Math Olympiad Question | Nice Algebra Equation | You should know this trick!!

[@leesweehuat]

If x & y are whole numbers & x + xy + y = 54, find x + y. x(1+y) + 1 + y = 55 (x+1)(y+1) = 5 * 11, try x+1, y+1 as different factors of 55 (1, 5, 11, 55) When x = 0, y = 54 When x = 4, y = 10 When x = 10, y = 4 When x = 54, y = 0. ==> x + y = 54 or 14. x + y = 14, if x & y are positive integers which exclude the value 0.

[@mookz34]

In the US, we are taught the natural numbers are the positive integers.

[@suraj7938]

@@mookz34 that's same in all coutries i think

[@thebeastslayer6329]

I searched google and it says 0 is a poitive integer

[@mraprender9074]

@@suraj7938what do you mean by “I think” , tats obviously the same everywhere in world. Mathematics don’t change from place to place, may be the method not concept…

[@suraj7938]

@@mraprender9074 yes correct... He was saying 'in the US' that's y I said😅

[@hillbillydeluxe27]

Many years ago, I was taking a grade 12 calculus exam and drew a complete blank on a question. Like I literally sat there staring at a question that I knew the answer to but couldn’t figure it out. A complete brain fart. So I wrote in “Atlanta Georgia “. Imagine my surprise when I got the test back with half a mark for my ridiculous answer. The teacher wrote alongside it “wrong but funny”. To this day when I don’t have an answer, I say Atlanta Georgia “.

[@PazLeBon]

and by default she inadvertantly lowered somenbody elses ranking, thats unfair to them

[@alexmorrison2360]

@@PazLeBonif a course isn’t marked on a curve than no one looses when you do well.

[@thechumpsbeendumped.7797]

@@alexmorrison2360 Then not “than” and loses not “looses”.

[@hillbillydeluxe27]

@@PazLeBon oddly enough everyone had a good laugh over the situation and no one was put out.

[@tarvoc746]

​@@PazLeBon ...isn't it great how everything in capitalism is a competition?

[@ChuckHenebry]

Far faster than this is to think of the problem geometrically: the term xy is a rectangle x long and y high (with area xy). Stack on top a single-high row x long (adding area x to the original xy) to make a rectangle x long and y +1 high. Then add a single-wide column y high to make a slightly irregular shape: a rectangle with a notch taken out of one corner. That completed shape has area xy + x + y (i.e. the left side of your equation). But it's also a rectangle x+1 wide and y+1 tall with a notch area 1 taken out of it. If you add back in area lost by the notch, that's total area 54+1 = 55. Now we know x and y are integers, so x+1 and y+1 must also be integers. The only two integers that multiply to make 55 are 5 and 11 (ignoring 1 and 55). So if x+1 = 5, x=4, and if y+1 = 11, y=10. This can all be done in one's head. No real algebra required.

[@SharonOnTheNet]

that's the way I did it

[@dre3951]

Genius solution. You win. Thanks so much for writing this out. This is why I read the comments in youtube.

[@ralfp8844]

Cool solution, i always prefered visualisations of algebraic problems. For me you are the winner. I applied polynomial division and shot that chicken with a bazooka 😂.

[@wilsonnkwan]

Ha, I posted the same solution and then I saw this. Great minds think alike.

[@Lolwutdesu9000]

And this is why I think most of these channels are utter bs because they provide a useless convoluted solution, with zero understanding nor elegance.

[@benperry8309]

My thought process behind this went like this. Looking at the equation we’re given we can immediately deduce that the 2 numbers we’re looking for must both be even. Odd + (Odd)(Odd) + Odd = Odd Odd + (Odd)(Even) + Even = Odd (and vice versa) Even + (Even)(Even) + Even = Even ✅ Then after that I noticed that 9 * 6 = 54. From this I know that the answer lies somewhere around these numbers. A small bit of guessing and checking with even numbers around 6 and 9 later and I came up with 4 and 10 which satisfies the equation. Therefore, x + y = 14.

[@ntayyan]

similar thinking. I will add that as the 2 numbers diverge from each other, their product will be less. This helps you guess faster

[@sivasudheendra6215]

Exactly.... Took 2-3 minutes of trail and error....not an effective way though..

[@vaibhav-kr1cx]

Bruh , i am an indian class 12 student , the first thing which strikes in my mind after seeing the product and sum together is to do +1 for factorisation... So took just 20 sec to get to the answer

[@cursedcat_art]

exactly same thought process flow bro

[@meowcat5596]

@@vaibhav-kr1cx Didn't have to mention your nationality, "bruh"

[@konyecstrength4life]

Its been about 20yrs since my mind was performing at a level that this would seem basic. Now i feel like i bruised my brain following. But every instinct is telling me there is an easier way. 🤓. In any case this unsolicited youtube video served its purpose. I’m back on the MATH again. 😂 Good stuff.

[@alpborakirte801]

It seems like there is a easier way because you can see it immediately that x and y must be 4 and 10 this is middle school level math it’s easier for us to

[@JuanDelaCruz-il9wy]

x+xy+y=54 For x as positive integers: If x=1 then 1+y+y=54, 2y=53 and y=26.5 in w/c the solution will be FALSE If x=2 then 2+2y+y=54, 3y=52 and y=17 1/3 in w/c the solution will be FALSE If x=3 then 3+3y+y=54, 4y=51 and y= 12.75 in w/c the solution will be FALSE If x=4 then 4+4y+y=54, 5y=50 and y= 10 in w/c the solution will be TRUE (x=4,y-10) HENCE x+y = 14

[@ThelntellectualTruth]

Best solution ! By integer substitution (very few case scenarios), it turns out the fastest method. Good job!

[@WikiBidoz]

@@ThelntellectualTruth best solution? He just tested there's one solution without proving it's the only one, and it can also only work with small numbers in the right hand

[@samueljehanno]

@@WikiBidoz 👍

[@rickdesper]

You also need to say something about uniqueness. I.e., that there is no other pair (x,y) that will solve the original equation with a different value of (x+y). It's particularly important since there are two solutions (4,10) and (10,4). But why only two?

