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1.

Consider a 5x3 rectangle- it is 5 unit squares wide and 3 unit squares tall, thus made up of 15 little squares. If you shade the outside squares, the border squares, 12 squares are shaded, with 3 in the middle left unshaded.

Can you make rectangle with its border squares shaded so that the number of shaded squares equals the number of squares in the center? If so, how many rectangles (including squares) like that can you make? What are the dimensions of those rectangles?

If you think for a bit, you will realize that the formula for the amount of squares on the edge of a rectangle is:

border squares shaded = 2x+2y-4

(the -4 to eliminate the 4 overlapped corner squares)

where x and y are the dimensions of the rectangle

2.

A guy was having a paper written in English translated into French. He got the assistance of a French translator named Jacques. At the bottom of the paper was the following: (written in French of course)

"Much thanks to my friend Jacques for translating the above paper into French."

and then:

"More thanks to my friend Jacques for translating the above sentence into French."

and then:

"More thanks to my friend Jacques for translating the above sentence into French."

At first glance, this would have to continue forever, for proper thanks to be due. But it doesn't need too... it can (and does) end right there. Why?

no hint, this is easy. Use your brain

:P
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5x12 [or 12x5]: 60 squares, 30 on the perimeter.

6x8 [or 8x6]: 48 squares, 24 on the perimeter. That's all.

If the first acknowledgment is written in French, Jacques must have translated it

Plus de mercis à mon ami Jacques de traduire la phrase ci-dessus dans le Français.

That required an acknowledgment, which Jacques must also have translated,

since the wording is slightly changed - Plus de became Beaucoup.

Beaucoup mercis à mon ami Jacques de traduire la phrase ci-dessus dans le Français.

That also required thanks. But since the writer chose to use the same words,

he could do without Jacques' help; so no further attribution was needed.

Beaucoup mercis à mon ami Jacques de traduire la phrase ci-dessus dans le Français.

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Let us consider the height= x units

the width= y units

border perimeter=2x+2y-4

area inside perimeter= (y-2)(x-2)

So, (x-2)(y-2)=2x+2y-4

=> xy-2x-2y+4=2x+2y-4

=> xy=4x+4y-8

Any x,y which meets the above equation can be the rectangle size. After a tiny matlab program:

x=1;

y=1;

for n=1:10000

x=n;

for m=1:10000

y=m;

cond=x*y-4*x-4*y+8;

if (cond==0)

X=x

Y=y

end

end

end

Result: X=5(or 12), Y=12(or 5)

X= 6 (or 8), Y=8(or 6)

Very interesting....these are the only possible solutions!!!

End of the paper??? no space for writing??

Edited by storm
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Posted (edited) · Report post

If the first acknowledgment is written in French, Jacques must have translated it

Plus de mercis à mon ami Jacques de traduire la phrase ci-dessus dans le Français.

That required an acknowledgment, which Jacques must also have translated,

since the wording is slightly changed - Plus de became Beaucoup.

Beaucoup mercis à mon ami Jacques de traduire la phrase ci-dessus dans le Français.

That also required thanks. But since the writer chose to use the same words,

he could do without Jacques' help; so no further attribution was needed.

Beaucoup mercis à mon ami Jacques de traduire la phrase ci-dessus dans le Français.

But the riddle says he was having a paper written in English translated into French. That sounds like he wrote the whole thing, including those last three lines, and then submitted it to Jacques. He couldn't have known the translation before hand to have included it.

Edited by Lausus
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But the riddle says he was having a paper written in English translated into French. That sounds like he wrote the whole thing, including those last three lines, and then submitted it to Jacques. He couldn't have known the translation before hand to have included it.

The OP is not clear about who wrote the original paper.

The action of the "guy" was to have Jacques translate the paper; for that, he wanted to credit Jacques.

Reasonably, then, the credits were written [in English] after the translation.

The credits then needed to be translated; and that required the additional credits.

Make sense?

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The OP is not clear about who wrote the original paper.

The action of the "guy" was to have Jacques translate the paper; for that, he wanted to credit Jacques.

Reasonably, then, the credits were written [in English] after the translation.

The credits then needed to be translated; and that required the additional credits.

Make sense?

I admit that makes the most sense. It's probably the answer. Though it seems a little unreasonable that the paper should be returned by the translator, a sentence added, and then given back. The OP implies the credits were written before the translator got hold of the paper.

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bonanova is correct :D

the "guy" wrote it in English, later getting the help of Jacques to translate it, and then leave a thank you message (and a thank you message for translating the thank you message, and then again, and so on... or not). Storm, the amount of paper doesn't matter- there's a logical solution as to why it's not an infinitely repeating line.

The answers and my methods:

5x12 and 6x8 are the only solutions

method:

number of border squares on a rectangle of x and y dimensions:

2x+2y-4

we're looking for an amount of border squares equalling half the area, so:

2x+2y-4 = xy/2

multiply everything by two:

4x+4y-8 = xy

subtract xy and then add 8:

4x+4y-xy=8

multiply by -1:

xy-4x-4y = -8

add 16:

xy-4x-4y+16 = 8

now we can factor:

(x-4)(y-4) = 8

whole number positive factor pairs of 8:

8,1

4,2

2,4

1,8

so these are pairs of x-4 and y-4, so add 4 to get x and y from the pairs:

12,5

8,6

6,8

5,12

obviously half of those are duplicates, but if the order of x and y matter, than there are 4 solutions. Otherwise just 5x12 and 6x8

algebra ftw :P

There's no "method" to this, I got it right away when I saw it (it was in a Martin Gardner book, you should check em out :P), though I changed it around a little bit I think (been a while since I read it).

The answer is that he may need Jacques to right "thanks to Jacques for translating the above sentence", but then he can copy that same sentence again, knowing what it means. Thus it ends the infinite sequence.

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5X12

6X8

0X0 --trivial answer

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EventHorizon, 0x0 wouldnt be a rectangle ;D it would be a single point. Like how 1x0 rectangle is actually a line.

A 0x3x3 rectangular prism isnt actually a 3d rectangular prism as one of its dimensions is 0- it's a 2-dimensional object, as geometry goes. If a dimension is 0 then it's not counted as a dimension in geometry- it has to have some value to extend the axis in that direction. We don't say that a square has 5 dimensions if its 0x0x0x2x2, know what i mean?

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1.

Consider a 5x3 rectangle- it is 5 unit squares wide and 3 unit squares tall, thus made up of 15 little squares. If you shade the outside squares, the border squares, 12 squares are shaded, with 3 in the middle left unshaded.

Can you make rectangle with its border squares shaded so that the number of shaded squares equals the number of squares in the center? If so, how many rectangles (including squares) like that can you make? What are the dimensions of those rectangles?

If you think for a bit, you will realize that the formula for the amount of squares on the edge of a rectangle is:

border squares shaded = 2x+2y-4

(the -4 to eliminate the 4 overlapped corner squares)

where x and y are the dimensions of the rectangle

2.

A guy was having a paper written in English translated into French. He got the assistance of a French translator named Jacques. At the bottom of the paper was the following: (written in French of course)

"Much thanks to my friend Jacques for translating the above paper into French."

and then:

"More thanks to my friend Jacques for translating the above sentence into French."

and then:

"More thanks to my friend Jacques for translating the above sentence into French."

At first glance, this would have to continue forever, for proper thanks to be due. But it doesn't need too... it can (and does) end right there. Why?

no hint, this is easy. Use your brain

:P

Any soloution where 4x+4y-xy-8=0

Once Jacques had translated the sentence once, the writer could copy that translation, saving Jacques from having to translate it again.

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Here's another riddle, I just thought this one up:

3.

In base x, a number is 10101. The same number is 273 in base x+6

What is x? (ie, the base of the first number)

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Statman, your #2 answer is correct, though for #1 you might be surprised to know there are only two answers (or 4 if you're picky). Look at my spoiler in the post above yours on the first page :P your answer is correct, but you only went halfway

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EventHorizon, 0x0 wouldnt be a rectangle ;D it would be a single point. Like how 1x0 rectangle is actually a line.

A 0x3x3 rectangular prism isnt actually a 3d rectangular prism as one of its dimensions is 0- it's a 2-dimensional object, as geometry goes. If a dimension is 0 then it's not counted as a dimension in geometry- it has to have some value to extend the axis in that direction. We don't say that a square has 5 dimensions if its 0x0x0x2x2, know what i mean?

yes, but I was simply finding the values could be substituted for the width and length to fulfill the "number of shaded squares equals the number of squares in the center." 0X0 can be argued not to be a rectangle, but it is still a degenerate case of a rectangle. So I guess I should have said "degenerate case" instead of "trivial answer." I still believe it could/should be included.

