A 1968 by 2006 rectangle is divided into unit sqaures.?

Question

How many of the unit sqaures does a diagonal of the rectangle pass through?

Answer 1

The first thing we need to do is work out the greatest common factor of 1968 and 2006, which is 2.
So the line goes from (0,0) up to (984, 1003) where it first hits an intersection point, and then we do the whole thing all over again as we go up to the top right.
So, lets find the answer for a 984*1003 rectangle, and double it.

Now, we go into a new unit square every time we cross over a vertical line or a horizontal line. So we count the initial square as 1, then we cross over 983 vertical lines and 1002 horizontal lines to get to the top right. That makes 1+983+1002 = 1986 squares.

So we double that to get 3972.

The reason we worked out common factors first was to make sure we never pass over a vertical line and horizontal line at the same time - because then you only enter one new square. However, once we've eliminated common factors, that never happens.

(So in general, if the sides are a and b and have greatest common factor g, the answer will be
g*(a/g + b/g - 1) = a + b - g.)

Answer 2

The answer is 3972. Two proofs:

1.- 2006 (base) + 1968 (height) + 2 (extremum) = 3972
The argument holds because:
factors{2006}= {2,17,59}
and
factors{1968}={2, 2, 2, 2, 3, 41}
and 1968/2006~1, so there are no combinations of prime factors leading to a low rational ratio.

2.-
2006
Sum ceil(m i) - floor(m i-1) = 3972
i=1
where m i = 1962/2006 i. You can verify that it is the exact quantity of unit squares marked by the diagonal of the rectangle, because m i is the upper part of the marked column and m i-1 is the lower part marked. The ceil and floor functions denote the operation of taking the upper and lower integer part.

Hope it helps you :)

Answer 3

Help is on the way:
(1968)^2+(2006)^2=(x)^2
x=?

Answer 4

it all depends upon how many squares you choose to divide it into. What are you dividing by?