b***z
发帖数: 921 | 1 找到一个最大的整数，这个整数用3，6，20来分割后余1或是2，也就是说这个整数不能
是3*m+6*n+20*k. |
a*******e
发帖数: 253 | 2 这个整数没有范围限制么？
譬如99+100*a. a=1,2,3...可以分解为: 20*(5a+4)+6*3+1
能满足要求不？
【在 b***z 的大作中提到】
: 找到一个最大的整数，这个整数用3，6，20来分割后余1或是2，也就是说这个整数不能
: 是3*m+6*n+20*k.
|
b***z
发帖数: 921 | 3 这个不满足要求。
99+100*a.
100*a是20的倍数，99是3的倍数。
这个整数要求不能被3，6，20来整倍分割。
【在 a*******e 的大作中提到】
: 这个整数没有范围限制么？
: 譬如99+100*a. a=1,2,3...可以分解为: 20*(5a+4)+6*3+1
: 能满足要求不？
|
g**3
发帖数: 3 | 4 你这分割是怎么定义的 不唯一吧 而且确定有最大值么 6是废的，3m+20k+1 m k取质数
的话不行吗 |
x******a
发帖数: 6336 | 5 3 and 20 are coprime... you can get any integer with 3m+20n if no other
condition on m and n than they are integers.
if m>=0, n>=0 are enforced, the largest I got is 37. |
n****e
发帖数: 2401 | 6 难的面试题有两种，一种是很难想出解法，考察逻辑思维，另一种是很难猜到没说完的
限制条件，考察交流技巧。
这题是第二种。对于第二种，每当你说出一个完全符合原题的答案，对方立即说不对，
因为还有个限制条件要补充一下。。。 |
r**a
发帖数: 536 | 7 are you sure 37 you got is the largest not smallest?
【在 x******a 的大作中提到】
: 3 and 20 are coprime... you can get any integer with 3m+20n if no other
: condition on m and n than they are integers.
: if m>=0, n>=0 are enforced, the largest I got is 37.
|
x******a
发帖数: 6336 | 8 I am not sure if it is the largest... But I am sure it is not the smallest
【在 r**a 的大作中提到】
: are you sure 37 you got is the largest not smallest?
|
r**a
发帖数: 536 | 9 I do not think it is the largest one and I even doubt the existence of the
largest one.
try 110
【在 x******a 的大作中提到】
: I am not sure if it is the largest... But I am sure it is not the smallest
|
b***z
发帖数: 921 | 10 可能这个实在考察沟通。怎么也想不出这题的答案。
【在 n****e 的大作中提到】
: 难的面试题有两种，一种是很难想出解法，考察逻辑思维，另一种是很难猜到没说完的
: 限制条件，考察交流技巧。
: 这题是第二种。对于第二种，每当你说出一个完全符合原题的答案，对方立即说不对，
: 因为还有个限制条件要补充一下。。。
|
|
|
b***z
发帖数: 921 | 11 m>=0 and n>=0 required.
How do you get 37?
【在 x******a 的大作中提到】
: 3 and 20 are coprime... you can get any integer with 3m+20n if no other
: condition on m and n than they are integers.
: if m>=0, n>=0 are enforced, the largest I got is 37.
|
s******y
发帖数: 416 | 12 你说的分隔，英文是divide么？
【在 b***z 的大作中提到】
: 找到一个最大的整数，这个整数用3，6，20来分割后余1或是2，也就是说这个整数不能
: 是3*m+6*n+20*k.
|
x******a
发帖数: 6336 | 13 I don't think i have to.
Because for any number p greater than 40, one of three numbers p, p-20 or p-
40 is divisible by 3.
correct me if I am wrong. Thanks!
【在 r**a 的大作中提到】
: I do not think it is the largest one and I even doubt the existence of the
: largest one.
: try 110
|
r**a
发帖数: 536 | 14 "for any number p greater than 40, one of three numbers p, p-20 or p-40 is
divisible by 3. " I agree.
However, I did not understand what conclusion you want to make here.
p-
【在 x******a 的大作中提到】
: I don't think i have to.
: Because for any number p greater than 40, one of three numbers p, p-20 or p-
: 40 is divisible by 3.
: correct me if I am wrong. Thanks!
|
r**a
发帖数: 536 | 15 Even if it is "divide", although it is totally different with LZ's 3m+6n+20k
, i think any number like 60k+1 will satisfy the condition.
Again, there must be something missing.
【在 s******y 的大作中提到】
: 你说的分隔，英文是divide么？
|
s******y
发帖数: 416 | 16 如果是divide，这是中国剩余定理。。。最大的没有，最小正或最大负解可以求到。解
的形式不是你说得这个样子。
20k
【在 r**a 的大作中提到】
: Even if it is "divide", although it is totally different with LZ's 3m+6n+20k
: , i think any number like 60k+1 will satisfy the condition.
: Again, there must be something missing.
|
x******a
发帖数: 6336 | 17 either p=20*0+ 3*n
or p=20*1+ 3*n
or p=20*2 +3*n
for some n>0
hence for any number greater than 40, it could be written in the form 20*m+6
*n+3*k for some nonnegative integers m, n, and k. therefore, the largest
number could NOT be written in this form should not be greater than 40.
Is this clear enough? Thanks!
【在 r**a 的大作中提到】
: "for any number p greater than 40, one of three numbers p, p-20 or p-40 is
: divisible by 3. " I agree.
: However, I did not understand what conclusion you want to make here.
:
: p-
|
b***z
发帖数: 921 | 18
p-
那么最大的满足条件的就是37. 因为任何大于40的数p, p, p-20或p-40都可以被3整除。
【在 x******a 的大作中提到】
: I don't think i have to.
: Because for any number p greater than 40, one of three numbers p, p-20 or p-
: 40 is divisible by 3.
: correct me if I am wrong. Thanks!
|
r**a
发帖数: 536 | 19 cool. thanks.
+6
【在 x******a 的大作中提到】
: either p=20*0+ 3*n
: or p=20*1+ 3*n
: or p=20*2 +3*n
: for some n>0
: hence for any number greater than 40, it could be written in the form 20*m+6
: *n+3*k for some nonnegative integers m, n, and k. therefore, the largest
: number could NOT be written in this form should not be greater than 40.
: Is this clear enough? Thanks!
|