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Here's another riddle, I just thought this one up:

3.

In base x, a number is 10101. The same number is 273 in base x+6

What is x? (ie, the base of the first number)

x=4

273 (decimal) in base 4 is 10101

the difference in bases between base 4 and decimal is 6.

edited to remove any confusion with bases

Edited by EventHorizon
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yes, but I was simply finding the values could be substituted for the width and length to fulfill the "number of shaded squares equals the number of squares in the center." 0X0 can be argued not to be a rectangle, but it is still a degenerate case of a rectangle. So I guess I should have said "degenerate case" instead of "trivial answer." I still believe it could/should be included.

Too bad my edit time expired...so here's a new post. Any degenerate case of a rectangle (either dimension is 0) will work.

So....how about in 3 dimensions....where each border cube is "shaded"?

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You got #3 right ;D

as for the 3d rectangular prism with border cubes shaded:

2x+2y+2z-8 = number of shaded cubes

2x+2y+2z-8 = xyz/2

*2

4x+4y+4z-16 = xyz

+16-xyz

4x+4y+4z-xyz = 16

*-1

xyz-4x-4y-4z = -16

hmm. Not sure how to make that factorable lol. There's *probably* a way, I'll think on it :P

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Why can we not use rectangles with a width of 1? there is an infinite amount of answers this way .. (1x2,1x3....1x4761) did i miss something?

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Why can we not use rectangles with a width of 1? there is an infinite amount of answers this way .. (1x2,1x3....1x4761) did i miss something?

Lets say the rectangle is 1Xx. Then there are x total squares, and all of which are shaded (all on the border). There are no unshaded squares. Therefore the number of shaded squares are not equal to the number of unshaded squares.

So rectangles of this sort will not work.

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Lets say the rectangle is 1Xx. Then there are x total squares, and all of which are shaded (all on the border). There are no unshaded squares. Therefore the number of shaded squares are not equal to the number of unshaded squares.

So rectangles of this sort will not work.

hi, im new. allow me to ask the dumb questions for a while

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hi, im new. allow me to ask the dumb questions for a while

Certainly. I'm fairly new to this site too, but like to explain things to people.

There are no dumb questions so long as you learn from them.

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Here's another riddle, I just thought this one up:

3.

In base x, a number is 10101. The same number is 273 in base x+6

What is x? (ie, the base of the first number)

x=4

To change both sides into base 10:

x4+x2+1=2(x+6)2+7(x+6)+3

Collecting same degree variables together, and factorizing

x4-x2-31x-116=0.

(x-4)(x3+4x2+12x+17)=0.

x=4 is one solution. The other solutions are -ve and fraction, so not suitable for a base.

Edited by brhan
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To brhan:

yep you got it :P

To eventhorizon's last post:

Exactly :P there are no dumb questions (well sometimes there are lol)

oh and I was wrong about my formula for shaded cubes. It's obviously not

2x+2y+2z-8 when you think about it for a second. It's all 6 faces, minus the shared strips

so that's

2xy+2yz + 2xz - shared strips

one sec on the shared strips ;D

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Exactly :P

oh and I was wrong about my formula. It's obviously not

2x+2y+2z-8 when you think about it for a second. It's all 6 faces, minus the shared strips

so that's

2xy+2yz + 2xz - shared strips

one sec on the shared strips ;D

Expand (x-2)(y-2)(z-2) (the unshaded cubes) and subtract it from x*y*z (total).

Thats the way I started the 2d one.

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Ah you did it from the inside out. I see. For the 2d one I just quickly saw it was 2x+2y-4, though it's not as simple for the 3d one. I'll do what you suggested! :P

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edited to remove any confusion with bases

x=4

273 (decimal) in base 4 is 10101

the difference in bases between base 4 and decimal is 6.

did anyone use math to solve this one? i mean break down the stuff. I came up with this

x^4 - x^2 - 31x - 116

however my math is a bit old so i could not figure out how to solve for x. but 4 seems to satisfy my equation.

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did anyone use math to solve this one? i mean break down the stuff. I came up with this

x^4 - x^2 - 31x - 116

however my math is a bit old so i could not figure out how to solve for x. but 4 seems to satisfy my equation.

you could use the quartic formula (not recommended), or just plot it on a calculator and then use synthetic division to remove roots that you find.

According to brhan's post (first on page 3) your equation is right.

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