mulla sadra

1) Find all integers n in the interval [100,1997] that divide 2^n+2.

2) Find all positive integers n that divide 2^n+1.

If you have x^2+a*x+b=0, solve it to get two roots a1<=b1.

Now, form another similar equation: x^2+a1*x+b1=0.

Solve this second equation to get two roots a2<=b2.

Continue this pattern.

Question: Prove that you can not produce more than five equations. That is, prove that you will reach, in the best case, up to x^2+a5*x+b5=0, which will not have solutions in real number system.

Note: a and b are nonzero real numbers.

number of unit cubes that the longest diagonal passes through?

Note: the longest diagonal is the line segment from one vertex to the farthest vertex in the rectangular box.

the rectangular box is just a box with width, length, and height or depth.

The box is divided in unit cubes in the same way a rectangle in two dimension is divided into a grid of unit squares.

I want the number of unit cubes the diagonal traverses and not its length.

This is a general question that needs accordingly a general answer in form of a procedure or a formula. The width, length, and height are all general and not specific.

However, you can try 3,4,5 initially

If you have x^2+a*x+b=0, solve it to get two roots a1<=b1.

Now, form another similar equation: x^2+a1*x+b1=0.

Solve this second equation to get two roots a2<=b2.

Continue this pattern.

Question: Prove that you can not produce more than five equations. That is, prove that you will reach, in the best case, up to x^2+a5*x+b5=0, which will not have solutions in real number system.

f(x) = 520*sqrt { x * ( [2 / (x+1)] ^ [ (x+1) / (x-1) ] ) }.

What is lim f(x) as x ----> infinity.

number of unit cubes that the longest diagonal passes through?

Note: the longest diagonal is the line segment from one vertex to the farthest vertex in the rectangular box.

the rectangular box is just a box with width, length, and height or depth.

The box is divided in unit cubes in the same way a rectangle in two dimension is divided into a grid of unit squares.

sqrt[x-sqrt(2x-1)]+sqrt[x+sqrt(2x-1)] = sqrt(2).

For example, if n=3, we have 000,111,010,011 as possible bit arrangements around a circle. All other combinations are in fact one of these. For example, 001,& 100 are the third one (010) but just shifted circularly.