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HELP~!!输血寂静第12题,没看懂答案意思,小菜抱拳啦~~

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楼主
发表于 2012-3-21 23:21:27 | 只看该作者 回帖奖励 |倒序浏览 |阅读模式
12. V1. 一个数除以53等,表达出来即x1=5a+3, x1=7b+4, x2=5c+3, x2=7d+4,然后是x1>x2,问x1,x2的最大公约数 (kdavid1945)

V2. 狗主回忆的版本信息貌似有误,下面是GWD原题,请大家看一下哈!
When positive integer x is divided by 5, the remainder is 3; and when x is divided by 7, the remainder is 4. When positive integer y is divided by 5, the remainder is 3; and when y is divided by 7, the remainder is 4. If x > y, which of the following must be a factor of x - y?

A.12 B.15 C.20 D.28 E.35
答案:x=18,53,88,123,...,35n+18,答案选E

小菜没有懂:35n+18哪里来的?还有前面的数是怎么出来的~~
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沙发
 楼主| 发表于 2012-3-21 23:21:37 | 只看该作者
芝麻来人
板凳
发表于 2012-3-21 23:30:31 | 只看该作者
嗯~呵呵~打了一群汉字~还是觉得自己的方法不太清楚的说~还是帮bob同学找个正规解法啊~
When positive integer x is divided by 5, the remainder is 3; and when x is divided by 7, the remainder is 4. When positive integer y is divided by 5, the remainder is 3; and when y is divided by 7, the remainder is 4. If x > y, which of the following must be a factor of x - y?
A. 12
B. 15
C. 20
D. 28
E. 35

When the positive integer x is divided by 5 and 7, the remainder is 3 and 4, respectively: x=5q+3 (x could be 3, 8, 13, 18, 23, ...) and X=7p+4 (x could be 4, 11, 18, 25, ...).

There is a way to derive general formula based on above two statements:

Divisor will be the least common multiple of above two divisors 5 and 7, hence 35

Remainder will be the first common integer in above two patterns, hence 18--> so, to satisfy both this conditions x must be of a type x=35m+18  (18, 53, 88, ...);

The same for y (as the same info is given about y): y=35n+18 ; x-y=(35m+18)-(35n+18)=35(m-n) --> thus x-y must be a multiple of 35.

Answer: E.

地板
发表于 2012-3-21 23:36:55 | 只看该作者
这题今天我也问过,木有人鸟我啊。
5#
 楼主| 发表于 2012-3-21 23:38:14 | 只看该作者
哎??这个月考啊??占楼。这个饭饭会。先占楼再做。
-- by 会员 泾渭不凡 (2012/3/21 23:30:31)

谢谢饭饭姐啦O(∩_∩)O
6#
发表于 2012-3-21 23:56:05 | 只看该作者
The difference must be the multiple of 35, which is LCM of 5 and 7.
1) In order for x and y to leave the same remainder when divided by 5, the gap between two numbers should be a multiple of 5.
2)In order for x and y to leave the same remainder when divided by 7, the gap between two numbers should be a multiple of 7.
But x and y leave the same remainders when divided by both 5 and 7...so the gap between x and y should be a multiple of 5 AND a multiple of 7 or simply it should be a multiple of 35, which is LCM (5,7).

The only number that is a multiple of 35 is E, hence E is an answer.


那个不太清晰就看这个~
7#
发表于 2012-3-21 23:57:38 | 只看该作者
这题今天我也问过,木有人鸟我啊。
-- by 会员 christete (2012/3/21 23:36:55)


要不要这样啊。。。饭饭不是晚上才来JJ区的咩。。。。
8#
 楼主| 发表于 2012-3-22 00:00:15 | 只看该作者
嗯~呵呵~打了一群汉字~还是觉得自己的方法不太清楚的说~还是帮bob同学找个正规解法啊~
When positive integer x is divided by 5, the remainder is 3; and when x is divided by 7, the remainder is 4. When positive integer y is divided by 5, the remainder is 3; and when y is divided by 7, the remainder is 4. If x > y, which of the following must be a factor of x - y?
A. 12
B. 15
C. 20
D. 28
E. 35

When the positive integer x is divided by 5 and 7, the remainder is 3 and 4, respectively: x=5q+3 (x could be 3, 8, 13, 18, 23, ...) and X=7p+4 (x could be 4, 11, 18, 25, ...).

There is a way to derive general formula based on above two statements:

Divisor will be the least common multiple of above two divisors 5 and 7, hence 35

Remainder will be the first common integer in above two patterns, hence 18--> so, to satisfy both this conditions x must be of a type x=35m+18  (18, 53, 88, ...);

The same for y (as the same info is given about y): y=35n+18 ; x-y=(35m+18)-(35n+18)=35(m-n) --> thus x-y must be a multiple of 35.

Answer: E.

-- by 会员 泾渭不凡 (2012/3/21 23:30:31)

哎呀,太感谢饭饭姐啦,这个解释太给力啦~!!抱拳啦~
9#
 楼主| 发表于 2012-3-22 00:01:01 | 只看该作者
这题今天我也问过,木有人鸟我啊。
-- by 会员 christete (2012/3/21 23:36:55)

亲,你可以去看饭饭姐的回复,那个是标准答案~
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