[@jaimesinclair9585]

@@rickdesper There is not but one solution for (x+y), which is 14. You're not required to submit x and y values separately, so this should be a valid answer.

[@Robertita988]

Rewrite as (x + y) + xy = 54 to recognize the symmetry: if a number does not work for x there is no need to try for y. Plunk in number 1,2,3... for x. 1 + 2y = 54? y = 53/2 not integer 2 + 3y = 54? y = 52/3 not integer 3 + 4y = 54? y = 51/4 not integer 4 + 5y = 54? y = 50/5 = 10 :-)

[@MaharshiRay]

my rugged solution - forget integer for a moment, take x=y, we have a quadratic equation x^2 + 2x = 54. It means one of the numbers < 7. Also note that both x and y must be even because x(1+y)= 54-y, so if y is odd then RHS is odd while LHS is even due to (y+1) factor (similarly with x). This leaves us with only 3 choices for the smaller integer : 2,4,6. out of which only 4 gives an integer solution attached with 10.

[@GurpreetSinghMadaan]

Excellent

[@ashokkhullar6650]

good thinking.

[@19Szabolcs91]

I like this solution, as well.

[@Joosher56]

My braindead, completely incorrect solution: 54/4 = 13.5, therefore x and y are both 13.5 lol

[@MaharshiRay]

@@Joosher56 that is exactly why they say don't drink and derive ;)

[@TheToledoTrumpton]

For budding students it is also very important to realize that these questions are usually very fair, so when you are solving them make sure you use every piece of information given. In this example the fact that the answer is an positive integer is so key. When you see not only that, but they want just the sum of (x + y) as an answer, you are almost led to the solution by the question. Don't panic, relax, and go where they are leading you.

[@transklutz]

Also, the word "integer", limiting the factors of the equated number.

[@Jeph629]

Very good advice! Better read it a few more times!!!!!!!

[@PlatuzCubing]

I was thinking the same thing while watching the video this is so true

[@david_allen1]

Note that 55 has only 2 prime factorizations: 1*55 and 5*11. Since (x+1)*(y+1)=55, and x & y are positive integers, we can set the individual factors equal and solve for x and y. We know that the 1*55 factorization will not yield a solution because it would force either x or y to be zero (the solution of x+1=1), so we use 5 and 11. Thus x+1=5 and y+1=11 gives x=4 and y=10; therefore x+y=4+10=14. This also proves that there is only 1 solution, a fact that brute force guessing does not prove.

[@renesperb]

A very simple approach is as follows: Solve the equation for y : y= (54-x)/ (x+1) . Now you plug in values of x ,such that y is a posive integer. You easily find x= 4 ,hence y = 10 ,i.e. x+y = 14.

[@leif1075]

That's still a lotnof aues to plug and check..54 to he exact .you didn't address perhaps if a way to narrowest down Like say if x is positive you knkw ybis negative etc..or you just guess and checked?

[@timolaitinen2858]

Last part was a bit odd. Why not just take one equation (x + 1 = 11 and y + 1 = 5 ) and move +1:s to right side of equations and you get x = 10 and y = 4 and sum it up x +y = 14. Now it was made a bit hard.

[@sirmixalot3332]

Far easier to follow and far more concise.

[@raviprabhala]

then when you replace the value of X and y in the original equation we get that 14+40 = 54

[@pietergeerkens6324]

Alternatively: Let s = x+y and p = xy so that p = 54 - s. Then x and y are solutions to x^2 - sx + (54-s) and, since x and y are positive integers, the discriminant must be a perfect square n^2 so that s^2 - 4(54-s) = n^2 or s^2 + 4s - 216 = n^2. Then, completing the square on the left hand side, (s+2)^2 = n^2 + 220 or, factoring as difference of squares, (s + 2 - n) * (s + 2 + n) = 220 = 2*2*5*11. Then both left side factors must be even, giving only 2 * 110 and 10 * 22 as acceptable factorings. Since the first would make s equal 54 and p equal zero, only the second is accepted; then s + 2 - n = 10, s + 2 + n = 22, and s = x + y = 14.

[@chad0x]

your way is *much* more complicated

[@pietergeerkens6324]

@@chad0x But more general.

[@trietphanminh6334]

this looks easier for asian🤣

[@gvc76]

@Robert Clive the founder of India That's essentially what the video described.

[@reconquistahinduism346]

@Robert Clive the founder of India white Norman and Anglo saxon barbarians could not even count. Never accepted zero and other numbers. The Hindus gave it to you. You racist fanatics were non existent till 1800's and you are an aberration. We Hindus are an eternal civilization. One off civilization. We Hindus built your looted world. We gave you everything that you have. You own nothing.

[@larrycornell240]

Completing the multi linear form is a cool trick. Add one to both sides then factor 55. The only divisor pair that works is (5,11)->(4,10), so x+y=14.

[@alandulusia]

When i saw the 4 at end 14 came to my mind I must think of numbers that add uo to 14 in order to figure out x and y

[@glasgowbrian1469]

@@ragnhildthingstad7458 ??? 4+4x5+5 = 29 as “xy” means multiply x and y. Sorry to disappoint.

[@supermatt-99]

@ragnhild thingstad x + y is 14 in this question, I don't know what math you're inventing, maybe it's meth

[@yaywippee]

@@ragnhildthingstad7458 LOL

[@mohammedzaid_IITR]

@ragnhild thingstad bruh. xy written here means X multiplied with Y and NOT any two digit number XY

[@bhoopendramasram6423]

Another approach could be X(1+y) + y = 54 X(1+y) = 54-y X = 54-y/1+y Add 1 to both side X+1 = 55/1+y Since x and y are positive integers therefore 55/1+y should be an integer to now we need factors of 55which are 11 and 5 ,hence y=4 corresponding x = 10 and y= 10, corresponding x = 4

[@nagajothikannan6538]

You done it cool🤟

[@GurpreetSinghMadaan]

Saw your solution after I had posted, almost same reasoning

[@karriem5666]

My exact same approach, and just plugged in a number for "y" that would yield a whole number for "x".

