Question

x,y and z are positive integers such that x<y,x+y=201,z−x=200. What is the maximum value of x+y+z?

Answer 1

x + y = 201

y = 201 - x

y > x that's why x < 201 / 2

x < 100.5

Nearest integer:

x = 100


x + y = 201

y = 201 - x

y = 201 - 100

y = 101


z − x = 200

z = 200 + x

z = 200 + 100

z = 300


x + y + z = 100 + 101 + 300 = 501

Answer 2

OR

x + y = 201

y = 201 - x


z − x = 200

z = 200 + x


x + y + z = x + 201 - x + 200 + x

x + y + z = 401 + x


x + y = 201

mean :

x < 201 / 2

x < 100.5

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For example :

If x = 101

y = 201 - x = 201 - 101 = 100

That not satisfie condition :

x < y


If x = 102

y = 201 - x = 201 - 102 = 99

That not satisfie condition :

x < y


If x = 103

y = 201 - x = 201 - 103 = 98

That not satisfie condition :

x < y

etc.

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That's why :

x < 201 / 2

x < 100.5

Largest integer less of 100.5 are 100

so

x = 100


x + y + z = 401 + x = 401 + 100 = 501