What would make this equation true??

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[ConManAU]

4:40 You could also note that since y-48 is odd, then that means that x-12 must be a multiple of 64 because all those factors of 2 have to go somewhere. But the only multiple of 64 that keeps x-12 < 100-12 = 88 is 64 itself, meaning that x = 76 is the only possibility.

[MizardXYT]

All positive solutions to 1/x + 4/y = 1/12, without the other constraints:
(x, y) ∈ { (13, 624), (14, 336), (15, 240), (16, 192), (18, 144), (20, 120), (21, 112), (24, 96), (28, 84), (30, 80), (36, 72), (44, 66), (48, 64), (60, 60), (76, 57), (84, 56), (108, 54), (156, 52), (204, 51), (300, 50), (588, 49) }

[ZonaALG]

I actually thought whether he was recording on a school or not the whole video. Great ending!!!

[goodplacetostop]

4:56 Michael’s homework
6:03 Behind the scenes... It doesnt seem to have enough space for backflips without touching the softbox 🤔

[RandomBurfness]

I found that without the first restriction, namely that x, y <= 100, you still only have three solutions! The solutions are

x = 76, y = 57
x = 204, y = 51
x = 588, y = 49

[udic]

4:53 there are also minus1, minus 3 and minus 9 that you should have "checked". Of course that will result in x which is negative ( and thus in Z instead of N) but none the less they should at least be mentioned just like 1 & 3.
There are other questions like this one, which negative factors might result in positive x and y.

[ItIsApachee]

y - 48 can also be -1, -3, or -9. In some cases that would add new solutions to the equation. It's not the case in this particular one, but it doesn't mean you shouldn't consider negative factors.

[nnaammuuss]

A little gap in the argument: the possibilities –1, –3, –9 for y–48 haven't been considered. viz. we are trying to solve x′y′=9×64 (x′=x–12, y′=y–48); x′, y′ integers, y′ odd; and –11≤x′≤88, –47≤y′≤52 —and this doesn't automatically rule out the negative values.

However, just as one eliminated the possibilities 1 and 3, one could argue that negative values for y’ will give us –x′≥64, landing us outside the range, thus ruling them out.

[manucitomx]

Very nice problem for a Sunday morning.
Thank you, professor.

[thephysicistcuber]

It's a nice coincidence that 12*48=576 (the value you added) and that the values of y and x when put side by side form the number 5776 (which BTW is 76^2).

[FaerieDragonZook]

Another way to solve it would be the following:
4/y = 1/12 - 1/x = (x-12)/12x
Count the powers of 2 in numerator and denominator. Since the denominator has at least 2^2, x-12 must have at least 2^4, and thus x has the same numer of powers of to as 12, 2^2. 12x thus has a factor of exactly 2^4 and (x-12) is thus divisible by 2^6 = 64. Since only 64 is divisible by 64 between 1-100, x = 12+64 = 76. y = (12×76)/16 = 57.

[HungNguyen-rjek]

I have another method that seems easier.
12(x+4y)=xy, since y odd then x is divided by 4.
Let x=4t, t<25, then we have y=48t/(t-3)=16*3t/(t-3). Y is odd and t<25 so t-3 has to be 16, then we have the solution.

By this method we can find any solution without the condition <100

[trelligan]

Very nice multiple microphone setup. No lavaliere, but good sound despite turning to the blackboard and speaking while writing. Just hope there's no bad chalk squeals!

[dneary]

Without the x, y<100 condition, there are only 6 integer solutions with odd y - (x, y)=(588, 49), (204, 51), (76, 57), (-564, 47), (-180, 45), (-52, 39). I'm not sure why you would need the 0<x, y<100 condition to keep the solution set manageable.

[MathZoneKH]

Thanks for sharing a good figure out!! 😊❤️

[kevinmartin]

The other solutions without the 100 limit are (x, y) = (588, 49) and (204, 51). The fact that there are only two of these makes the 100 limit almost pointless.

On the other hand, the requirement that y be odd removes 6/7 of the 21 solutions to the unconstrained problem and so probably is a reasonable restriction to keep the length of the solution small.

The general solutions are (x, y) = (12 + 2^(6-j)*3^(2-k), 48 + 2^j*3^k) where 0<j<6 and 0<k<2, the presented solution is for j=0, k=2 and the other solutions with odd y have j=0, k=0 or 1.

[tomatrix]

Man! That snapshot of your setup is looking top notch Michael!

[شبلالإسلام-ظض]

Problem of the day,
Proove that for all a in ℂ\ [-1, 1] there exists a unique complex number b such that ( 2a = b + 1/b and |b| > 1 )

[handanyldzhan]

According to the equation, xy = 12y + 48x => y = 12y/x + 48.

Then x = 12y/(y-48). Since y-48 is also odd, 3y/(y-48) is also a natural number. It can be written as 3 + 144/(y-48), so 144 must be divisible by y-48. Thus, y can be 51 or 57, but 51 gives too large an x. The answer is x=76, y=57.

[vhsy]

set y=2n+1 (y is odd)

1/x + 4/(2n+1) = 1/12
0<x<100

x=76, n=28, y=57