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Marty McFly

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Marty,

Go to this site, GRAPH SKETCHER, and key in 2*x/(x-2) to see a graph of the function. It shows that it's a hyperbola, with the x (or y) approaching 2 as y (or x) approaches infinity. This is why the only positive integer pairs that work are (4,4) and (3,6) (or (6,3)).

I think you ought to review some definitions and concepts, so that you don't confuse your son. The fact that you say that 8/3 is not a real number, and not a rational number, shows that you either have very little mathematical training, or you have forgotten what you did have. Look up the definitions of real number, rational number, integer, whole number, irrational number, imaginary number. You need to use these terms the same way they are traditionally used in mathematics, or your son will be very confused. Also, 2.66666 repeating is just as precise as 4.0000 repeating. Any real number is just a precise as any other. Think about a right triangle with sides of length 1 and 2. The hypotenuse has a length of exactly the square root of 5, even though that's an irrational number.

oh, no no no, I never said it was not a REAL or RATIONAL number, only it's not a number you can really measure with. I mean take a unit: a unit or 4 units of the same size - basically a whole number or even not a whole number, but a number that does not have an infinite decimal. the answer IS 3,6 and 4,4. these ARE the only ones that work. but why is that? WHY?

and I don't agree that 4.0000 is the same as 2.66666 because an infinite number of zeros is still zero. and 4 will be on the fourth dot of the ruler, or exactly four candies or clap 4 times. but 2.666666666666666666666666666666666666666666666666666 if you look at the hyperbula, it's like 0.999999999999999999999999999999999999999999 it almost gets to 1 but it never really gets there. it never really touches that point. it keeps getting closer, closer, closer, but NEVER touches that point.

edit: now tha tI read all my previous posts, you are right, I did say that these # are not real or rational. sorry, as Steve said, I mis labled these numbers. what I meant was Whole Numbers. so the only whole # that work ARE 3x6 and 4x4.

Edited by Marty McFly
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Marty, I think the notion of a number that "one cannot measure with" is a bit vague. I can understand if you mean numbers that "a 9-year old would (or would not) be expected to consider" (i.e. some type of pedagogical notion rather than a mathematical one). Even there, the repeating decimal would not be the defining factor. For instance, 9 year olds would be expected to think of 1/3rd, even though it is 0.3333...

Edited by softwareNerd
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Think about a right triangle with sides of length 1 and 2. The hypotenuse has a length of exactly the square root of 5, even though that's an irrational number.

I always had a hard time agreeing with this a squared times b squared equals c squared.

if it's a right triangle with both sides 2 or 3 or 4 etc. the hypotenuse will be 2 or three or 4 according to this theory, but that CAN'T be. can it? so I never liked this theory :dough::confused::( as you can tell I am not a professor

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Again, false. You're being thoroughly misled essentially by the fact that we have ten fingers. As I already mentioned, it's quite easy to trisect a line segment, which means that you can divide a ruler into thirds, ninths, twenty-sevenths, and so on right down as far as you want with as much precision as you want (there's nothing special at all about dividing a ruler into tenths, hundredths, and so on instead or thirds), so in fact 2.6666.... is a number you can measure with just as readily as 4.000... The thing is, 3 is not a divisor of 10, so its decimal expansion is infinitely repeating. (And it's because it's infinitely repeating, roughly, which is what "rational number" means, that you can base a ruler on units of three. The case is different if they're infinitely many non-repeating digits.) But it still corresponds to a definite length precisely measurable by trisecting a unit if that unit is precisely measurable (though not one that will match up with any of the lengths ticked off on a ruler by subdividing it forever and ever by tenths, which is what the infinitely repeating threes means geometrically); otherwise you're in the position of saying that only some lengths can be precisely and meaningfully trisected, all others being meaningless lengths (essentially any lengths not formed from the original unit by combinations of repeated subdivisions by halves and fifths)--and this by simple virtue of the fact that you've chosen one essentially arbitrary length as your basic unit of length. If you choose a different unit, then many of these unmeasurable lengths will in general suddenly become measurable.

that's interesting. I am so distressed by this though. because .33333333333333333333333333 is such an impercise number, but on the other hand, it's totally ligitimate when it's divided in thirds..... (I have a head-ache now)

This isn't meant to give you a hard time. Rather, it's worth going through because all of this has been an issue in the philosophy of mathematics since the Ancient Greeks, though in a slightly different form. How do mathematical entities connect to reality? How does abstract geometry relate to actual measurements? A common Greek reply was essentially Platonic, that they are pure forms or what-not imperfectly reflected in reality. (Though at least one Greek, Protagoras, took essentially your approach, which is to say that only what is concretely real is true--thus, he argued that because in physical reality all circles and lines have finite thickness, then no lines or circles can really intersect in only one point.)

ok, I like Protagoras (how do you know all this btw?)