[@user-rf9me7xm1w]

That’s what I did, much quicker.

[@leif1075]

But WHybadd 1 to.both side s after dividing that's a neaky dirty trick I don't think most ppl would think of. So why not do it without that part?

[@henrikstenlund5385]

This can be solved in terms of generating a differential equation for y(x). That can be solved to give x=(C-1-y)/y where C is an integration constant. We substitute this to the original equation to get y=(C-1)/(56-C). Then one looks at this relation to see when it can give positive integers at some value of C. Integer values of C giving this are 45 and 51. This gives two solutions x=4 and y=10 and the second solution x=10 and y=4. The x+y is thus 14. There may exist other solutions, this was not a proof. Also noninteger values of C may provide integer values of x,y.

[@الباروني-ش2ث]

you are a lazy student, there is two solutions 14 (x=4 & y=10) and 54 (x=0 & y=54)

[@henrikstenlund5385]

@@الباروني-ش2ث I am not a student. I also indicated that this is not a unique solution and I left for the other people to find the rest.

[@londonviking3801]

Worked that out in my head in 1min. x = 10, y= 4.

[@prithvidhyani2002]

@@الباروني-ش2ث your second solution is invalid because the question clearly asks for positive integers...

[@wilsonnkwan]

Actually if both x and y are positive integers, I think we can go with a simpler approach by drawing it out. which is a rectangle with x and y sides, a rectangle with x and 1 sides, a rectangle with y and 1 sides and at the corner a square with sides of 1. So arranging those, you get a rectangle of (x + 1) and (y + 1) sides and the total area will be (54 + 1) = 55 given that 55 has only 1 solution that both numbers at least 2 to make sure x and y is not zero, there is only 5 x 11 which means x + y = (x + 1) - 1 + (y + 1) - 1 = 5 + 11 - 2 = 14. I know this sounds like basically what the lady wrote in equation form, but some times it doesn't hurt to illustrate it in diagrams how mathematicians did eons ago when they were solving for the general solution of quadratics and cubic equations.

[@nivirautela]

hey there! i really liked the way you answered this question but im having a bit trouble getting my mind around it could u please help me out and give a further detailed explanation?

[@wilsonnkwan]

@@nivirautela ========== = | X * 1 |1 | ========== = | |Y | | X * Y | * | | | 1| | | | ========== = Rectangle with (X + 1)(Y+1) = XY + X + Y + 1 = 55 = 11 x 5 = (10+1) x (4 +1) X + Y = 14

[@lumberjackdreamer6267]

Yes! And congrats on the ascii graphics!

[@ZRogers91]

Idk if this is an “elegant” solution, but here’s how I got it. x+y=54-xy Did the factors of 54, everything other than 6*9 seemed too big to add up. So I chose 6*9. When I multiplied it had to be less than that and knew the numbers probably weren’t far off from 6 and 9. It seemed reasonable that x+y had to be even. Tried 5*8, which was 13=54-40, so 13=14. Thus 4 and 10 are the answers.

[@felipegonzalez6139]

Es la lógica que muchos tenemos porque nos molesta hacer mucho trabajo para obtener muy simples y nada servibles términos, pero recuerda que está lógica no la quieren muchos profesores porque requiere del método más extenso posible... 🤔😔

[@sirmixalot3332]

Dr. Johnson, PHD in math, was the absolute best teacher of math I will have ever met. He could explain nearly every equation like three different ways. At least one approach would always turn your light on. A surgeon of numbers the late Great Dr. Johnson of KCMO! Many thanks to this humble man gifted with the art of teaching in addition to mastery of numbers. Best wishes to his family. A treasure indeed to countless people who had the fine pleasure of having met and been taught by him. Southwest Highway School 1980’s

[@wxrlp]

where is this school?

[@SurajSingh-wi9vd]

Me who solved it without calculation 10,4 or 4,10

[@sirmixalot3332]

Kansas City, Missouri

[@MasterYoist]

As a retired math teacher, I had been taught to teach at least three methods to solve each type of problem. All math teachers should do this as not all math students comprehend the same process with the same level of ease.

[@Tomohiko_JPN_1868]

Q: Related question. What kind of numbers (x,y) meet the condition, below ? ・x+y = Natural number A ・xy = Natural number B i find x,y = (2, 3), (5+√3 , 5-√3). So we know we can build Natural numbers with some of (x,y) in "Natural numbers to irrational numbers". But i can't find any of (x,y) in Non-natural Rational numbers. Why ?

[@INSEARCHOFPURPOSE23]

I solved it intuitively.The question asks what are two numbers who's sum of Addition and multiplication gives 54.Obviously both numbers cannot be in double digit as multiplication goes higher than 54.So i chose multiplication near 10 so as to go from large to small.Started multiplying with with 1,2,3,4 and came to 40 then added the 10 and 4 and got the answer.I always found math difficult as i never understood what the equation was asking in simple language.But this is trial and error method and we cannot rely on it for more complex equations.

[@19Szabolcs91]

Trial and error is nice and all, but it doesn't really prove that this is the only solution. Of course if you find a reasonable boundary, it can work. For example, here at least one of the numbers must be smaller than 7, because it both are at least 7, then x+y+xy must be more than 7+7+49 = 63 which is more than 54. Now you just have to try x=1, x=2, x=3, x=4, x=5 and x=6 whether they give any integer solution for y, and only x=4 does. (You can assume x is the smaller number, as the equation is symmetrical for x and y)

[@radhakrishnanpp1122]

I am 80 years old and this problem and solution still fascinates me

[@LKLogic]

🙏🏻🥰⭐️

[@danielgautreau161]

In Canada (where I am), and in America, most high-school graduates could not solve this. In Europe it would be on a 9th grade test.

[@brandonwu8353]

its a nice algebra problem

[@deadbush2812]

@@danielgautreau161 i live in america and learned this in seventh grade ☠

[@davidhart5426]

And we suffer for it...... I thought it harsh when a European colleague thought our education system was 'a joke' . ..... kinda hard to defend given the evidence ...

