find the number of solutions to the equation 2x+5y=2015 where x and y are positive integers?
回答 (3)
2015-09-05 10:32 am
✔ 最佳答案
2x = 2015 - 5yif y = 1, 2x = 2010 => x = 1005
if y = 3, 2x = 2000 => x = 1000
if y = 5, 2x = 1990 => x = 995
...
if y = 401, 2x = 10 => x = 5
if y = 403, 2x = 0 => x = 0
number of solutions = (401 - 1)/2 + 1 = 201
2015-09-06 2:52 am
2x+5y=2015
5y=(2015-2x)
y=(2015-2x)/5
y=403-(2x)/5
let x=5k
y=403-2k
x>0 => k>0
y>0 => 403-2k > 0
2k < 403
k < 201 1/2
=> k = 1 to 201
201 numbers
5y=(2015-2x)
y=(2015-2x)/5
y=403-(2x)/5
let x=5k
y=403-2k
x>0 => k>0
y>0 => 403-2k > 0
2k < 403
k < 201 1/2
=> k = 1 to 201
201 numbers
2015-09-05 10:06 am
2x + 5y = 2015
2x = 2015 - 5y
x = (2015 - 5y)/2
y is an integer.
When x is an integer :
[(2015 - 5y) > 0] and [(2015 - 5y) = an even number]
[5y < 2015] and [5y = an odd number]
[y < 403] and [y = an odd number]
Hence, y = 1, 3, 5, ......, 401
Number of solutions
= [(401 - 1)/2] + 1
= 201
2x = 2015 - 5y
x = (2015 - 5y)/2
y is an integer.
When x is an integer :
[(2015 - 5y) > 0] and [(2015 - 5y) = an even number]
[5y < 2015] and [5y = an odd number]
[y < 403] and [y = an odd number]
Hence, y = 1, 3, 5, ......, 401
Number of solutions
= [(401 - 1)/2] + 1
= 201
收錄日期: 2021-04-18 00:10:36
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150905014647AAmL6vn