With the adoption of decimal notation, we moderns have additional questions--how do numbers in general relate to the integers, and more generally how do algebra and geometry connect up with each other? The way it's handled in modern mathematics is to use bare numbers--integers and pure fractions--to indicate what is countable and hypothesized to be precisely measurable, decimal expansions to indicate measurements to a certain degree of precision; and the pure, abstract mathematics is tied to real measurements by indicating the degree of precision in any subsequent measurement in terms of the precision of the first measurement.

like the hypotenuse of a triangle?

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I always had a hard time agreeing with this a squared times b squared equals c squared.

if it's a right triangle with both sides 2 or 3 or 4 etc. the hypotenuse will be 2 or three or 4 according to this theory, but that CAN'T be. can it? so I never liked this theory :dough::confused::( as you can tell I am not a professor

Marty, You misunderstand the theorem. There's nothing to like or dislike in the Pythagorean theorem; it's simply a fact, not open to debate.

The theory actually says: sqr(a) + sqr(B) = sqr(HYP)

so, with both sides of 2, sqr(2) + sqr(2) = 4 + 4 = 8 = sqr(HYP)..... so HYP = Square-root of 8 = 2.82 (approx.)... not 2

The link I provided has various proofs of the theorem.

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that's interesting. I am so distressed by this though. because .33333333333333333333333333 is such an impercise number, but on the other hand, it's totally ligitimate when it's divided in thirds..... (I have a head-ache now)

ok, I like Protagoras (how do you know all this btw?)

like the hypotenuse of a triangle?

I just spent an hour replying to this, and the damn computer ate it. Sorry, I'm not up to doing it again.

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Okay, I can still give a stripped-down version of what I wrote while it's still fresh.

that's interesting. I am so distressed by this though. because .33333333333333333333333333 is such an impercise number, but on the other hand, it's totally ligitimate when it's divided in thirds..... (I have a head-ache now)

You should look at it this way--there's a numerical view of numbers (in which numbers are written in digits and decimals) and a geometrical view (in which numbers are represented by lengths), and each has its strengths. The numerical view is very convenient for arithmetic and is very easily suited to treating questions of precision naturally--for one thing, the number of digits you list tells you how precise the measurement is. The problem is that numbers have to be written out in base notation, corresponding to the number of subdivisions you make to represent the fraction geometrically, and there is no base in which all fractions terminate (have a finite number of digits), essentially because there are infinitely many prime numbers. Changing the base allows you to represent some of these fractions as terminating, but changing the base is a bit of a pain compared to just choosing a different subdivision for your unit of length in geometry. (Plus, if you've been taught math using our decimal notation, base 10, the other bases are not at all natural to handle.)

Then there's the question of numbers with infinitely many non-repeating digits, which are called irrational numbers. Rational numbers are those that are equivalent to a ratio of two whole numbers, or more precisely of integers* (and have either a finite number of decimal places or else repeat infinitely), while irrational numbers are not. Irrational numbers are not too much of a problem for the numerical view of numbers, especially for all practical uses, since they are exactly parallel to rational numbers--the rules and procedures for handling them in arithmetic are completely unchanged for rational and irrational numbers, and you just have to write them out to the precision needed so the question of how all later digits act is irrelevant. (They're not so convenient as rational numbers for number theory or just plain curiosity, however, since you have to calculate later digits than you might have written down with a formula--they're not as easily specified as rational numbers, in other words.) [* "Counting numbers" are 1,2,3... (Some people also call these the "natural numbers," which is what I know them as, while apparently others equate natural numbers and whole numbers.) "Whole numbers" are the counting numbers and 0. "Integers" are the whole numbers and the negatives of the counting numbers: ...-3,-2,-1,0,1,2,3...]

However, irrational numbers are a bit troublesome philosophically for the geometrical view because they can't be measured in the same way as rational numbers. To represent a fraction, you subdivide a unit and add up some of the subdivisions--in other words, you reduce measurement to straightforward counting. 3/8 would be rpresented by dividing the unit into equal eights and taking the length of three of them, for example. You can't do that with irrational numbers. Take a square of side 1 and consider the diagonal across it; the length of that diagonal is irrational. What that means is that there's no finite subdivision of the side of the square that you can add up so many times to give the length of the diagonal (more precisely, add to the length of the side, since it's greater than 1), but it's quite easy to construct that length geometrically in other ways. (That is, the two lengths are said to be "incommensurable"; neither one can be measured by a finite number of finite subdivisions of the other.) What this means for the numerical view of numbers is that there is no base, no subdivision of numbers by thirds or fifths or hundred-thirteenths or what-not, in which an irrational number will be infinitely repeating--it's always infinitely non-repeating.

ok, I like Protagoras (how do you know all this btw?)