[@godiswatching8110]

Y=(54-x)/(1+x)=55/(1+x)-1, therefore 1+x is either 5 or 11 and x is either 4 or 10, then by symmetry y is either 10 or 4.

[@youssef5814]

very fast & very good solution.

[@avinashbalakrishna1927]

Same thing, as explained in the Video.. Nothing different 🤷

[@martinbennett2228]

That is what I did too, the video seemed to be making it more complicated.

[@panjirizki2976]

Did I just watch math problem as entertainment?

[@sadhanasurve6002]

So did I!

[@gheffz]

Brilliant. Thank you. (I think you had it solved in the second last step, either x = 10, y = 4 or conversely x = 4 and y = 10 ... therefore x + y is 14. Loved your use of expanded substitution to solve this. I did have a go first before watching and didn't solve it in 3 minutes of trying.)

[@odylpierre7212]

The math is not good.

[@shortaybrown]

I just did it by looking at it and came up with 14. I wish I knew the real way

[@mikevanderwolf8575]

I am no math major, but, given the challenge, why not take x+xy+y=54 and change it to xy=54-x-y, given that only a certain range of numbers could apply to this equation, it doesn’t take long to discover that 10*4=40 and 54-10-4=40, thus x+y is as suggested, 14.

[@Jako19800126]

Yeah but this way you did not prove that these are the only solutions.

[@carstenlarsen8144]

i fpound 10 and 4 the same way for a start

[@hammerain93]

I’m a programmer and that’s how I solved this in my head

[@khalilfuller4939]

Because you didn’t explain how you got 10 & 4 when somebody could’ve picked 8 & 2

[@kularathnehasindu1575]

Maths is not about guessing tho

[@mfsolutions]

I am an engineering instructor and not a mathematician... I regularly teach my students how to solve multivariate equations like this in Excel... make 3 columns... x , y, and xy+x+y ... start with x = 1 and y = 1+1 ... copy the rows down and in 68 rows you have x=4, y= 10 and 54 and the alternate solution is obviously x=10, y=4 ... you can save so much time with Excel especially solving projectile problems where you need to find an angle and an initial velocity to hit a target...

[@solfeinberg437]

This is the poor man's solution. I thought of this too! I have no motivation on how to proceed with a solution, but I can see okay, we're talking about positive integers, there aren't too many possibilities, let's write them out.

[@MonkeyAndChicken]

@@solfeinberg437 Don't sell guess-and-check short by calling it "poor". The solution that wins in Math Olympiad is the quickest one, not the most general one. Guessing and checking is the most efficient way to approach this particular problem, because 54 is a small number, and the number of possible integers that would satisfy the conditions is quite few. If the equation were x + xy + y = 760,877 a more general approach would be quicker. But having the intuition to choose the best approach to solve a particular problem is just as important as knowing "tricks" like the one in this video. Math stops being about the mechanical manipulation of basic concepts, and more about intuition and experience as early as Calculus, and there's a lot more math "above" that than "below" it.

[@brettstembridge2507]

Besides the video could have been better rehearsed. This is a Math Olympiad question, therefore, in addition to the constraints that x and y are integers, it's paper and pencil only. Calculators are not allowed in the Math Olympiads. Certainly computers and spreadsheets are not allowed. (Although that's exactly how I solved it and in doing so found the answers for numbers other than 54. And found that there are numbers such as 50, 51 and 52 that there are not integral solutions and more.)

[@peter1423ka]

I took the same approach. The paper solution is also based on guessing, so why not use a computer for that?😁😁

[@PSLPtyLtd]

I'd like to try another way. Using odd or even grouping can be used to solve this problem. (x + y) + xy = 54 so both (x + y) and xy must be even which means (x + y) or xy must end with 0, 2, 4, 6, or 8. Set up a table, list all possible combinations: if (x + y) ends with 0 then xy must end with 4 or vice versa. If (x + y) ends with 2 then xy must end with 2. (x + y) ends with 6 then xy must end with 8, or vice versa, etc. This leads to 4 & 10 being the pair we're after.

[@rickymort135]

They don't both have to be even, two odd numbers added together is also even

[@theprofessor5625]

@@rickymort135 But when you multiply two odd numbers, you get an odd number. An odd number plus an even number will give you an odd number. So they both have to be even.

[@MizzSparkle90]

This is how I did it! The video is over complicating it imo

[@amishmittal2954]

​@@MizzSparkle90lmao not at all

[@vbanksd]

I did it this way in my head in about 20 seconds.

[@gerrykan3437]

The trick is really to recognize you can add the 1 to both sides and factor out the equation such that (x+1)(y+1) = 55, and the solution can be obtained by inspection (i.e., 4 and 10). I personally don't like this kind of questions because in the end the algebra only works particular conditions (RHS=54 in this case) which gives it an artificial feel.

[@alexkarlos3151]

At the end, it is way more logical and easier to just solve the for x and y and then add the two solutions. In both cases (11x5 / 5x11) we would get x + y is equal to 14. Case #1: X = 10, Y= 4 Case #2: X = 4, Y = 10.

[@jaimesinclair9585]

these can't be 2 odd values, because the result of the equation would be odd. The same applies when we take 1 odd value and 1 even value. So they must be 2 even values. Once you realize that, it's pretty simple to find 4 and 10 without having to follow all of these steps. If the result of the equation was a big number, that would be way trickier than that.

[@AthyskTFM]

The 2 values being odd does not imply that the result is odd, if you add up 3 to 7 you get an even number..

[@johnaron9819]

You got it immediately.

[@EgoTrip42]

so much easier to understand

[@FuriousMaximum]

If just an answer is required, then yea, this easy to brute force / trial-an-error your way through but, it gets this involved if proof is required.

[@ck.youtube]

What a great trick! One equation, two unknowns, I never knew it can be solved. Thanks for sharing!

