https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Shubhsuper&feedformat=atom AoPS Wiki - User contributions [en] 2022-01-25T03:32:17Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_2&diff=86203 2017 AIME I Problems/Problem 2 2017-06-28T19:29:24Z <p>Shubhsuper: /* Solution */</p> <hr /> <div>==Problem 2==<br /> When each of &lt;math&gt;702&lt;/math&gt;, &lt;math&gt;787&lt;/math&gt;, and &lt;math&gt;855&lt;/math&gt; is divided by the positive integer &lt;math&gt;m&lt;/math&gt;, the remainder is always the positive integer &lt;math&gt;r&lt;/math&gt;. When each of &lt;math&gt;412&lt;/math&gt;, &lt;math&gt;722&lt;/math&gt;, and &lt;math&gt;815&lt;/math&gt; is divided by the positive integer &lt;math&gt;n&lt;/math&gt;, the remainder is always the positive integer &lt;math&gt;s \neq r&lt;/math&gt;. Find &lt;math&gt;m+n+r+s&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let's tackle the first part of the problem first. We can safely assume: &lt;cmath&gt;702 = xm + r&lt;/cmath&gt; &lt;cmath&gt;787 = ym + r&lt;/cmath&gt; &lt;cmath&gt;855 = zm + r&lt;/cmath&gt;<br /> Now, if we subtract two values: &lt;cmath&gt;787-702 = 85 = 17\cdot5&lt;/cmath&gt;<br /> which also equals &lt;cmath&gt;(ym+r)-(xm+r) = m\cdot(y-x)&lt;/cmath&gt;<br /> Similarly, &lt;cmath&gt;855-787 = 68 = 17\cdot4; (zm+r)-(ym+r) = m\cdot(z-y)&lt;/cmath&gt;<br /> Since &lt;math&gt;17&lt;/math&gt; is the only common factor, we can assume that &lt;math&gt;m=17&lt;/math&gt;, and through simple division, that &lt;math&gt;r=5&lt;/math&gt;.<br /> <br /> Using the same method on the second half: &lt;cmath&gt;412 = an + s&lt;/cmath&gt; &lt;cmath&gt;722 = bn + s&lt;/cmath&gt; &lt;cmath&gt;815 = cn + s&lt;/cmath&gt;<br /> Then. &lt;cmath&gt;722-412 = 310 = 31\cdot10; (bn+s)-(an+s) = n\cdot(b-a)&lt;/cmath&gt; &lt;cmath&gt;815-722 = 93 = 31\cdot3; (cn+s)-(bn+s) = n\cdot(c-b)&lt;/cmath&gt;<br /> The common factor is &lt;math&gt;31&lt;/math&gt;, so &lt;math&gt;n=31&lt;/math&gt; and through division, &lt;math&gt;s=9&lt;/math&gt;.<br /> <br /> The answer is &lt;math&gt;m+n+r+s = 17+31+5+9 = \boxed{062}&lt;/math&gt;<br /> <br /> ~IYN~<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Shubhsuper