He was one of the Sophists before Socrates; his most famous statement is "Man is the measure of all things." I know all this because I read a lot about the history and philosophy of science and also like to play around with number theory. (That's the study of integers, prime numbers, fractions, decimal representations, rational and irrational numbers, and so on; like geometry, it's a fine hobby if you want a mathematical hobby because you can go from simple basics to very interesting results without having to make a career out of it.)

like the hypotenuse of a triangle?

Yes, though a better, simpler example is the area of a right triangle--cut a rectangle in two across one pair of diagonals. The area is A=x*y/2; the 2 would never have a decimal because it's a pure number (essentially it results from counting one of the two halves of the rectangle, if you want to look at it that way). In pure math, that's all you need to say, but if you're talking about the area of a real triangle whose sides you've measured, you have to take a bit more trouble. First, you can only measure the length of each side to a certain precision; if the shortest division on your ruler is d, then you only know for sure that the real length of each side is within half that distance d from the measured value (x and y). (Well, in many cases that's the only significant imprecision--if the hatch marks on the ruler are about the same size as the division between them, that adds a comparable source of imprecision, and there's also the question of how straight the ruler edge is and how accurate the measurement of the right angle is. Sometimes these have to be taken into account in the final result, sometimes not. And then if you're doing calculations on a computer you have round-off error as well. No doubt SoftwareNerd or one of the other regulars could tell you some tales about that.) So, x is no longer than x+d/2 and no shorter than x-d/2, and the same with y, so the largest area the triangle can have is (x+d/2)*(y+d/2)/2, and the smallest (x-d/2)*(y-d/2)/2. (All of those 2's are pure numbers because they result from the geometry of the situation, not from measurement.) If you expand the products, you find that the area can be no more than (x+y)*d/4+d^2/8 larger than A (calculated from the x and y you measured), and no more than (x+y)*d/4-d^2/8 smaller than A. Then, if d is very much smaller than x and y, d^2 is very much smaller still, so you can ignore the term in d^2 (if it's the last step in your math) to that degree of precision and say that the precision (to be exact, the maximum error) in the area is (x+y)*d/4. You can do the same thing for the error formula for the length of the hypotenuse of a right triangle from the Pythagorean Theorem, but you'd have to know the series expansion for the square root of a sum to follow it, which I suspect you're not familiar with. In the geometrical view you don't have to worry about all that, and to get the result in pure math, you just set d=0.

Edited by Adrian Hester
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Marty, You misunderstand the theorem. There's nothing to like or dislike in the Pythagorean theorem; it's simply a fact, not open to debate.

The theory actually says: sqr(a) + sqr(:confused: = sqr(HYP)

so, with both sides of 2, sqr(2) + sqr(2) = 4 + 4 = 8 = sqr(HYP)..... so HYP = Square-root of 8 = 2.82 (approx.)... not 2

The link I provided has various proofs of the theorem.

oh, shit! I am so embarrassed, I would hide in a burning building! I must have been drunk last night, because what I did was a squared TIMES b squared equals c squared! OMG I am such an idiot! totally disregard what I wrote before! Oh, I wish I could delete this post! (the shame, the shame!)

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Marty, I think the notion of a number that "one cannot measure with" is a bit vague. I can understand if you mean numbers that "a 9-year old would (or would not) be expected to consider" (i.e. some type of pedagogical notion rather than a mathematical one). Even there, the repeating decimal would not be the defining factor. For instance, 9 year olds would be expected to think of 1/3rd, even though it is 0.3333...

But then again, the Greeks themselves had a fondness for finding integer solutions to equations (Diophantine equations)--essentially solving problems involving countable things. And when you think about it, it's pretty gruesome to consider the real-world significance of non-integer solutions to word problems about the number of cattle in two herds, say. On the other hand, there's nothing gruesome or unnatural about rectangles with non-integer sides.

Edited by Adrian Hester
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oh, shit! I am so embarrassed, I would hide in a burning building! I must have been drunk last night, because what I did was a squared TIMES b squared equals c squared! OMG I am such an idiot! totally disregard what I wrote before! Oh, I wish I could delete this post! (the shame, the shame!)

You seem to be overly excitable regarding math as this is not the first mistake you have made (in this thread). Might I suggest you not play math games with your child if you don’t know what your talking about.

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First, when X and Y are not constrained to be integers there are an infinite number of rectangles with different proportions that fit the question. There is a solution to Y for every single X, except X=2.
You meant that there is a solution for every x that is greater than 2?

Heck, I suspect that the teacher wanted to be given a bunch of only 4x4 answers and would get a tad flustered by students who found 3x6, but I could be being too cynical.
I agree, especially if a student asked why only those two solutions worked.

I plan to try this one on my own 9 year old, to see how he tackles it. Will report back later.
Tell us how it goes :)

the most important question that follows that, is what my son hadn't yet figured out, is WHY are these the only possible answers? why can no other rectangle equal the perimiter # with the area #?
Area A= x*y

Perimeter P= 2(x+y)

Like John McVey said, there's no solution, whole number or not, if x<=2 or y<=2. At 2, the square's perimeter would equal 2(A/2 + 2)=A+4 i.e. the perimeter's going to equal more than the area. All the more so for any number less than 2.

And while I don't have a definitive quickeasyfast solution to show why no other numbers besides 4x4 and 3x6 work, I used

A/x + x = P/2

to test some values for x. (If A=P, then A/x=y.)

At very low (e.g. at or below 2) values of x, the x makes the perimeter>area (units aside). And "high" values of x make A/x significantly lower than A/2, low enough that adding x isn't enough to prevent the area from being greater than the perimeter.

</geekspeak>

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You seem to be overly excitable regarding math as this is not the first mistake you have made (in this thread). Might I suggest you not play math games with your child if you don’t know what your talking about.

I was about to respond to Adrian (who is a smart, well educated person) and just couldn't help to respond first to this one. it's not true what you say about not playing math games with my son. I am not giving him false information. I am simply giving him puzzles that he might enjoy (I know I enjoy them) and none of you answered the question WHY NO OTHER RECTANGLE HAS area = to perimiter other than 4,4 and 3,6? so no matter how uneducated I am, I'm still smart enough to answer this question

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I was about to respond to Adrian (who is a smart, well educated person) and just couldn't help to respond first to this one. it's not true what you say about not playing math games with my son. I am not giving him false information. I am simply giving him puzzles that he might enjoy (I know I enjoy them) and none of you answered the question WHY NO OTHER RECTANGLE HAS area = to perimiter other than 4,4 and 3,6? so no matter how uneducated I am, I'm still smart enough to answer this question

It has been answered already, in case you missed it, in Laure's posting up the thread, and expanded on by hunterrose. First, there are other rectangles that meet the given requirement, an infinite number of them, in fact. (For example, the rectangles with sides x=2.1, y=42; x=2.5, y=10; and x=2.8, y=7.) If you mean why are there no other rectangles with integer sides, that too has been abnswered. For x between 0 and 2, the formula gives y<0, which is physically meaningless. For x=2, y is undefined (division by 0). For x>4, you get the same results as for x between 2 and 4, just with the roles of x and y reversed; thus, for integers x>4, you get the same rectangle as for a value of x between 2 and 4, just rotated 90 degrees. So, the only integers for which you can have distinct rectangles are for x=3 and x=4. In this case, both possibilities work.

Edited by Adrian Hester
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Adrian, there is an answer much more visual and easier for a child to find, no one on this forum has yet reached that conclusion. no child can understand calculus, so there is a very simple reason that this is the only 2 rectangles that have area=perimiter. no one was able to answer that - not even math teachers. can any of you see it? I've seen it, and am hoping my son will. (I will not give the answer here because I want those of you who are lovers of math or geomitry to try and figure it out on your own)

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Adrian, there is an answer much more visual and easier for a child to find, no one on this forum has yet reached that conclusion. no child can understand calculus, so there is a very simple reason that this is the only 2 rectangles that have area=perimiter. no one was able to answer that - not even math teachers. can any of you see it? I've seen it, and am hoping my son will. (I will not give the answer here because I want those of you who are lovers of math or geomitry to try and figure it out on your own)

You mean algebra, not calculus, but anyhow, I think I have it.

Divide a rectangle up into its unit squares. Each square contributes one unit to the area. Ones on an edge have one side on the perimeter, and contribute one unit to the perimeter. Ones on corners have two sides on the perimeter, and contribute two units to the perimeter. Interior squares have no sides on the perimieter. So the n squares that are actually on the perimeter of the rectangle yield n+4 unit lengths of perimeter. To cancel these out you need exactly four squares in the interior that make no contribution to the perimeter, so that there are n+4 total squares in your rectangle.

So now focus on the interior four squares--they can either be in a horizontal line, a vertical line, or themselves arranged in a square. Then surround them with other squares to serve as the perimeter, and you get a 3x6, 6x3 or 4x4 arrangement of squares.

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You mean algebra, not calculus, but anyhow, I think I have it.

Divide a rectangle up into its unit squares. Each square contributes one unit to the area. Ones on an edge have one side on the perimeter, and contribute one unit to the perimeter. Ones on corners have two sides on the perimeter, and contribute two units to the perimeter. Interior squares have no sides on the perimieter. So the n squares that are actually on the perimeter of the rectangle yield n+4 unit lengths of perimeter. To cancel these out you need exactly four squares in the interior that make no contribution to the perimeter, so that there are n+4 total squares in your rectangle.

So now focus on the interior four squares--they can either be in a horizontal line, a vertical line, or themselves arranged in a square. Then surround them with other squares to serve as the perimeter, and you get a 3x6, 6x3 or 4x4 arrangement of squares.

YES!!!! you got it you got it!!! you deserve a prize! 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? :lol:

I am so happy you got it!! NO ONE else was able to figure it out!! and I was going crazy keeping it to myself!! because when I figured it out (And I, like a 9 year-old, I have no education, really) I wanted everyone else to do the same.

Edited by Marty McFly
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NO ONE else was able to figure it out!!

That's completely false. There were multiple explanations given, just none that fit the requirements you gave after the answers were already given. So, NO ONE else was able to figure out both the answer to the question and the restriction you had yet to put on the answer.

It's a very important skill of a rational person to know the limits of his knowledge in a particular field. Much as I would not try to tell an oil tycoon that he's digging in the wrong places based on my very rough knowledge of geology, I recommend that you don't try to tell mathematicians what parts of mathematics are rational based on your (self-admitted and evidenced by your posts on this topic) very rough knowledge of mathematics. You lack the understanding of the context behind such concepts as numbers that do not rise from ratios (which is what "irrational" means, literally) and the difference between the notation used to represent a number and the quantity actually represented by that number, and as such you're offering up ridiculous straw-man arguments or simply confusing statements.

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YES!!!! you got it you got it!!! you deserve a prize! 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? :lol:

I am so happy you got it!! NO ONE else was able to figure it out!! and I was going crazy keeping it to myself!! because when I figured it out (And I, like a 9 year-old, I have no education, really) I wanted everyone else to do the same.

Actually that was the first time you said you wanted a visual solution. And it took a while for you to make it clear you wanted whole numbers.

I am not uniquely brilliant (or alternatively--the only not-stupid person in this thread); I just happened to be the first to see your question in its final form.

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Actually that was the first time you said you wanted a visual solution. And it took a while for you to make it clear you wanted whole numbers.

I am not uniquely brilliant (or alternatively--the only not-stupid person in this thread); I just happened to be the first to see your question in its final form.

Ho, ok, now I'll slide back to my little hole. BTW, Adrian, I really appreciate your lesson. I read it a few times, and can see how much you enjoy the subject.

there is a rectangle in existence ( I wish i remembered which measurement it was) that when it's whole it has one area and when dissected by a diagnal making two triangles, it loses a square. one day I'll find it for you.

also, speaking of different bases, how can you tell if a base 2 number (1, 10, 11, 100 etc.) is divisible by 3 in base 10 witout first changing the number into base ten?I guess all the math teachers in my son's school are stupid, though. because it didn't take long for you guys to solve it.

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also, speaking of different bases, how can you tell if a base 2 number (1, 10, 11, 100 etc.) is divisible by 3 in base 10 witout first changing the number into base ten?I guess all the math teachers in my son's school are stupid, though. because it didn't take long for you guys to solve it.

Forget base 10. Change it into a base 3 number. If the low order digit (base 3) is zero, then and only then is the number divisible by 3.

Bob Kolker

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there is a rectangle in existence ( I wish i remembered which measurement it was) that when it's whole it has one area and when dissected by a diagnal making two triangles, it loses a square. one day I'll find it for you.

As I recall, that particular problem relies on the diagonal not being quite the same as the line on the re-assembled triangles. Off by just a hair.

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how can you tell if a base 2 number (1, 10, 11, 100 etc.) is divisible by 3 in base 10 witout first changing the number into base ten?

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).

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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).

ok guys! I love you *hugs* can you give me a puzzle to mull over? it won't take me only a day, it might take a month, but it will be fun.

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