[@carstenlarsen8144]

it is only bc 55 is only multp of 11x5- as integers. otherwise several x y

[@reiniernn9071]

It's not one equation. The extra is: X and Y are positiv integers. The equation still gives us a line... And the line has only 2 crossings with 2 positiv integers (X and Y ax).

[@dharmadattadash397]

go n practice math if u wonder at this

[@rickdesper]

Sometimes this kind of problem can be solved: if the original equation is invariant under a switch of variables, then a function that is invariant under a switch of variables might also be solvable.

[@epaminondas4106]

@@reiniernn9071 The equation, plotted on the x- plane is not a stright line, it is a curve...

[@invorokner282]

usually when the values are small then it's just easy to go for the 'guessing' approach, you'll just have to find the 2 values that will satisfy the equation. so, the logic is simple, which x and y will give the answer 54, for the equation on the video. so what I did, is trying to find a value close to 54 by just simply multiplying my 2 values, what I found is that is i multiply 4*10 and then add 10+4 I will get 54, and then the answer will be 14. just tried with a few different numbers till I got to the 10 and 4

[@vanessas2454]

Me too. Took about 3min. First, I figured out that neither x nor y could be 1, because if it was, the result of the equation had to be an odd number and therefore couldn´t be 54. Then, starting with two, I slowly went up trying 2&2, 2&3, 3&3, 3&4 etc. for x and y respectively to work my way up to 54. The first attempts yielded way too low results (for 2&3 for example: 2 + 2*3 + 3 = 11), so may as well jump a few steps. 6&6 got me to 42 (6 + 6*6 +6 = 42), while trying 6&7 was already too high (6 + 6*7 + 7 = 55). I didn´t need to go any higher with this pattern. Next, I took my closest result (6&7) and started lowering one variable while raising the other. 5&8 : 5 + 5*8 + 8 = 53 4&9 : 4 + 4*9 + 9 = 49, getting lower, so raising one side 4&10: 4 + 4*10 + 10 = 54. Bingo. x = 4, y = 10, therefore x + y = 14.

[@PazLeBon]

yeah that how i did it in micro seconds mentally, started with like a 5 and 4, realised it was out, adjusted to 10 and then was clear. Took literally moments

[@stephengardner763]

just noticed your reply.Mine is above.Took me about 20 seconds with that approach.

[@kathyg8535]

I did it this way too. Just randomly trying different numbers. Took 2 minutes

[@thearcheologist6679]

I also made a guess that one of them would be 10 and work from there. Not a robust approach but practical for this question. Even so, I learned a lot from the lady.

[@ralfp8844]

I just substituted x+y=z, eliminated y and got z=(x^2+54)/(x+1). After a short polynomial division you get z=x-1+55/(x+1),so x is 0, 4, 10 or 54, due to the fact that each part has to be integers, which leads to contradictions for 0 and 54. Without any further ado you can see, that either x is 4 and y is 10 or the other way round due to symmetry.

[@ralfp8844]

@@TheXmabax It's not an obvious problem and 54 or 0 would be a fair solution, if not being excluded by assumptions. And nonlinear equations in more than one unknown are tricky.

[@marcospark2803]

I liked very much the solution and the clear way it's explained. I didn't like the ridiculous vertical camera.

[@stuartsafford3927]

I think the brute force method is pretty quick to begin with. You get y = (54-x)/(x+1). It is quick, at that point, to substitute integers in for x so the numerator decreases from 53 as the denominator increases from 2. The only two whole number solutions are 10 and 4. It isn't as fast as recognizing that this is (x+1)(y+1) = 55, but it requires no pondering or inspiration.

[@Osirion16]

haven't participated in any math olympiads but i'm quite sure that to get your marks you need an actual algebraic solution, not trial and error Trial and error does work and I used to do that as a kid, but ultimately it was only to get the answer from which I could work out the problem backwards and find an actual solution

[@Bleighckques]

Only two positive solutions. (-6, -12) works as well

[@AcePhotoSverige]

@@Osirion16 but even this method ends with trial and error, bcs 1, 55 doesnt work

[@markzed66]

I solved it, but there was pondering involved.

[@dre3951]

I solved this based on the youtube thumbnail, which did not have the restriction to positive integers. So for that ... By symmetry of the equation, I concluded x = y could be a solution. Simplified the equation and solved it. x = y = sqrt(55) - 1 So x+y = twice that. I'll let someone else crunch the numbers on that one. Of course restricting to positive integers (as the video clearly does at the very beginning) nukes this solution. But anyway ... When I saw the integer restriction, I then tested a few numbers around 6-7, quickly concluded they both needed to be even, and quickly found 4 and 10 to work.

[@marti6532]

if it hadn't to be integers it could simply be 26,5 and 1 since 26,5 + 26,5 · 1 + 1 = 26,5 + 26,5 + 1 = 54 If it hadn't to be positive it could even be 54 and 0 (which is what I thought of first) since you can do 54 + 54 · 0 + 0 = 54 + 0 + 0 = 54. I think there would be many options but the intention is to reduce the possibilities by adding that limitation

[@chakravarthy68]

It has been 38 years since I worked on any math question such as this. Thank God I can still do it in my head without any tools and mistakes.

[@magicman-sv3jm]

We can start by factoring the left side of the equation: x + xy + y = x(1+y) + y = (x+1)(y+1) - 1 So the equation becomes: (x+1)(y+1) - 1 = 54 Adding 1 to both sides: (x+1)(y+1) = 55 Now we need to find two factors of 55 that differ by 2 (since x and y differ by 1). The factors of 55 are 1, 5, 11, and 55. If we try x+1=5 and y+1=11, then x=4 and y=10, which satisfies the original equation. Therefore, x+y=4+10=14.

[@desmondaubery9621]

Excellent. Thank you. Observation: If you use Desmos on x+xy+y=54, x+y=14, Two solutions emerge: (x, y) =(4,10) and (10,4)

[@abhityagi6562]

Method of substitution works good with such linear equations, where x and y are both declared as positive integers.

