What is the last two digits of this number ?
16^198
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