Congruent numbers

**xXxSANJIxXx** · August 8th 2006, 08:45 PM

what is the last 2 digits of 3 raised to 1994 and 7 raised to 1994??

Plz help me this is urgent

**Soroban** · August 9th 2006, 06:34 AM

Hello, xXxSANJIxXx!

I found two different methods for these problems . . .

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

Since $3^{1994} \:=\<img src=$ 3^2)^{997} \:=\:9^{997} \:=\

10 - 1)^{997}" alt="3^{1994} \:=\

3^2)^{997} \:=\:9^{997} \:=\

10 - 1)^{997}" />, consider the binomial expansion:

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

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

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

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

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

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

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

**ThePerfectHacker** · August 9th 2006, 07:38 AM

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,
$\gcd (3,100)=1$ you have,
$3^{\phi(100)}\equiv 1 (\mbox{mod } 10)$
To find, $\phi(100)$ prime factorize 100,
$100=2^4\cdot 5^2$
Thus,
$\phi(100)=100\left(1-\frac{1}{2}\right)\left(1-\frac{1}{5} \right)=\frac{100\cdot 1\cdot 4}{2\cdot 5}=40$
Thus,
$3^{40}\equiv 1 (\mbox{mod }100)$
Raise both sides to the 50th power,
$3^{2000}\equiv 1 (\mbox{mod }100)$
Express as,
$3^{1994}\cdot 3^6 \equiv 1 (\mbox{mod }100)$
Let, $x=3^{1994}$ then,
$3^6x\equiv 1(\mbox{mod }100)$
$729x\equiv 1(\mbox{mod }100)$
Solve for $x$ ,
$(729 - 700)x\equiv 1 (\mbox{mod }100)$
$29x\equiv 1(\mbox{mod }100)$
$x\equiv 69$

**xXxSANJIxXx** · August 11th 2006, 01:50 AM

Thank you guys... i kinda get it....

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