[@ANunes06]

In general, this can jam you up when there are multiple valid solutions. But yeah. Doesn't take long to land on 10,4

[@simens8646]

I think you are overcomplicating the last part of the calculation (at 6:15). You simply have x=10 and Y=4 (or vice versa). Whichever way you add them up you get 14. No need to do the elaborate step of substituting the x+1 and y+1 values in the equation.

[@ungureanutiberiu7072]

Wonderful, did you pull 4 and 10 out of the hat?

[@simens8646]

@@ungureanutiberiu7072 , at 6:15 she has already determined that there is a solution with x+1=11 and y+1=5, hence at x=10 and y=4. Therefore x+y is 14. For some inexplicable reason she spends 2 minutes and uses an overcomplicated approach with substitution to arrive at this answer.

[@HarbourPoland]

You can even go a little bit further. From this: x+xy+y=54 you can make this: x+y=54-xy, and from the second equation it looks like 4 and 10 (or 10 and 4 :-P) will be applicable, because 10+4=54-(10*4). So you can say what numbers x and y should exactly be.

[@TarekTawfik]

This is how I initially thought. Not sure why it got convoluted; if I can solve it in two lines

[@aaabbb-py5xd]

​​@@TarekTawfik Because you're not solving it, you're guessing

[@sirmixalot3332]

One can be very capable in math and yet be a poor teacher of math.

[@mattdebyl8321]

​@@aaabbb-py5xdno, it's not.

[@aaabbb-py5xd]

@@mattdebyl8321 lol really, since when did plugging in numbers to guess become "solving"

[@zervasfan]

Get equation to : y = (54-x)/(1+x) , y needs to be integer. Probe values of x starting from 1....at x=4, y = 10. Done... ( factorizing 55 is O(n) complexity whilst 4 divisions and assignments are Θ(n) )

[@TheNadnerb]

I tried to solve this based solely on the thumbnail and since that didn't mention positive integers I immediately went for x=0 and y=54, thus x+y=54. Without that extra detail that's provided only in the video, it was a valid solution.

[@roeydaz]

Great….I’ve never been good at Math but am still fascinated by it!

[@vimmivimmi3173]

Same here

[@flaviojoaquim2660]

I dont understand The math, but my english is improving.

[@SHANKAR-Nitrr]

Nice explanation 👍 but in competition exam in india there is no time to waste to much for single question .I will go for value putting to satisfy first equations

[@rickdesper]

If pressed for time, I would just start plugging in values for x: 1,2,3,4...and done!

[@christofferanthony1950]

I managed a solution that's almost the same as what's shown... had to try to figure it out first, before watching, of course! After a little unsuccessful fiddling with possible factors (mostly roots), I noticed that (1+x)*(1+y) contains original expression - - x + xy + y - - only it adds 1... to get 1 + x + xy +y. From that I added 1 to 54 on the other side, and reasoned as she does that 11*5 (or 5*11) was the best choice to get to positive integers to multiply to 55... and followed the rest as shown. Great fun on a rainy day like today!!

[@maslina10]

It's important to require that x and y must be positive integers. Also, the accurate language is "Find x+y" instead of "Solve x+y". Not sure why the solution requires such a lengthy explanation. Write (x+1)(y+1) = x + xy + y + 1 = 54 + 1 = 55. 55 admits only one prime factorization: 5*11. Therefore, either (x+1) = 5, (y+1) = 11, or vice versa (interchange x and y). If one allows negative integers, then (x+1) and (y+1) must be both negative. In any case, (x+1) and (y+1) must be of the same sign.

[@ksmyth999]

There is a much more straightforward way to do this. Observe that (x+1)(x+y) = xy + y + x + 1. So we can rewrite the equality. (x+1)(x+y) = 55. 55 has factors 55 & 1 or 5 & 11. 55 and 1 gives a solution with one term = 0 which is not a positive integer. So assume (x+1) = 5 and (y+1) = 11. By simple elimination you arrive at x+y = 14. Actually I wrote this before watching your video which I have now skipped through. You are more or less doing the same thing but in my opinion in an unnecessarily complicated way.

[@SenthilKumar-pb3nu]

Yes this is what i did

[@48Ballen]

where is the X**2 term?

[@Vardaris]

@@48Ballen He made a typo. He means observe that (x+1)(y+1) = xy+x+y+1, so we can write the problem's equation like that: xy+x+y=54 => xy+x+y=55-1 => xy+x+y+1=55 => (x+1)(y+1)=55.Since x&y are positive integers only possible combination is 5*11=55. So x+1 is either 5 or 11 so x is either 4 or 10. If x is 4 then y must be 10 and vice versa. So x+y=14.

[@bvanpelt8]

That's what I did. No need for even a writing utensil this way.

[@khalilfuller4939]

She didn’t do a trick or shortcut, but explained how she got the integers 10 & 4

[@larsprins3200]

Both x and y must be even. Otherwise x+xy+y would be odd. At least one of x, y must be smaller than 8. Otherwise xy would already be 64 or larger. So one of x, y is 2, 4 or 6. For x=2 we get 3y = 52 which has no integer solution. for x=4 we get 5y = 50 so y = 10, which is a solution. For x=6 we get 7y = 48 which has no integer solution. So 4, 10 is the only solution. Her solution is more practical for larger numbers though.

[@timeonly1401]

How did you know ahead of time (before finding the range & types x- & y-values to try) that both x & y have to be even, that "otherwise x+xy+y would be odd"? In trying to prove this, I did the factoring and +1 to both sides just as the person in the video did, to get (x+1)(y+1)=55, which tells us that x+1 & y+1 are both odd (the only way for the product of 2 numbers to be odd is if both are odd), which means that x & y must both be even (odd +/-1 is, of course, even). Did you do all that, or did you intuit this without all the algebra? Or was there some other thing that proved to you that "they both must be even"? Thx in advance.

[@larsprins3200]

@@timeonly1401 If x is odd and y is even, then x+xy+y is odd + odd.even + even = odd + even + even = odd. The same happens when x is even and y is odd since their roles are interchangeable. If both x and y are odd, then x+xy+y = odd + odd.odd + odd = odd + odd + odd = odd. So both x and y must be even for x+xy+y to be even, i.e., to be 54 as given.

