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

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

JohnD
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Stefan Hanssen
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    FWIW, you can also just look at the exact answer to find that it is in fact divisible by $10,000$. – geometrian Nov 22 '14 at 22:57
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    @imallett How about $4^{10^{1000000}} - 6^{10^{50000000}}$? ;) – Thomas Nov 23 '14 at 01:26
  • @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)! – geometrian Nov 23 '14 at 04:02
  • @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) – Thomas Nov 23 '14 at 04:28
  • @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. – geometrian Nov 23 '14 at 04:50

6 Answers6

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

Adhvaitha
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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 $4^{1000}-6^{500}$?

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

Dr. Sonnhard Graubner
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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.

Nick Matteo
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Elliot Gorokhovsky
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$4^2 \equiv 6 \mod 10$

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

Neil W
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5

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.

Community
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geometrian
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