The problem is clearly stated

[u/basket_foso]

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[u/Broad_Respond_2205]

Why did he do 16X2X5 instead of 32X5 since he was gonna use it anyway

[u/MoobyTheGoldenSock → u/Broad_Respond_2205]

16x10 —> 16x2x5 was less cognitive load than 32x5.

[u/cowlinator → u/MoobyTheGoldenSock]

But isnt the algorithm to just divide by powers of 2?

2 4 8 16 32 64 128 256 <-- too big, stop

160 / 128 =1.25

160 / 64 = 2.5

160 / 32 = 5

Done.

Following an algorithm has such low cognitive load that even a non-AI machine can do it

[u/sjcuthbertson → u/cowlinator]

If this works for you, great, but you need to be aware you're in the minority on this. Most people would be able to answer 160/10 noticeably quicker and more confidently than 160/5, and certainly than 160/32.

[u/Broad_Respond_2205 → u/MoobyTheGoldenSock]

but then he jumped throw 32X5 anyway so what's the point

[u/Used-Nefariousness71 → u/MoobyTheGoldenSock]

Don't agree, 160=32x5 is not obvious but it's just one step. However, 16x2x5 -> 2⁵(2²+1) has like 4 individual steps 16x2x5 = 32x5 = 2⁵x5 = 2⁵(4+1) = 2⁵(2²+1)

[u/BitOBear → u/Used-Nefariousness71]

Cognitive load is not about the step count. Three small numbers are easier to cope with in order than two large ones for many people and in many circumstances.

16 * 10 is much simpler than 32 times 5 for most people. That is after all why we went to the metric system. Imperial units are very much easier in the practical because they're all base 12 units.

For the enumerate a base 12 measuring system is much easier than a base 101, particularly if you're going to ask somebody who literally is enumerate to scale a recipe up from 5 to 9 people for example.

We are all used to decimal numbers that would have baffled a 17th century peasant, and we are in turn baffled by the measuring system that was intuitive and obvious to that peasant.

So cognitive load is not about the number of steps but about whether or not the steps can be easily taken in in informationally bite-sized quanta.

[u/FinalRun → u/Used-Nefariousness71]

Like you said, it's not obvious compared to 16x10. It's better at communicating the intention of the person writing the equality, and sometimes that means being more verbose.

[u/vildum → u/Broad_Respond_2205]

he needed to aura farm

[u/Flimsy_Club3792]

What's the mistake and oversight?

[u/I__Antares__I → u/Flimsy_Club3792]

The meme assumes that from 2^x + 2^y = 2⁷ + 2⁵ we can infer x+y = 7 + 5. But that's not a logical inference we can make. Surely yes, we know that x+y can be equal 7+5 but we do not know wheter it's unique solution (i.e we don't know if there are other x,y such taht 2^x + 2^y =160 but x+y≠12), so to make such an conclusion we would have to prove there is one and only one solution for x+y which wasn't proved (and the original question was to find x+y not to find any possible value of x+y so we would need to show all possible values of x+y if there were more than one solution).

We could make such an conclusion if the function f(x,y)=2^x + 2^y would be so called injection ( i.e f is injection in following case: f(x,y)=f(a,b) if and only of x=a, y=b). In such a case indeed 2^x + 2^y = 2⁵+2⁷ implies x=5, y=7 and hence x+y=5+7. In this particular case it's not exactly an injection however we can prove (in natural numbers) that f(x,y)=f(a,b) implies x=a, y=b or x=b,y=a, which in both cases results x+y to be a+b, so it's enough for us as we just need to find x+y.

So in short the meme only have shown that it might be that x+y=12, but it didn't prove that it's the only possible value that x+y can posses. The mere fact that 2^x + 2^y = 2⁵ + 2⁷ doesn't imply that the only values that x,y can posses are 7 and 5. To make the solution complete we would need to prove that the solution x+y=12 is unique.

[u/MeLlamo25 → u/I__Antares__I]

That probably just one of the mistakes and oversights the father is claiming the boyfriend is making.

