prep 129 一道关于余数的数学求助

guoshuyu

129.    13872-!-item-!-187;#058&010142
If t is a positive integer and r is the remainder when t2 + 5t + 6 is divided by 7, what is the value of r ?

(1) When t is divided by 7, the remainder is 6.

(2) When t2 is divided by 7, the remainder is 1.
【答案】A
【思路】题目t2 + 5t + 6=t(t+5)+6
(1)   + 5t + 6=t(t+5)+6,表示t(t+5),可以被7整除,the remainder is 6 =r
(2)  =7y+1,当t=6 ,  +5t + 6=36+30+6=72, the remainder=2=r
               t=8,  +5t + 6=64+40+6=110, the remainder=5=r
               在这个情况下,无法得知the value of r, insufficient

怎么也想不明白（1），求助啊！

guoshuyu

为什么选A啊，真是搞不明白

xiaochen1221

t除以7余数是6，就假设A是一个7的整倍数的数，那么t=A+6，t的平方就等于A平方+12A+36，前面两项都是7的倍数，最后一项36除以7余数是1,5t=5A+30，余数是2,1+2+6=9，除以7余数是2=r，所以（1）可以推出r。
（2）只是给了t方与7的倍数的关系，没法确认，举出两个反例就给否了，所以选A吧……

我瞎扯的……欢迎拍砖啊

gaoyitian

先看这个帖子，楼主讲得很明白
http://forum.chasedream.com/GMAT_Math/thread-437516-1-1.html
然后分析题目
令S=t2+5t+6=(t+1)(t+5)
由A能推出t+1除以7余数为零，故S除以7余数为零——充分
由B能推出t2除以7余数为1或-1，一定不要忘了有等于-1的情况，具体为什么我只能写个式子给你，具体要自己先看明白之前那个帖子，我给你的式子如下：t2 mod 7=1  推出  (7x+1)2 mod 7=1或(7y-1)2 mod 7=1，即t除以7要么余1要么余6
那么余1时，S除以7余数为2*6=12，再用12mod7=5
余6时同条件A，S除以7余数为0，无法判断是哪一种情况，故不充分
综上所述，本题选A


严重建议看懂了那个帖子再来想，要是看懂了还是不会做，再看我给你的思路，不然直接看估计看不懂

easttiger2010

先看这个帖子，楼主讲得很明白
http://forum.chasedream.com/GMAT_Math/thread-437516-1-1.html
然后分析题目
令S=t2+5t+6=(t+1)(t+5)
由A能推出t+1除以7余数为零，故S除以7余数为零——充分
由B能推出t2除以7余数为1或-1，一定不要忘了有等于-1的情况，具体为什么我只能写个式子给你，具体要自己先看明白之前那个帖子，我给你的式子如下：t2 mod 7=1  推出  (7x+1)2 mod 7=1或(7y-1)2 mod 7=1，即t除以7要么余1要么余6
那么余1时，S除以7余数为2*6=12，再用12mod7=5
余6时同条件A，S除以7余数为0，无法判断是哪一种情况，故不充分
综上所述，本题选A


严重建议看懂了那个帖子再来想，要是看懂了还是不会做，再看我给你的思路，不然直接看估计看不懂

-- by 会员 gaoyitian (2012/5/8 1:30:33)

这样推（1）的sufficiency可能更好理解点：


1）t=7m+6
2)  原题变成了求（7m+8）(7m+9)   ，注：t^2+5t+6=(t+2)(t+3)


显然余数是唯一确定。




t=

GTI

先看这个帖子，楼主讲得很明白
http://forum.chasedream.com/GMAT_Math/thread-437516-1-1.html
然后分析题目
令S=t2+5t+6=(t+1)(t+5)
由A能推出t+1除以7余数为零，故S除以7余数为零——充分
由B能推出t2除以7余数为1或-1，一定不要忘了有等于-1的情况，具体为什么我只能写个式子给你，具体要自己先看明白之前那个帖子，我给你的式子如下：t2 mod 7=1  推出  (7x+1)2 mod 7=1或(7y-1)2 mod 7=1，即t除以7要么余1要么余6
那么余1时，S除以7余数为2*6=12，再用12mod7=5
余6时同条件A，S除以7余数为0，无法判断是哪一种情况，故不充分
综上所述，本题选A


