https://artofproblemsolving.com/wiki/index.php?title=2008_iTest_Problems/Problem_47&feed=atom&action=history2008 iTest Problems/Problem 47 - Revision history2024-03-28T10:24:52ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2008_iTest_Problems/Problem_47&diff=96143&oldid=prevRockmanex3: Solution to Problem 472018-07-12T23:33:10Z<p>Solution to Problem 47</p>
<p><b>New page</b></p><div>==Problem==<br />
<br />
Find <math>a + b + c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are the hundreds, tens, and units digits of the six-digit integer <br />
<math>123abc</math>, which is a multiple of <math>990</math>. <br />
<br />
==Solution==<br />
<br />
Because <math>990 = 10 \cdot 9 \cdot 11</math>, the integer <math>123abc</math> is a multiple of <math>10</math>, <math>9</math>, and <math>11</math>. That means <math>c = 0</math>, <math>6 + a + b \equiv 0 \pmod{9}</math>, and <math>2 - a + b \equiv 0 \pmod{11}</math>. The last two congruences can be rewritten as <math>a + b \equiv 3 \pmod{9}</math> and <math>a - b \equiv 2 \pmod{11}</math>. After testing values of <math>a</math> and <math>b</math>, we see that <math>a = 7</math> and <math>b = 5</math>, so <math>a+b+c=\boxed{12}</math>.<br />
<br />
==See Also==<br />
{{2008 iTest box|num-b=46|num-a=48}}<br />
<br />
[[Category:Introductory Number Theory Problems]]</div>Rockmanex3