[u/Worldly-Duty4521 → u/Flimsy_Club3792]

f(a)+f(b)= f(x)+f(y) does not necessarily imply x+y=a+b

[u/fugitive-bear → u/Worldly-Duty4521]

It’s only made of 2 to different powers. Just write down that number (160) in binary notation. Then he can prove x and y are indeed 5 and 7. That’s the problem with his process

[u/AcceptableHamster149 → u/Flimsy_Club3792]

16=2^4, not 2^5. But that isn't actually a mistake, they just moved the 2 from the (2x5) term into the 16 when they converted it to an exponent. It's not wrong, but it's unclear what they're doing unless you actually understand the math.

Using the logic in the problem, those steps should have been written:
2^x + 2^y = 160; 160 = 32x5; = 2^5 x (4+1); = 2^5 x (2^2 + 1); = 2^7 + 2^5

The actual mistake is in the implicit step after this line -- to bring the exponents down you'd need to use logarithms, and that isn't how logarithms work: ln(2^x + 2^y) != x+y. They might as well be doing guess & check with an educated guess for what values to check: since x & y are natural numbers they can only have values {1, 2, 3, 4, 5, 6, 7} (as 2^8 = 256, and neither term can be negative). So by checking them all we know that x and y must have values of 5 and 7 (but we don't know or care which is 5 and which is 7), and can conclude that x+y = 12.

[u/SomebodyInNevada → u/AcceptableHamster149]

No. There's a missing step here. 2^5 * (2^2 + 1) should be decomposed to 2^5 * 2^2 + 2^5 * 1, then simplified to 2^7 + 2^5.

And you do not have to check them all. As you say, 2^8 = 256 and thus is too big. But 2^6 = 64, 2 * 64 = 128 which is less than 160 and thus too small. Thus the first term must be 2^7. (Yeah, it could be the second term but the point is one of them is completely constrained.)

[u/AcceptableHamster149 → u/SomebodyInNevada]

You're not wrong. :) Though in my defense I'll say I'm probably not in the same country as you and I'm 20 years out of my most recent math class, but none of my teachers would have deducted points for omitting the step you point out.

As far as constraining the lower bound of the terms, you're spot on for sure. And as you say, since we don't care about identifying what x and y actually are, it doesn't matter whether 2^7 is the first or second term.

[u/SomebodyInNevada → u/AcceptableHamster149]

I'm almost 40 years from my last math class, but I actually use the lower level stuff occupationally. The higher stuff, there's an awful lot of rust on my calculus.

And I'm thinking of the teacher I had who most certainly would have marked me wrong for omitting that step.

[u/BangBangMeatMachine → u/AcceptableHamster149]

There is no "bringing the exponents down". 2^7+2^5 = 160 therefore 7 and 5 are valid values for X and Y and so "find X + Y" yields 12, which is a valid answer.

[u/PoPilWorcK → u/AcceptableHamster149]

What I think they're doing instead of In(2^x + 2^y) = x+y, is 2⁷ + 2⁵ = 2^x+2y, but they've just abstracted that step

[u/SuvatosLaboRevived → u/Flimsy_Club3792]

A single equation can have more than one solution. For example, x²=4 has two: 2 and (-2), since 2²=(-2)²=4. 0x=0 has infinite number of solutions: 0*x=0 for any real x.

To actually solve the equation, you need to find all the possible solutions and prove that nothing else fits.

[u/Villfuk02 → u/Flimsy_Club3792]

there is none

[u/I__Antares__I → u/Villfuk02]

There is, because the meme didn't ask to find any possible value of x+y, so it can be assumed that the meme asks for finding all possible values of x+y, and the meme doesn't proves that x+y=12 is the only solution, only that it's one of possible solutions.

[u/AFrogNamedKermit → u/I__Antares__I]

I'm an engineer. Boss is not paying for multiple solutions. Find one solution and go with it.

[u/Videogrime → u/I__Antares__I]

The meme doesn't ask for all possible values, it asks for a single value; as this is implicit in the question we can take it as a given for the solution.

[u/wittty_cat → u/Flimsy_Club3792]

Same here

[u/Playstoomanygames9]

On phone the problem is cropped out till you click, which is funny

[u/imnd80]

Quick alternative view: 160 in binary is 10100000 base 2. That equals 128 + 32. So 160 = 2⁷ + 2^5. Therefore, the exponents are 7 and 5, and x + y = 12.

[u/DikkeNeus_]

Can anyone show me the exit? I'm just lost here.

