https://artofproblemsolving.com/wiki/index.php?title=2003_JBMO_Problems/Problem_1&feed=atom&action=history2003 JBMO Problems/Problem 1 - Revision history2024-03-29T07:42:08ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2003_JBMO_Problems/Problem_1&diff=97432&oldid=prevRockmanex3: Solution to Problem 1 -- kinda like 1998 JBMO Problem 12018-08-28T04:14:36Z<p>Solution to Problem 1 -- kinda like 1998 JBMO Problem 1</p>
<p><b>New page</b></p><div>==Problem==<br />
<br />
Let <math>n</math> be a positive integer. A number <math>A</math> consists of <math>2n</math> digits, each of which is 4; and a number <math>B</math> consists of <math>n</math> digits, each of which is 8. Prove that <math>A+2B+4</math> is a perfect square.<br />
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==Solution==<br />
<br />
Using the definition of base 10, we know that<br />
<cmath>\begin{align*}<br />
A &= 4(10^{2n-1} + 10^{2n-2} \cdots 10^1 + 10^0) \\<br />
&= \frac{4 \cdot (10^{2n} - 1)}{9} \\<br />
B &= 8(10^{n-1} + 10^{n-2} \cdots 10^1 + 10^0) \\<br />
&= \frac{8 \cdot (10^{n} - 1)}{9}.<br />
\end{align*}</cmath><br />
Thus, we have<br />
<cmath>\begin{align*}<br />
A+2B+4 &= \frac{4 \cdot 10^{2n} - 4 + 16 \cdot 10^n - 16 + 36}{9} \\<br />
&= \frac{4 \cdot (10^n)^2 + 16 \cdot 10^n + 16}{9} \\<br />
&= \left(\frac{2(10^n + 2)}{3}\right)^2.<br />
\end{align*}</cmath><br />
Since we know that <math>A+2B+4</math> is an integer, we confirm that <math>A+2B+4</math> is a perfect square.<br />
<br />
==See Also==<br />
{{JBMO box|year=2003|before=First Problem|num-a=2|five=}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]</div>Rockmanex3