mbox{mod }100)$
Let, $\displaystyle x=3^{1994}$ then,
$\displaystyle 3^6x\equiv 1(\mbox{mod }100)$
$\displaystyle 729x\equiv 1(\mbox{mod }100)$
Solve for $\displaystyle x$,
$\displaystyle (729 - 700)x\equiv 1 (\mbox{mod }100)$
$\displaystyle 29x\equiv 1(\mbox{mod }100)$
$\displaystyle x\equiv 69$
• Aug 11th 2006, 01:50 AM
xXxSANJIxXx
THx
Thank you guys... i kinda get it.... :confused: