Thread: number theory

What is the last two digits of this number ?

16^198

Originally Posted by mathaddict

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 ....

sorry , i still don get u

Originally Posted by mathaddict

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.

Hello,

Here is another way :



etc...

More generally :


etc...

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

So ends with a 6, whatever positive integer 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.









etc...

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

Wonderful , thank you for the very precise explaination