【求助】本月鼠穴狗68题

jessjuejue · 发表于 2010-4-29 09:51:34

68. m能够表示成a+b+c（a,b,c是连续正整数）也能表示成x*y*z（x,y,z也是连续正整数）
问m被5除余几
（1） a被5除余1
（2） x被5除余1

有什么思路吗，谢谢

flyingbunny · 发表于 2010-4-29 10:51:33

条件1，
a=5t+1
b=5t or 5t+2
c= 5t-1 or t5+2 or 5t+3
therefore insuff

条件2，
same as 1), insuff

1+2;
a=5t+1,
b=5t,
c=5t-1,

x=5k+1,
y=5k,
z=5k-1,

m mod 5=0

a=5t+1,
b=5t+2,
c=5t+3，

x=5k+1,
y=5k+2,
z=5k+3,

m mod 5=1

therefore insuff

选 E

jessjuejue · 发表于 2010-4-29 11:11:19

谢谢flyingbunny，思路清楚了，不过感觉X*y*z算MOD很奇怪

canols · 发表于 2010-4-29 15:27:05

看不过去了，楼主的还说明白思路了
我认为选D，都是余1
条件1推出
a=5t+1, so b=5t+2, c=5t+3
m=a+b+c=15t+6=5(3t+1)+1
条件2推出
x=5t+1, so y=5t+2, z=5t+3
m=xyz=(5t+1)(5t+2)(5t+3)=125t^3+150t^2+55t+6=5(25t^3+30t^2+11t+1)+1

correct me if i am wrong

条件1，
a=5t+1
b=5t or 5t+2
c= 5t-1 or t5+2 or 5t+3
therefore insuff

条件2，
same as 1), insuff

1+2;
a=5t+1,
b=5t,
c=5t-1,

x=5k+1,
y=5k,
z=5k-1,

m mod 5=0

a=5t+1,
b=5t+2,
c=5t+3，

x=5k+1,
y=5k+2,
z=5k+3,

m mod 5=1

therefore insuff

选 E

-- by 会员 flyingbunny (2010/4/29 10:51:33)

flyingbunny · 发表于 2010-4-29 15:55:34

看不过去了，楼主的还说明白思路了
我认为选D，都是余1
条件1推出
a=5t+1, so b=5t+2, c=5t+3
m=a+b+c=15t+6=5(3t+1)+1
条件2推出
x=5t+1, so y=5t+2, z=5t+3
m=xyz=(5t+1)(5t+2)(5t+3)=125t^3+150t^2+55t+6=5(25t^3+30t^2+11t+1)+1

correct me if i am wrong

条件1，
a=5t+1
b=5t or 5t+2
c= 5t-1 or t5+2 or 5t+3
therefore insuff

条件2，
same as 1), insuff

1+2;
a=5t+1,
b=5t,
c=5t-1,

x=5k+1,
y=5k,
z=5k-1,

m mod 5=0

a=5t+1,
b=5t+2,
c=5t+3，

x=5k+1,
y=5k+2,
z=5k+3,

m mod 5=1

therefore insuff

选 E

-- by 会员 flyingbunny (2010/4/29 10:51:33)

-- by 会员 canols (2010/4/29 15:27:05)

这个so 是 so 不出来的.

a is not necessarily the first 1 among these 3 consecutive numbers.

canols · 发表于 2010-4-29 15:58:32

我认为你这个思路不是GMAT的考查点

看不过去了，楼主的还说明白思路了
我认为选D，都是余1
条件1推出
a=5t+1, so b=5t+2, c=5t+3
m=a+b+c=15t+6=5(3t+1)+1
条件2推出
x=5t+1, so y=5t+2, z=5t+3
m=xyz=(5t+1)(5t+2)(5t+3)=125t^3+150t^2+55t+6=5(25t^3+30t^2+11t+1)+1

correct me if i am wrong

条件1，
a=5t+1
b=5t or 5t+2
c= 5t-1 or t5+2 or 5t+3
therefore insuff

条件2，
same as 1), insuff

1+2;
a=5t+1,
b=5t,
c=5t-1,

x=5k+1,
y=5k,
z=5k-1,

m mod 5=0

a=5t+1,
b=5t+2,
c=5t+3，

x=5k+1,
y=5k+2,
z=5k+3,

m mod 5=1

therefore insuff

选 E

-- by 会员 flyingbunny (2010/4/29 10:51:33)

-- by 会员 canols (2010/4/29 15:27:05)

这个so 是 so 不出来的.

a is not necessarily the first 1 among these 3 consecutive numbers.

-- by 会员 flyingbunny (2010/4/29 15:55:34)

flyingbunny · 发表于 2010-4-29 16:06:21

没有给的条件就不要去假定.
simple as that.

summer529 · 发表于 2010-4-29 16:15:31

数学菜鸟，但是今天请教了一个同学，他觉得是D

jessjuejue · 发表于 2010-4-29 16:17:43

我看了切尔西的讨论稿里也选E，不管怎么样，谢谢各位

canols · 发表于 2010-4-29 16:18:55

如果你说abc是连续整数，那么就是c>b>a，你翻翻OG看看
you can pick E in real test, if you want

没有给的条件就不要去假定.
simple as that.

-- by 会员 flyingbunny (2010/4/29 16:06:21)

【求助】本月鼠穴狗68题

所属分类: GMAT考试

正在浏览此版块的会员 ()

浏览过的版块