严重建议看懂了那个帖子再来想，要是看懂了还是不会做，再看我给你的思路，不然直接看估计看不懂

-- by 会员 gaoyitian (2012/5/8 1:30:33)

首先其实应该是S=(t+2)(t+3),
关于条件B，因为t2被7除余1，所以t被7除的余数设为a，a2 mod 7=1，即a2=7x+1,x为常数。

所以a2-1=7x, (a-1)(a+1)=7x, 所以x只能等于5或9。a只能等于6或8 。

但由于代入6或8时，2者的最终结果不同，各自最终答案余数为2和1 。所以无法确定。

GTI

纠正，最终答案为2或5 。

0412501

以下摘自曼哈顿网站，有兴趣的鞋童们可以看原帖。偶觉得这个帖子说得非常清晰易懂。我把最重要的部分贴出：

http://www.manhattangmat.com/forums/if-t-is-a-positive-integer-and-r-is-the-remainder-when-t-2-t4697.html

one fact that's pretty cool, and which happens to apply to this problem, is that you can do normal arithmetic with remainders, as long as all the remainders come from division by the same number. the only difference is that, if/when you get numbers that are too big to be authentic remainders (i.e., they're equal to or greater than the number you're dividing by), you have to take out as many multiples of the divisor as necessary to convert them back into "legitimate" remainders again. you can think of the remainders as on an odometer that rolls back to 0 whenever you reach the number you're dividing by.

so with statement (1), all the remainders are upon division by 7, so we can do normal arithmetic with them:
if t gives a remainder of 6, then t^2 = t x t gives a remainder of 6 x 6 = 36 --> this is more than 7, so we take out as many 7's as possible: 36 - 35 = 1.
if t gives a remainder of 6, then 5t gives a remainder of 5(6) = 30 --> this is more than 7, so we take out as many 7's as possible: 30 - 28 = 2.
and finally, 6 itself gives a remainder of 6.
therefore, the grand remainder when t^2 + 5t + 6 is divided by 7 should be 1 + 2 + 6 = 9 --> take out one more seven --> remainder will be 2.
sufficient.

by the way, much more generally (and therefore perhaps more importantly), the patterns in remainder problems will always emerge fairly early when you plug in numbers. therefore, if you don't IMMEDIATELY realize a good theoretical way to do a remainder problem, you should get on the number plugging RIGHT AWAY.

with statement (1), generate the first 3 numbers for which the statement is true: 6, 13, 20.
try 6: 36 + 30 + 6 = 72, which yields a remainder of 2 upon division by 7.
try 13: 169 + 65 + 6 = 240, which yields a remainder of 2 upon division by 7.
try 20: 400 + 100 + 6 = 506, which yields a remainder of 2 upon division by 7.
i'm convinced. (again, remember that PATTERNS EMERGE EARLY in remainder problems. 3 examples may not be enough for other types of pattern recognition, but that's usually pretty good in a remainder problem.)

with statement (2), as a poster has already mentioned above, find the first two t^2's that actually do this, which are 1^2 = 1 and 6^2 = 36.
if t = 1, then 1 + 5 + 6 = 12, which yields a remainder of 5 upon division by 7.
if t = 6, then 36 + 30 + 6 = 72, which yields a remainder of 2 upon division by 7.
insufficient.

if you know about 'modulo arithmetic' (see this post), then you can shortcut this sort of problem with it.

0412501

继续上面的回复：

一个学生问：I thought divisibility rules work for multiplication as well. So if t^2 gives a remainder of 1. I assumed t would give a remainder of 1. 1 * 1 = 1. Why does the rule not work here?

答：it doesn't work because a remainder is not the original number. i.e., in your example, 1 is a single number, but there are infinitely many numbers that will give remainders of 1 upon division.

for instance (considering division by 10 instead of division by 7), the remainder of 19/10 is 9, but the remainder of 19^2/10 =361/10 is 1