[@timeonly1401]

@@larsprins3200 Thx!

[@johndoe3092]

very clever to add 1 to both sides 👍

[@carstenlarsen8144]

of course- that comes if you turn the left side to 2 polynomes.. and then it is obvious that only 5 x 11 can =55..bieng integers end therefor x=14 and y=5... She is making it wayyyy to complicated- but the video can be looooonger.. ?

[@dharmadattadash397]

thats the process mennn🤣🤣

[@mayankgupta4520]

We can also solve this question by assuming three possibilities of positive integers 1. Both are odd 2. One odd and one even 3. Both even From equation it is clear that first two conditions can't be true so we are left with only third possibility Where both numbers are even Starting with x=2, we get y=52/3(not possible) Now x=4, we get y=10 So x+y=14

[@schrodingcheshirecat]

Wanted to try it before seeing technique used. => x(y +1) = 54 - y =>. x = (54 -y)/(y + 1) y+1 must be a multiplicative factor of 54-y. => y +1 into -y + 54 =>. -1 + 55/(y+1) ( y+1) must be {1,5,11,55} giving => (y,x) pairs { (0,54),(54,0),(4,10)(10,4) } Yeah, so this took maybe 3 minutes including deciding how to approach it.

[@logicus3402]

I personally think you should explain the tricks that you apply, even if they appear obvious for you. Not all nations share the same math proficiency that you do. =)

[@ramsaybolton9151]

not knowing the tricks literally makes the whole lesson pointless.

[@jmar7631]

It's not necessarily a matter of proficiency but the way in which finding the solution to a problem is explained. In this case, I don't think she explained her method all that well. In fact, I think her method was a tad bit "convoluted" at the end.

[@PazLeBon]

it just doesnt need tricks at all, its childs play with those numbers

[@morgana8037]

@@PazLeBon oh yeah? I'm probably an extremely dumb shit then.... either that or 26 years old counts as newborn. probably the first, I was ALWAYS behind the whole class with algebra. WAY MUCH behind. I watched the video and I still don't understand a thing.

[@PazLeBon]

@@morgana8037 because they made it ten time more complicated that it actually is :) none of us really needed it anyway lets face it

[@rajarshiraychaudhuri2351]

Please also mention the level ( standerd of exam) - ie class / middle std/ higher secondary etc - it seems this question is a very basic one ( may be- class/standerd- VI)

[@MyOneFiftiethOfADollar]

It is easy, but it is common practice for channel owners to falsely claim problem is Olympiad/difficult in an attempt to get more page views. Some channels rely heavily on the ad revenue and will do anything to get views even though in the long run it is bad for channel growth.

[@dorian_cthulhu]

I actually solved this in under 3 sec. Just tested 10 and 4 and got the result so 14. Idk y I feel so proud of myself for solving such a basic problem but the feeling’s all I need ^^

[@intotheunknown8386]

On this occasion it’s a sum you can do in your head in under a minute. Look at the XxY…….think of the X as being the bigger number. What number multiplied by a smaller number comes to well under the 54. Obvious place to start is 10. So subtract 10 from the 54….leaves 44 So (10xY)+Y=44 The number 4 jumps out. 10+(10x4)+4=54 X+Y=14 Simple on this occasion ,methodology is better for more complex questions.

[@GurpreetSinghMadaan]

The general equation x + xy + y = N can be written as (x+1)(y+1)=(N+1) In other words, the Factors of (N+1) will give you the values of (x+1) and (y+1), and hence x,y. Qed

[@jim2376]

10 and 4 by inspection. x + y = 14

[@jmack619]

What is inspection.. to you?

[@jim2376]

@@jmack619 Just looking at it and recognizing the answer quickly and without crunching through steps.

[@JohnSmith-nx7zj]

But how do you know those are the only solutions?

[@jim2376]

@@JohnSmith-nx7zj How do you know they aren't? And why would I care?

[@nathanielomorogiuwa124]

Is really cool, I love the way you take you to your time to explain it. Thanks a lot it's really helpful.

[@devenderkumar2597]

Both variables are still unknown but we do know their sum

[@davidosterberg1643]

X + XY + Y =54 ; X (1+Y) + Y =54 ; X(1+Y) + 1+Y = 54+1 ; X(1+Y) + (1+Y) = 55; Let (1+Y) = A; AX+A = 55 . With the given constraints A=5 and X=10 but A= Y+1 so Y=4 and X+Y =14

[@SethKhon]

The factors are just 1x55 and 5x11; the order makes no difference, so we don't need to list them twice. Also, I don't understand how you can eliminate the factors with the 1. We are not told and cannot derive that x or y must be greater than 1. It's true that y+1>=2 and x+1>=2, but we never established that x, y >=2. So, although we now know the solution, we would not have been able to eliminate the factors with the 1 before trying it out. What did I miss?

[@winnebagus4476]

We can solve the equation x + xy + y = 54 by factoring out the common factor of y from the first two terms: x + xy + y = 54 y(x + 1) = 54 - x y = (54 - x) / (x + 1) Since y must be a positive integer, we can try different values of x and see if the corresponding value of y is a positive integer. If x = 1, then y = (54 - 1) / (1 + 1) = 26.5, which is not a positive integer. If x = 2, then y = (54 - 2) / (2 + 1) = 17.33, which is not a positive integer. If x = 3, then y = (54 - 3) / (3 + 1) = 13.875, which is not a positive integer. If x = 4, then y = (54 - 4) / (4 + 1) = 10, which is a positive integer. Therefore, (x+y) = 4 + 10 = 14 is a valid value that satisfies the equation x + xy + y = 54 when x and y are positive integers

[@ysfjzn8538]

That's fantastic ❤️👏

[@LKLogic]

Thank you! 😊

[@kapa4163]

I don’t understand

[@naninolovyou6388]

I used to ask my mother to help me with these and she would say “Nani, it’s already done it equals 54” 😂❤

[@johnchampion7819]

It's 40+ years since I did algebra. Unfortunately, I lost it at between step 3 and step 4 when the equation changed from adding 2 factors to multiplying 2 factors. Can anyone explain?

