What is the last two digits of this number ?

16^198

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- Sep 30th 2008, 07:25 PMmathaddictnumber theory
What is the last two digits of this number ?

16^198 - Sep 30th 2008, 07:27 PMmr fantastic
- Sep 30th 2008, 11:44 PMmathaddict
sorry , i still don get u

- Oct 1st 2008, 12:22 AMmr fantastic
- Oct 1st 2008, 11:32 AMMoo
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

$\displaystyle 198=5 \times 39+3$ - Oct 2nd 2008, 08:05 AMmathaddict
Wonderful , thank you for the very precise explaination