It didn't take long to get a quick correct answer posted as a response to my question of the remainder when 4⁴⁴⁴⁴  is divided by 7.  Check the response to the previous note with the correct solution provided by Greg.

So  I decided to ask two more.

      What are the last two digits of 3¹⁴¹⁵⁹ ?  ...of 7⁷⁷⁷⁷⁷ ?

The rules are the same as for the last bonus question.

Sometime over the weekend I will post a note with an attachment containing the questions that I want you to work on during Monday's class.  Make sure that you make a copy.

I will also have your next homework assignment ready sometime next week.

              --CFW

87 and 49

3^14159 = 87

and 7^77777 = 49

I found the solution by finding a combination of the last two digits. I found one combination for the last digit which consisted of 3971 repeatedly occuring. I found which one of the those four was the last digit by solving for 14159 = kmod4. Whatever the remainder was was that digit. For example: if K = 1 the final digit would be 3.

The second digit to the left went into a combination of 20 different numbers which are the following: 00, 284, 286, 844, 426, 026, 860. Now solving for 14159 = kmod20. I found K = 18. Which is 8 according to the list given.

That is how I found the last two digits of the first problem.

The second problem the last digit combination was 7931. Solving for 77777 = kmod4. I found K = 2. Now for the second to last, I found a combination of 0440. Solving for 77777 = kmod4 equaled 2. And that according to the order of the combination of numbers = 4.

I hope this is true. Have a good weekend Dr. Waizeres.

I know this wasn't asked, but here is how one can tell if it is possible to easily find the last digits.

The way you can tell if you can easily find the last two digits of any number is if, when you raise it to the powers, the last digit cycles through 1.

For example, 3 works because the the last digit cycles 3,9,7,1 (you can check it easily). 3^4=81, which means 3^41mod10. And we know that we can now raise it to any power and still be x1mod10 (where x is any power of 3^4). 
It can be quickly realized that when we raise both sides to a power, 3^4 because 3^(4*power). For example, if we raised both sides by 2, it becomes 3^(4*2)1^2mod10.
This tells us 3 to the power of any multple of 4
aka, 3^(4*2), 3^(4*3), 3^(4*4) ...
will always have a 1 as the last digit. In fact, they will by the only powers that will have 1 as the last digit

But we want the last two digits. It isn't too hard to find them now that we know which powers of 3 to check. So now we can simply use some trail and error until we find one with 01 as the last digits. We will find that 3^(4*5) will give a number with 01 as the last digit (note: I did that in a roundabout way in my first post)

So now it is simple to find the last two digits.

We can not, however, find the last two digits using ths method for, for example, 4. Because whatever power of 4 you have, it will always be even, and therefore not 1. So you will never have 4^x1mod10, so consequently, we can not find the last digit using this method
(it is easy to find the last digit once you notice that 4 to a power always has a last digit that cycles between 4 and 6. But it is not as easy for, say, the last two digits of any power of 4)

For the same reason you can't find it for 6

You can with 7 because 7^4 has 1 as the last digit (the last two digits are actually 01 in fact)

You can also do this with 13 because 13^4 has a last digit of 1.

Hopefully this makes things a bit clearer, but perhaps not...

Anyway, if for some reason people feel like doing a problem, solve this one.

Find the last two digits of 13²⁴⁹³⁹ (hint: you wouldn't be able to figure this out on your portable calculator. Use one on a computer, window's built in one will do fine)

jnehring got both of the questions that I posted yesterday.  He also gave a very nice explanation of how he found the solutions.

Maybe someone else can answer the question that he asked about the last two digits of 13 raised to a power.  He gave a pretty good hint.

          --CFW

Your Response

My answer is 89 hopefully its right :-/.

Last two digits of 3¹⁴¹⁵⁹

^{To find, for example, the last digit of a number, what you want is what the remainder is when you divide by 10 (for example 34235 when divided by 10 is 3423(10)+5, 5 being the last digit). The last two digits would be the remainder when divided by 100, and so on.}

Since we want the last two digits, we want to know what 3^14159_mod100 is.

3^4=81

3^481mod100

raise both to the 5th power

81^5=3486784401
(I picked this number because the remainder when divided by 100 is 1. I used trial and error to get it, but 81 is a good place to start because the last digit will always be 1, which is what we want)

So, 3^(4*5)3486784401mod100, which equals 3^201mod100

707*20=14140, so raise both sides by 707

3^141401mod100

Multiple both sides by 3^19 (3^19 equals 1162261467)

3^141591162261467mod100

3^1415967mod100

Answer is 67

Because the remainder when diveded by 100 is 67, which is the same as the last two digits

Last two digits of 7⁷⁷⁷⁷⁷

For reasons in my last post, we want to find what 7^7777mod100 is.

7^4=2401 (which is the smallest power that has "01" as the last digits)

So 7^42401mod100

So 7^41mod100

1944*4=7776

So raise each side to the 1943 power

7^(4*1944)1mod100

7^77761mod100

multiply both sides by 7

7^77777mod100

So the last two digits are 07

I just realized that I answered for 7^7777, not 7^77777, so let me do that again. We know that

 7^41mod100

4*19444=77776

So raise each side by the 19444 power

7^(4*19444)1mod100

7^777761mod100

Multiple both sides by 7

7^777777mod100

So the answer is actually still the same, it is 07