the rectangle

[Robert J. Kolker]

Errrr... yairse.

Werpsidaisy

I was going to say that I couldn't believe this thread was still going on, but I see the topic has shifted somewhat.

JJM

Next stop: Diophantine equations of degree three or less.

Bob Kolker

[hunterrose]

lol NOW do you guys see how easy it is for a 9 year old to see it especially if he has graph paper at his disposal?

So long as the problem is limited to whole numbers, it certainly is a very simple, effective, and intuitive solution.

Take a number in base 2 like 1110111001. First, add up the digits in boldface, that is, alternate ones starting with the last digit; call that sum A (here=3). Then add up the other digits; call that B (here=4). Then add A and two times B (here giving 11). If this sum is divisible by three, then the original number is as well. In this case, 11 is not divisible by 3, so neither is the original number (which is, if my addition is right, 953).

I was thinking in a similar vein. I used determining whether A-B (or B-A) is divisible by 3.

[Steve D'Ippolito]

So long as the problem is limited to whole numbers, it certainly is a very simple, effective, and intuitive solution.

I was thinking in a similar vein. I used determining whether A-B (or B-A) is divisible by 3.

It's analagous to the method you use to determine whether a base 10 number is divisible by 11. I suppose you could use it to determine whether any base n number is divisible by n+1.

I did end up proving for a homework problem in college that the method for determining if a base 10 number is divisible by 3 (add up all the digits, if the answer divides by 3, the original number divides by 3. If you add up all the digits and still get a huge number, you can add up all the digits in *that* number, as well; and repeat as often as necessary) was useful for other bases. If the number (in this case 3) has a multiple that is one less than the base (3x3=9 is one less than the base, 10), this trick works for that number in that base.

[brian0918]

Hi, people. I presented my 9 year old son with this question. he couldn't answer it. he went t ohis math teacher who couldn't either answer it. NO ONE in his school could answer it, not even the eighth grade teacher. Why do you think?

the question:

Find the area of any rectangle then find the perimiter. can the area ever equal to the perimiter? if it does, which rectangles would equal? why those particular ones?

my son (a fourth grader) was coming home with homework like "find the area of this, find the perimiter of that" the teacher should have known the answer, no? he's TEACHING the subject!

whadaya think?

The area can never equal the perimeter because they are not in the same units (meters versus square meters). However, if you just look at it as numbers without units... After some quick math, a rectangle with a height equal to 2/(1-2/base) would have the same area as its perimeter. So, if the base is 3, and the height is 2/(1-2/3) = 6, then the perimeter is 18 and the area is 18.

Edited February 11, 2008 by brian0918