What is the last two digits of this number ?
16^198
Hello,
Here is another way :
$\displaystyle 6^1=6$
$\displaystyle 6^2=36$
$\displaystyle 6^3=216$
etc...
More generally :
$\displaystyle (10k+6)^1=10k+6$
$\displaystyle (10k+6)^2=10(10k^2+2k)+6$
etc...
Any power of something ending with a 6 ends with a 6
So $\displaystyle 16^{n}$ ends with a 6, whatever positive integer $\displaystyle n \ge 1$ is.
Now, what are the two last digits ?
16 is divisible by 4. So 16^(198) is obviously divisible by 4.
We know that a number is divisible by 4 iff its two last digits form a multiple of 4.
There are only 5 two-digit numbers ending with a 6 and divisible by 4 :
16,36,56,76,96.
That's for giving you an idea.
$\displaystyle 16^1=\bold{1}6$
$\displaystyle 16^2=2\bold{5}6$
$\displaystyle 16^3=40\bold{9}6$
$\displaystyle 16^4=16 \times 4000+16 \times 96=16 \times 4000+1536=\dots\bold{3}6$
$\displaystyle 16^5=\dots=\dots \bold{7}6$
$\displaystyle 16^6=\dots=\dots \bold{1}6$
$\displaystyle 16^5=\dots=\dots \bold{5}6$
etc...
You can see that it repeats every 5 powers. So if the power is in the form $\displaystyle 5k+1$, the last 2 digits are $\displaystyle 16$.
If in the form $\displaystyle 5k+2$, it'l be $\displaystyle 56$ and so on
$\displaystyle 198=5 \times 39+3$