find the 10 digits of 3^2007.
how many 4-digit numbers whose digits are all odd are multiples of 11?
I will use Congruences and Euler's Generalization of Fermat's Little Theorem..
We note that,
$\displaystyle \gcd(3,100)=1$
And, $\displaystyle \phi(100)=100\left(1-\frac{1}{2} \right)\left(1-\frac{1}{5}\right)=100(1/2)(4/5)=40$
Thus,
$\displaystyle 3^{40}\equiv 1(\mbox{mod }100)$
Bring both sides to the power of 50 thus,
$\displaystyle 3^{2000}\equiv 1(\mbox{mod }100)$
Note that,
$\displaystyle 3^7\equiv 87(\mbox{mod }100)$
Thus, their multiplication yields,
$\displaystyle 3^{2007}\equiv 87(\mbox{mod }100)$
Since it is congruent to 100 it shows the last two digits. The second to last digits it 8.
You have a 4 digit number,
$\displaystyle ABCD$ with $\displaystyle A\not = 0$
und,
$\displaystyle 0\leq A,B,C,D\leq 9$
The necessary and suffient condition for the divisibility of 11 to exist we need that the Alternating sum is a divisible by 11. Thus,
$\displaystyle D-C+B-A=0 \mbox{ or } 11 \mbox{ or } -11$
Thus,
$\displaystyle D+B=A+C$
$\displaystyle D+B=A+C+11$
$\displaystyle 11+D+B=A+C$
Basic trial and error will produce all solutions.
The smallest 4 digit number divisible by 11 is 91*11, and the largest
4 digit number divisible by 11 is 909*11. Therefore there are 819 such
numbers divisible by 11. Of these (819-1)/2 are even and the rest are odd
so there are 410 odd numbers 4 digit divisible by 11.
RonL