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#21 brhan

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Posted 04 March 2008 - 11:30 PM

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, 04 March 2008 - 11:36 PM.

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#22 unreality

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Posted 04 March 2008 - 11:30 PM

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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#23 EventHorizon

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Posted 04 March 2008 - 11:33 PM

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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#24 unreality

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Posted 04 March 2008 - 11:50 PM

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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#25 lord

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Posted 05 March 2008 - 12:01 AM

Spoiler for answer


edited to remove any confusion with bases


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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#26 EventHorizon

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Posted 05 March 2008 - 12:25 AM

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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#27 unreality

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Posted 05 March 2008 - 02:22 AM

Yep, and to brhan: there are number bases based off of irrational numbers, but they're kind of complicated lol. My favorite is phinary, just cuz the Golden Ratio (and fibonacci numbers) are so amazing

anyway, how I would mathematically solve #3 if I didnt already know the answer:

10101 in base x is, in other words:

x0 + x2 + x4 in base 10
or
1 + x2 + x4 in base 10

changing 273 from base x+6 into base 10 is also easy:

2*(x+6)log(100) + 7*(x+6)log(10) + 3*(x+6)log(1)

or:

2*(x+6)2 + 7*(x+6)1 + 3

now that they are both in base 10, they can equal each other:

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

subtract 1 from both sides

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

(x+6)2 is (x+6)(x+6), which is x2+12x+36, all multiplied by 2 is 2x2+24x+72

7*(x+6)1 is 7x+42

x2 + x4 = 2x2+24x+72 + 7x + 42 + 2

now we have a ton of like terms on the right side to merge:

x2 + x4 = 2x2 + 31x + 116

subtract x2 from both sides

x4 = x2 + 31x + 116

subtract x4

0 = -x4 + x2 + 31x + 116

*-1

0 = x4 - x2 - 31x - 116

factor

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

From there, what can easily turn that into 0. The clear answer is 4, cuz 4-4=0, which would negate the rest of it. If you try that, 10101 in base 4 is 273 in base 10 (4+6)

I believe this was brhan's method. This is the mathematical/algebraic way to do it.

The logic way to do it, which is probably what I would have really done, is:

* x+6 cant be 7 or smaller, cuz a 7 appears in the number so it would carry over as a 10 if x+6 was 7 or lower. Therefore x+6 is 8 or more
* which means x is 2 or more
* x+6 could be more than 10, but no symbols were used in 273 so we're not sure if x+6>10 yet, though it's unlikely, since no symbols were used in 273, so it's not showing the part of the base, and what symbols would you use, letters..? To see if that's an option, figure out if x was higher than 4 and see if that's a possible solution. However if x was higher than 4, for example 5, than 1+5^2+5^4 = 651 in base 10, or, in base 11, 542, already much too big than 273
* so our range is x=2 or 3 or 4
* if x was 2, simple binary, then x^4+x^2+x^0 is 16+4+1 = 21 in base 10, which is (16=20)+5 = 25 in base 8 (8=x+6 if was x was 2). Clearly 25 is much less than 273.
* I could try x=3, but I can already see that 1+9 is 10 in base 10, or 11 in base 9, so 3^4 would have to be 262 in base 9, but I know that 3^4 is 9^2 and 9^2 is 81 in base 10- which is very cleanly 100 in base 9. 100+11=111 in base 9, not 273. Not even close. So I dont even need to check base 3 fully
* the only thing left is x=4
* 1+16+256 is 273. Yay! The number! But wait- dont you have to convert it to base 10, so it will change *sinking feeling* But not! Because if x is 4, than x+6 is 10. 273 is in both base 10 and base x+6.

Therefore x=4

:P
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