What is the last two digits of this number ?

16^198

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- Sep 30th 2008, 08:25 PMmathaddictnumber theory
What is the last two digits of this number ?

16^198 - Sep 30th 2008, 08:27 PMmr fantastic
- Oct 1st 2008, 12:44 AMmathaddict
sorry , i still don get u

- Oct 1st 2008, 01:22 AMmr fantastic
- Oct 1st 2008, 12:32 PMMoo
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

- Oct 2nd 2008, 09:05 AMmathaddict
Wonderful , thank you for the very precise explaination