hi,
got stuck on a rather easy looking question. How many natural numbers satisfy the inequations
1) x+y >= 7
2) (x^2)+(y^2) <= 30
thanks
I suggest trying combinations of numbers and seeing what works. "Natural numbers" are positive integers, although some people define them to include the number 0 as well. So let's try:
Case 1: Let x+y = 7. Possible combinations for x & y are: 0 & 7, 1 & 6, 2 & 5, 3 & 4, 4 & 3, 5 & 2, 6 & 1, 7 & 0. Of these, which ones (if any) have the property that x^2 + y^2 <=30?
Case 2: now try x + y = 8. Posible combinations for x & y are: 0 & 8, 1 & 7, 2 & 6, 3 & 5, 4 & 4, etc. Again, which if any of these have the property that x^2 + y^2 <=30?
You should be able to see that the sum of the squares is smallest if x=y, which means for x+y = N that x and y both equal N/2. So once N gets large enough that (N/2)^2 > 30 there's no point in checking larger numbers.
Hello, nikhil!
How many natural numbers satisfy these inequations?
. . $\displaystyle x+y \:\ge\:7$
. . $\displaystyle x^2+ y^2\:\le\:30$
We are seeking points on or above the line $\displaystyle x + y \:=\:7$
. . and points on or inside the circle $\displaystyle x^2+y^2\:=\:30.$
Since $\displaystyle x$ and $\displaystyle y$ are natural numbers,
. . we are seeking "lattice points" only.
I found four points: .$\displaystyle (2,5),\,(5,2),\,(3,4),\,(4,3).$
x + y => 7
x^2 + y^2 <= 30
If x or y is > sqrt(30) > sqrt(25) = 5, then y^2 will exceed 30. Thus, consider points (x, y) where 1 <= x, y <= 5
(1, 1) (2, 1) (3, 1) (4, 1) (5, 1)
(1, 2) (2, 2) (3, 2) (4, 2) (5, 2)
(1, 3) (2, 3) (3, 3) (4, 3) (5, 3)
(1, 4) (2, 4) (3, 4) (4, 4) (5, 4)
(1, 5) (2, 5) (3, 5) (4, 5) (5, 5)
Eliminate the ones that don't satisfy your equation.