Find the last digit of the sum of two numbers with large exponents.

[johnnyICON]

Hi, I am trying to solve this question in my textbook. It asks me to find the last digit of the following expression:

$$222^{555} + 555^{222}$$

Now, I have tried using modular arithmetic to solve this question, but I'm stuck.
And I'm not all too sure if took the right approach or not, any hints?

 

[shmoe]

You should be able to do the two terms seperately, can you find 555^222 mod 100? 222^555 mod 100?

It might make the arithmetic a little easier if you consider powers of 2, 5, and 11 mod 100 seperately but this isn't mandatory.

 

[johnnyICON]

I did something like that. I used mod 11 for both numbers. Found 222 congruent to 2 mod 11, and 555 congruent to 5 mod 11.

So I made the statement that $$222^{555} \equiv 2^{555} mod 11$$and $$555^{222} \equiv 5^{222} mod 11$$

Then, I did successive powers of both 2 and 5 and found their congruencies until I found a pattern.

I determined that after five successive powers of 5, the congruencies would repeat. And after ten successive powers of 2, the congruencies would repeat.

So I divided the the exponents 222 and 555 by 10 and 5, respectively, to find another congruency. And that's where I get stuck.

 

[shmoe]

Sorry, I was misleading in my last post. For some reason I thought you wanted the last two digits, not just the last one. To find the last digit you need only find 222^555 and 555^222 mod 10, which will be easier than finding them mod 100. This also means the bit about 11 is superfluous.

 

[johnnyICON]

Maybe you can take a look at this example:

Example. Find the last digit of $$7^{100}$$.
We need to evaluate $$7^{100}$$ (mod 10). Since $$7^2$$ = 49$$\equiv$$ −1 (mod 10) we have $$7^{100} \equiv (72)^{50} \equiv$$ (−1)$$^{50} \equiv 1$$ (mod 10).Thus the last digit is 1.

Is the last digit 1 because of the last congruency?

 

[johnnyICON]

I am trying to apply this process that I found online to solve my problem.

I get stuck at the point where I divide the exponent by the length of the pattern.

http://mathcentral.uregina.ca/QQ/database/QQ.09.04/landon1.html

 

[shmoe]

Yes, the last digit of 7^100 is 1 because they've shown that the remainder you get when you divide 7^100 by 10 is 1.


So you look at the powers of 2 and find the pattern is 4 digits long. So 2^1, 2^5, 2^9, 2^13, ...2^105, ... are all congruent to 2 mod 10. Essentially you can add a 4 to the exponent and not affect the result. Same with 2^2, 2^6,... In the language of congruences, if $$a\equiv b\ mod\ 4$$ then $$2^a\equiv 2^b\ mod\ 10$$. There is a good reason why this is true, a power of 2's congruence class mod 10 is completely determined by its congruence class mod 5 (Chinese remainder theorem if you've met it) and we know $$2^4\equiv 1\ mod\ 5$$.

In the link you provided, they have $$a\equiv b\ mod\ 4$$ implies $$7^a\equiv 7^b\ mod\ 10$$ (we actually know here $$7^4\equiv 1\ mod\ 10$$). So to find $$7^{358}\ mod\ 10$$ notice $$358\equiv 2\ mod\ 4$$ so $$7^{358}\equiv 7^2\equiv 9\ mod\ 10$$.

 

[johnnyICON]

Wow, that makes a lot of sense now. And thanks for relating to the link I provided. I was really uncertain about this part::

notice $$358\equiv 2\ mod\ 4$$ so $$7^{358}\equiv 7^2$$

When I divided the exponent by the length and found its congruence, I wasn't sure if it meant that $$7^{358}\equiv 7^2$$ or if it meant $$7^{358}\equiv 7^{2 (mod 4)}$$.

So when dealing with congruencies, I can replace a with b by symmetry right? That is, $$358\equiv 2\ mod\ 4$$ is equivalent to $$2\equiv 358\ mod\ 4$$.

 

[shmoe]

When I divided the exponent by the length and found its congruence, I wasn't sure if it meant that $$7^{358}\equiv 7^2$$ or if it meant $$7^{358}\equiv 7^{2 (mod 4)}$$.

$$7^{358}\equiv 7^2\ mod\ 10$$ would be the correct one. You don't usually put "mod" up in an exponent.

So when dealing with congruencies, I can replace a with b by symmetry right? That is, $$358\equiv 2\ mod\ 4$$ is equivalent to $$2\equiv 358\ mod\ 4$$.

Yep. The "modular equivalence" works in many ways like your usual equal sign. It's symmetric, transitive and reflexive.

 

[johnnyICON]

Great! Thanks a lot shmoe, you've helped a lot! Too bad I couldn't rate your help.
10/10