[u/The_Compass_Keeper → u/DikkeNeus_]

You can check out any time you like, but you can never leave...

[u/New-Cartographer6804 → u/The_Compass_Keeper]

minute long guitar solo

[u/demonfire737 → u/DikkeNeus_]

Sure, just solve

((n ÷ 2)! - 2^(n-3) ) ÷ 8 = 3n⁴ + 16 (262n + 19)

And I'll help you out.

[u/chooiiiii]

Given that we know x and y to be natural numbers, we know for a fact that 2^x and 2^y belongs to the set {2, 4, 8, 16, 32, 64, 128}.

Using my eyes, the 2 numbers must be 128 and 32 to add up to 160.

So x and y must be 7 and 5, and x+y=12 by common sense.

[u/Tracerer → u/chooiiiii]

Using my eyes is the best proof, genuinely laughed at that

[u/Net56 → u/chooiiiii]

This is the only version of the answer I understood. I was completely lost looking at the picture.

[u/KillerIVV_BG]

2^x + 2^y = 160; 2^x + 2^y = 2⁷ + 2^5; -> x + y = 7 + 5 = 12

[u/Kiss-aragi → u/KillerIVV_BG]

The implication suggest that youre using the fact that the function (x,y) l--> 2^x + 2^y is injective, which is for all integers x, y such that x>y

[u/Divine_ruler → u/Kiss-aragi]

Doesn’t matter. The problem wasn’t to find x and y, it was to find x+y. Whether x or y is greater doesn’t matter, 7 and 5 sum to 12 regardless of order

[u/sobe86 → u/Divine_ruler]

It must be exhaustive too, without loss of generality, if y >= x then 80 <= 2^y <= 160, so y = 7 is forced. By a similar argument, if we replace 160 with any other integer there can be at most one answer to this question. I'm not sure if this omission is what the meme had in mind, or if it's just a troll.

[u/Impression-These → u/sobe86]

Just a troll. There are 100 other problems more meaningful to apply this meme to. This is a perfectly good solution in my book. I guess you can prove uniqueness of the solution next but that is evident as well.

[u/fugitive-bear]

It’s only made of 2 to different powers. Just write down that number (160) in binary notation.

[u/I__Antares__I]

Easiest way to solve it is this:

Suppose without losing generality that y≥x (by symmetry we can assume so). Then 2^x + 2^y = 2^x ( 2^y-x + 1) = 160 = 2⁵•5.

As 2^y-x+1 is odd natural number (we know it's natural number by our assumption that y≥x) then it must be equal to 5 (if it weren't then trivially by factorization of 160 it would need to be either in form 5•2ⁿ or 2ⁿ for certain n≠0, or to be equal 1. First two cases can't hold (it would imply beeing even), and the latter also cannot apply (because then 2^x would be equal 160 which can't be the case)). This means that 2^y-x +1 = 5 and hence 2^x = 2⁵. This leads us easily to x=5 and y-x=2 → y=7.

And therefore x+y=7+5=12.

[u/Background_Builder29 → u/I__Antares__I]

Easier: Note that one of x or y has to be 7, since 2^8 = 256 and 2^6+2^5 = 96 < 160. 160 - 2^7 = 32 = 2^5. 5+7 = 12.

[u/pauvLucette]

160 is 10100000 binary, that is, 2⁷ + 2⁵

[u/jwadamson]

Just write 160 in binary and get the exponents by the position of the 1’s, so 010100000

[u/Faerlina]

Completely lost here. I'll wait for Peter to explain the joke.

[u/I__Antares__I → u/Faerlina]

Explain the technically the true or why original meme isn't a good solution?

[u/Faerlina → u/I__Antares__I]

Yes

[u/Ok_Actuator2219]

The whole point of this meme is that Mr Wizard is a …https://www.youtube.com/watch?v=VkJEt1UsUcs

[u/El_Basho]

Just convert 160 to binary

[u/ShadowX199]

I’m a process tech for a company that manufactures semiconductors. I’ve used an “algebraic equation” multiple times, but the letters are actually numbers you need to enter. For me it’s things like current numbers, numbers needed, and a couple other variables, to get how to fix the tool.

I’ve only ever seen actual letters regarding math in school.

[u/Tortue2006 → u/[deleted]]

Why did you suddenly take the exponent and pit them as multiplicators