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枫下家园 / 望子成龙 / Find all n for which n²+2n+4 is divisible by 7. 解答思路见内。希望能帮到准备竞赛的同学们Find all n for which n²+2n+4 is divisible by 7
第一秒一看有点棘手,有平方还不能因式分解。第二秒马上想到分解常数让有n的部分可因式分解,常数部分可被7整除。第3,4,5秒想到(n+3)(n-1)+7, 因为n²+2n需要的常数是0,1,-3. 第三个就是想要的。剩下就简单了。常数被7整除,所以只要(n+3)(n-1)被7整除。n+3 or n-1 被7整除就行。
-xiaohui(xiaohui);
2020-4-6
{311}
(#12837249@0)
+7
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not impressed. I'll give a more proper answer.
-less_is_more(-=+);
2020-4-6
(#12837334@0)
+2
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Here is my solution to this question:n²+2n+4 is divisible by 7
Let n=7p + q (0 <= q < 7) Don't ask me why, you are supposed to be smart enough.
n²+2n+4 = (7p + q)²+2(7p + q)+4
= 49p² + 14pq + q² + 14p + 2q + 4
= (49p² + 14pq + 14p) + (q² + 2q + 4)
If n²+2n+4 is divisible by 7
Then q² + 2q + 4 (0 <= q < 7) is divisible by 7
Enumerating all the q in its domain, you'll get q=1 and q=4 are qualified.
So the answer would be any integers by the forms of 7p+1 or 7p+4 (p ∈ Z).
□
-less_is_more(-=+);
2020-4-6
{448}
(#12837446@0)
+2
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💯
-tuteng(闻风而动);
2020-4-6
(#12837476@0)
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我觉得楼主的推理比你这个简单多了。
-ding_ding(丁_丁);
2020-4-6
(#12837483@0)
+11
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+1!
-datura(带刀山贼);
2020-4-6
(#12837539@0)
+1
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The OP's solution is to convert a question to another without providing any final answer (an exact set of domain or range).
-less_is_more(-=+);
2020-4-6
(#12837569@0)
+2
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我觉得楼主的办法取巧,只适用于非常聪明的学生,你这个办法看似笨拙,但可以解决更多更复杂的问题。
-897102(⑧⑨⑦⑩②);
2020-4-6
(#12837573@0)
+1
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Do I have to say the OP's "solution" won't be rewarded any credits in any contests?
-less_is_more(-=+);
2020-4-6
(#12837582@0)
+4
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楼上是在中国上过奥数班的,有一套方法,中国的教学就这个毛病,扼杀了创造性。
-ding_ding(丁_丁);
2020-4-6
(#12837584@0)
+1
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Never knew about "奥数班" when I was young. But I became a youth member of the Shanghai Association of Mathematicians at the age of 14
-less_is_more(-=+);
2020-4-6
(#12837604@0)
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你以前学过木匠吗?
-cricketkiller(白牙青);
2020-4-6
(#12837747@0)
+3
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???
-less_is_more(-=+);
2020-4-6
(#12837836@0)
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测量一下世界有多大,看来你不是小木匠,也可能我们说的是两个地方。
-cricketkiller(白牙青);
2020-4-6
(#12837966@0)
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LZ的方法是多伦多的一些奥数班教过的trick,楼上用的是比较正统的解决类似问题的方法。不信的话,可以找些上过竞赛培训班的人印证一下。
-tuteng(闻风而动);
2020-4-6
(#12838003@0)
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看这个就头痛,原题是n²+2n+4,你一圈下来整了个q² + 2q + 4 (0 <= q < 7),像不像拖裤子放P
-cricketkiller(白牙青);
2020-4-6
(#12837727@0)
+2
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That's just a coincidence. This method is used to shrink the domain of the input variable by transformation.
-less_is_more(-=+);
2020-4-6
(#12837732@0)
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国内教学就这样啦,没办法,管他三七二十一,上来就套路。。。
-ding_ding(丁_丁);
2020-4-6
(#12837759@0)
+2
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If I say I never saw such kind of questions before, would you change your theory? { I don't think so. }
-less_is_more(-=+);
2020-4-6
(#12837767@0)
+2
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How about this one: Find all n for which n²+2n+4 is divisible by 19教授出题很有技巧,小聪明可解,蛮力也行。关键是蛮力丢掉了宝贵的时间。10题的数学竞赛最后两题要花很多时间。小聪明是平时花时间的体现。
-xiaohui(xiaohui);
2020-4-6
{130}
(#12837509@0)
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n=19k+3 or 19k+14, k is any integer
-tuteng(闻风而动);
2020-4-6
(#12837650@0)
+1
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照楼主的分解法,就是n+5或者n-3能被19整除。
-littlepony2002(山茶花);
2020-4-6
(#12837676@0)
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Then, what is n?
-tuteng(闻风而动);
2020-4-6
(#12837688@0)
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就差一步嘛,19K+3, 19K-5 (也就是19K+14)
-ding_ding(丁_丁);
2020-4-6
(#12837693@0)
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What you try to promote {19K-5 (也就是19K+14) } is just a coincidence.
-less_is_more(-=+);
2020-4-6
(#12837710@0)
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That's not very useful. If the mod is not a prime, the OP's method would be doomed.
-less_is_more(-=+);
2020-4-6
(#12837697@0)
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我个人这样做(n+1)^2+3被7整除 从而得出1,4,8,16,32...
-palwang2000(重在修心);
2020-4-6
(#12837714@0)
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谢谢。捅破窗户纸就非常容易了。
-wstl(武水拖蓝);
2020-4-6
(#12839435@0)
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上面少即是多用的方法从余数的基本性质出发,没有使用什么技巧就把问题解决了,我觉得更有通用性
-tuteng(闻风而动);
2020-4-7
(#12840535@0)