# How to know if a term is divisible by 10 [closed]

I have some difficulties to solve this easy problem, could someone help me?

Is $4^{1000}-6^{500}$ divisible by $10$?

• FWIW, you can also just look at the exact answer to find that it is in fact divisible by $10,000$. Nov 22, 2014 at 22:57
• @imallett How about $4^{10^{1000000}} - 6^{10^{50000000}}$? ;) Nov 23, 2014 at 1:26
• @Thomas: certainly! By examining these numbers, both of which end in "$\cdots1,787,109,376$", it's simple to discover that your expression is also divisible by $10,000$ (and indeed a great deal more)! Nov 23, 2014 at 4:02
• @imallett Yes, "both of which end in", which implies you don't need to compute the entire number to look at the last digits, which is what your first comment was implicitly suggesting (and why I gave the tongue in cheek reply of submitting unimaginably large numbers to force you to consider only the last digits :p) Nov 23, 2014 at 4:28
• @Thomas . . . clearly. And for the same reason, I answered tongue in cheek.¶ The larger point here is that for certain kinds of problems, the obvious approach should not be overlooked for a clever one. Since the easiest thing works yet the OP hasn't said whether he cares about a general approach, I presented it as a comment in case he wants to amend the question. Since it's been 5 hours, I think I'll post it as an answer anyway. Nov 23, 2014 at 4:50

Your expression can be written as $16^{500}-6^{500}$. Recall that $a^n-b^n$ is divisible by $a-b$ for $n \in \mathbb{N}$.

HINT: What is the right-most digit of $4^{1000}$? What is the right-most digit of $6^{500}$? What does that tell you about the right-most digit of $4^{1000}-6^{500}$?

we have $4^{1000}-6^{500}\equiv 6-6=0\equiv 0 \mod 10$

We can see that $4^1 = 4$, $4^2 = 16$, $4^3 = 64$, etc. It starts out with 4 as the last digit, which when multiplied by 4 gives you a 6, which when multiplied by a 4 gives you a 4, setting up the pattern of 6 as the last digit for even powers. Therefore $4^{1000}$ has 6 as its last digit. Clearly all powers of 6 have 6 as their last digit because if you multiply something that ends with 6 by 6 the product will end with a 6 since $6\times6=36$. Since the last digits are both 6, when you subtract you get 0 as the last digit, which means the difference is divisible by both 5 and 2 and is therefore divisible by 10.

$4^2 \equiv 6 \mod 10$

$\Rightarrow 4^{1000} \equiv 6^{500} \mod 10$

It is unclear from your question whether you require a general method. I've upvoted three of the other four answers here since they are all good. In particular the current best is the slickest trick and the next two are good generalizations.

That said, as I mentioned in the comments, the obvious approach should not be overlooked for a needlessly clever one. If you don't care about generality, then you can just evaluate the expression:

114813069527425452423283320117768198402231770208869520047764273682576626139237031385665948631650626991844596463898746277344711896086305533142593135616665318539129989145312280000688779148240044871428926990063486244662594025981424571409593728603035337632150388437182293551503447920102344948592677163033795451303978737969476432059651388725310756060866840299888538186292656743914465551536060049704715010157731117606858818025434780563119782817848700370999189150586174225945806076652256660690364965642083240783804920125367468611373661642590443113055089150622765944210552490560361847080855944387010578350080000

From here, it's pretty clear that it is divisible by $10$. I hesitate to post this as an answer, since it's by far the dumbest--but I think it's important to remember that elegant solutions aren't always needed to solve simple problems. Simple problems sometimes just have simple answers.