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2011-02-01T00:00:01
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<p style="text-align: center;"><mdo:image alt="Is the number 3 to the power 444 plus 4 to the power 333 divisible by 5?" height="145" src="big-powers.gif" width="219"></mdo:image></p>
<p style="text-align: center;"> </p>
<p style="text-align: left;"><a href="http://nrich.maths.org/public/viewer.php?obj_id=6176&part=">Click here for a poster of this problem</a>.</p>
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<span class="editorial">In the main, successful solvers of this
problem went about it in one of two ways.</span> <br></br>
<br></br>
<span class="editorial">One way was to `break it down' and study
the end digit for a set of powers. This is what Samantha and Zoe of
Maidstone Girls' Grammar School and Angela from Hethersett High in
Norfolk did.</span> <br></br>
<br></br>
<span class="editorial">Another way was to look at some remainders
when dividing by $5$. This is what Avishek, James, Martin, Thomas,
Kintel and Marcus from Simon Langton Boys' School did.</span>
<br></br>
<br></br>
<span class="editorial">This is Angela's solution:</span> <br></br>
<br></br>
<div class="editorial" style="font-style: normal;"></div>
<div>You have to think about how you distinguish numbers that are
divisible by five, they are numbers that end in a five or a zero.
So you know that for this number to divide by five it has to end in
a five or zero and it is on the basis of this knowledge that we can
work out this problem.</div>
<br></br>
<div>There is a pattern in the units digits for powers of
$3$:</div>
<br></br>
<div>$3$</div>
<br></br>
<div>$3\times 3=9$</div>
<br></br>
<div>$3\times 3\times 3=27$</div>
<br></br>
<div>$3\times 3\times 3\times 3=81$</div>
<br></br>
<div>$3\times 3\times 3\times 3\times 3=243$</div>
<br></br>
<div>$3\times 3\times3\times 3\times 3\times 3=729$</div>
<br></br>
<div>From this pattern we know that $3^{444}$ is going to end in a
$1$ because $444$ is divisible by $4$. $4^{333}$ works in much the
same way. The units digit for powers of $4$ are alternately $4$ and
$6$. Using these results we see that $4^{333}$ will end in a
$4$.</div>
<br></br>
<div>So now combine all the information we have so far which
is:</div>
<br></br>
<div>$3^{444}$ will end in a $1$</div>
<br></br>
<div>$4^{333}$ will end in a $4$</div>
<br></br>
<div>and $1+4=5$.</div>
<br></br>
<div>So $3^{444}+4^{333}$ will end in a $5$ and so it is divisible
by $5$.</div>
<br></br>
<div class="editorial" style="font-style: normal;"></div>
<div class="editorial" style="font-style: normal;">There is an
assumption here that these patterns in the units digits continue to
hold for ALL positive powers of $3$ and $4$ and you might like to
take up the challenge of proving this.</div>
<div class="editorial" style="font-style: normal;">
</div>
<div class="editorial" style="font-style: normal;"></div>
<div class="editorial" style="font-style: normal;"></div>
<div class="editorial" style="font-style: normal;"></div>
<div class="editorial" style="font-style: normal;"></div>
<div class="editorial" style="font-style: normal;">Students who
have been introduced to some Advanced Level Mathematics might be
interested in reading the following solution sent in by
Giridhar:</div>
<div class="editorial" style="font-style: normal;"></div>
<div style="font-style: italic;"></div>
<div style="font-style: italic;"></div>
<div style="font-style: italic;"></div>
<div style="font-style: italic;"></div>
<div style="font-style: italic;"></div>
<div style="font-style: italic;"></div>
<br></br>
<div>$$3^{444} + 4^{333} = (3^4)^{111} + (4^3)^{111} = a^{111} +
b^{111}$$ where we define $a = 3^4$ and $b=4^3$.</div>
<br></br>
<div>Now, any expression of the form: $a^n + b^n$, has $(a+b)$ as a
factor, when $n = 1, 3, 5 \dots$</div>
<br></br>
<div>That is, we can write $$a^n + b^n = (a+b )( \dots)$$ We know
that: $$ a + b = 3^4 + 4^3 = 81 + 64 =145,$$ and since $145$ is
divisible by $5$, $3^{444} + 4^{333}$ must be divisible by $5$
too.</div>
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At first glance, a challenging problem; but no algebra is required
to justify the solution.<br></br>
<br></br>
Students who meet this problem for the first time may need a
significant amount of support in structuring a solution so it is
useful to be able to find similar tasks to which they may apply
their new-found understanding.<br></br>
<br></br>
You can find a number of similar problems on the NRICH site - for
example:<br></br>
<a href="http://nrich.maths.org/public/viewer.php?obj_id=847&part=847">Power crazy</a> <br></br>
<a href="http://nrich.maths.org/public/viewer.php?obj_id=2024&part=2024">What an odd fact(or)</a> <br></br>
<br></br>
It is important to be able to justify any pattern. How can you be
sure it continues?<br></br></mdoxml>
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What determines whether a number is divisible by $5$ or not?<br></br>
<br></br>
Find: $3^{1}, 3^{2}, 3^{3}, 3^{4}, 3^{5}, 3^{6}, 3^{7}, 3^{8},
3^{9} \ldots$<br></br>
<br></br>
What do you notice?<br></br>
<br></br>
Find: $4^{1}, 4^{2}, 4^{3}, 4^{4}, 4^{5}, 4^{6}, 4^{7}, 4^{8},
4^{9} \ldots$ <br></br>
<br></br>
What do you notice? <br></br>
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Note equation 2 can be proved by the Remainder Theorem for $a=-b$
<br></br>
[That is $(a+b)$ is a factor of $a^n +b^n$ if $a^n +b^n = 0$ for
$a=-b$]<br></br></mdoxml>
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Big Powers
Three people chose this as a favourite problem. It is the sort of
problem that needs thinking time - but once the connection is made
it gives access to many similar ideas.
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