实例解析GMAT数学求通项问题
<p> <span word="GMAT">GMAT</span>考试大家对于数学的复习还是比较顺利的，但是呢我们知道<span word="GMAT">GMAT</span>数学也是有难题的。求通项是难题里出现次数比较多的，很多人对此不是特别的了解，小编下面就详细的给大家分享一些做通项题的<span word="GMAT">GMAT</span>考试技巧：</p><p> <span word="GMAT">GMAT</span>数学简便方法一：</p><p> 看到过一堆堆问通项如何求的帖子啦，这里说一个一招搞定的做法：</p><p> 通项<span word="S">S</span>，形式设为<span word="S">S</span>=<span word="Am">Am</span>+<span word="B">B</span>，一个乘法因式加一个常量</p><p> 系数<span word="A">A</span>必为两小通项因式系数的最小公倍数</p><p> 常量<span word="B">B</span>应该是两个小通项相等时的最小数，也就是最小值的<span word="S">S</span></p><p> 例题：4-<span word="JJ">JJ</span>78.<span word="ds">ds</span>某数除7余3，除4余2，求值。</p><p> 解：设通项<span word="S">S</span>=<span word="Am">Am</span>+<span word="B">B</span>。由题目可知，必同时满足<span word="S">S</span>=7<span word="a">a</span>+3=4<span word="b">b</span>+2</p><p> <span word="A">A</span>同时可被7和4整除，为28</p><p> <span word="B">B</span>为7<span word="a">a</span>+3=4<span word="b">b</span>+2的最小值，为10</p><p> 所以<span word="S">S</span>=28<span word="m">m</span>+10</p><p> <span word="GMAT">GMAT</span>考试数学简便方法二：</p><p> 129 <span word="DS">DS</span></p><p> <span word="x">x</span> 除8余几?</p><p> <span word="x">x</span>除12余5</p><p> <span word="x">x</span>除18余11</p><p> <span word="E">E</span></p><p> 条件1，令<span word="x">x</span>=12<span word="m">m</span>+5, <span word="m">m</span>=8<span word="k">k</span>,8<span word="k">k</span>+1,8<span word="k">k</span>+7</p><p> <span word="hang">hang</span>13:由1，<span word="X">X</span>=5时候除8余5，<span word="X">X</span>=17时候除8余1，不确定</p><p> 由2，<span word="X">X</span>=11时候除8余3，<span word="X">X</span>=29时候除8余5，不确定</p>
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