# number theory

• September 30th 2008, 07:25 PM
number theory
What is the last two digits of this number ?

16^198
• September 30th 2008, 07:27 PM
mr fantastic
Quote:

What is the last two digits of this number ?

16^198

16^198 = (2^4)^198 = 2^792.

Now consider powers of 2 and look for the pattern in the last two digits ....
• September 30th 2008, 11:44 PM
sorry , i still don get u
• October 1st 2008, 12:22 AM
mr fantastic
Quote:

sorry , i still don get u

2^2 =
2^3 =
2^4 =
2^5 =
2^6 =
2^7 =
2^8 =
2^9 =
2^10 =
2^11 =
2^12 =
2^13 =
2^14 =

Look for a pattern in the last two digits of the value. Use that pattern to answer your question.
• October 1st 2008, 11:32 AM
Moo
Hello,

Here is another way :
$6^1=6$
$6^2=36$
$6^3=216$
etc...

More generally :
$(10k+6)^1=10k+6$
$(10k+6)^2=10(10k^2+2k)+6$
etc...

Any power of something ending with a 6 ends with a 6

So $16^{n}$ ends with a 6, whatever positive integer $n \ge 1$ 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.

$16^1=\bold{1}6$
$16^2=2\bold{5}6$
$16^3=40\bold{9}6$
$16^4=16 \times 4000+16 \times 96=16 \times 4000+1536=\dots\bold{3}6$
$16^5=\dots=\dots \bold{7}6$
$16^6=\dots=\dots \bold{1}6$
$16^5=\dots=\dots \bold{5}6$

etc...

You can see that it repeats every 5 powers. So if the power is in the form $5k+1$, the last 2 digits are $16$.
If in the form $5k+2$, it'l be $56$ and so on

$198=5 \times 39+3$
• October 2nd 2008, 08:05 AM