[@zeravam]

She added 1 to y. You can see 1+y multiplying to 1+x and 1 ((1+y)*1). 1+y is the common factor to 1+x and 1. I hope I explained in an easy way to you

[@Greebstreebling]

just as in step 2, x is a common factor of x +xy and is taken out to give x(1+y), the same rule applies in step 3 where the common factor is (1+y) so tyake it out to give (1+y) (x+1).

[@The14Some1]

I've skipped the first part of calculations by imagining that xy is a rectangle, and two additional x and y adds up to sides, forming a bigger rectangle (x+1) by (y+1) without one in the corner. Thus (x+1)(y+1) = 54 + 1 Does this make sense for you?

[@smgdfcmfah]

I can explain it: You're old and no longer grasp simple concepts... or maybe that's just me! 😁

[@smgdfcmfah]

@@cufflink44 It's a joke - and one that had MYSELF at the butt of the joke. Trigger a little too easy, snowflake?

[@Vermiliontea]

That was a methodical solution. Which is the value of it. However, due to the smallness of the example, you see the solution very quickly. X and Y are interchangeable, their roles being identical. Let's find the smallest, and just for now assume it's Y. X = (54-Y)/(1+Y). X = 53/2 doesn't work, 52/3 doesn't work, ...but hey, 50/5 will work, meaning that X=10 and Y=4. It could be the other way round (44/11), of course, but what's asked for is X + Y, so 10 + 4 = 14.

[@kevinstreeter6943]

I have a BS in math. it is rare that I find an Algebra video like this one that impresses me.

[@ДетелинаИлиева-ч2е]

Обясненията към тази задача да са по-кратки и динамични.

[@berachtdorian6191]

Da

[@jeremywilliams5107]

To be honest, doing numerical substitution is as fast - 54 =9x6, we have an xy that means that this combination is too large. Start with x=6 anyway (6 + 7y =54, no integer solution for y but close), halve it (Gauss method) x=3 gives 3+3y=54 so y=17, x=3, x+y=20.

[@rosnalrajan9826]

Good solution. Logical way would be since x + xy + y = 54 , x + xy has to be equal to 50. Hence y = 4 . Thus substituting 5x + 4 = 54 giving x = 10. So x + y = 14.

[@theerthalaRahul3705]

I love the way you have explained and your voice so good to hear

[@LKLogic]

Thank you so much 🙂

[@michaelgarcia812]

It would be nice to introduce yourself (and see you) at the beginning of your videos. Very nice videos, thank you!

[@samueleverson8088]

The above was solved in my head. Sorry if it’s out there. After the factorization, we know: x(y+1)+y=54 Let z be the integer such that z=y+1. A byproduct being that y=z-1. By the previous statement, xz+z-1=54. Therefore after an inverse operation and factoring, z(x+1)=55. The only two factors of 55 are 11 and 5. Therefore x+1=11, and so x=10 Is a reasonable assumption. Given that the equation is symmetrical for variable values from the getgo, you can reason y=4.

[@Novel49h]

x + xy + y = 54 x(1+y) + y = 54 x(1+y) = 54 - y x = (54 - y)/(1+y) Now, we can substitute this expression for x into the original equation: (54 - y)/(1+y) + y(54 - y)/(1+y) + y = 54 Simplifying this equation, we get: (54 - y) + y(54 - y) + y(1+y) = 54(1+y) 54 - y + 54y - y^2 + y + y^2 = 54 + 54y 108y = 54 y = 1/2 Substituting this value of y into the equation for x, we get: x = (54 - 1/2)/(1 + 1/2) = 35/3 Finally, we can find x+y: x+y = (35/3) + (1/2) = (70/6) + (3/6) = 73/6 Therefore, x+y = 73/6.

[@richardcabrera5483]

Hi everyone. The condition that the two numbers x and y must be integers leaves to apply the teached trick. But if the two numbers were not integers but real numbers then the solution becomes to infinite solutions corresponding to the x plus y pair. Thanks for uploading this exercise.

[@LarrySiden]

This is great for falling asleep. I’m going to put it in a loop.

[@davidn4125]

The equation can be written as x=(54-y)/(1+y). Then let y=1,2,3,4 and 4 works with x=10.

[@billcook4768]

Since we are given that x and y are positive integers, it’s pretty obvious that trial and error will be faster than solving. Noting that both have to be *even* positive integers makes guessing the right answer that much quicker.

[@rmpdasilva]

x = (54 - y) / (y+1) Making z = y + 1, x = 55/z -1 In this form, the result is very intuitive, as y and x have to be positive integers, thus z >= 2, you can immediately see that z=5(or 11), thus x=10 and y=4

[@MultiSenna12]

Without guessing factors one can produce two equations x + y = 54- xy and (x+y)^2 = 54 + xy solving these two lead to a quadratic equation for x+y : (x+y)^2 + (x+y)- 108 = 0 this gives a value of close to 20 for x+y This does not have the restriction that x and y have to be integers

[@slchance8839]

thank you for posting. adding 1 to both sides after factoring x+1 on the left was clever and deceptively simple. Once I did your Step 2, I could solve the rest.

[@abumounir7615]

Interesting! However we can make it simpler from the point where you got X+1=11, and y+1 = 5. From here we get x=10 and y=4. Therefore xy = 40. Now given that x+xy+y =54. Substituting the value of xy with 40 we get x+40+y= 54. Therefore x+y = 54-40 =14. (Answer)

[@sateesha7432]

This question has 2 variables and one equation so we can put any value in place of one variable. Put y=10 in the equation X+10X+10=54 11X=44 X=4 X+Y=10+4= 14

[@janaksharma6105]

now the ques in the exam will be : prove that, x+y(a+b+c) >= tan 2cos (cot 90°) in which a, b, c, x and y are not alphabets.🙂

[@aasz1978]

Solved in 5 minutes by making the equation X(1+Y)=54-Y, then trial and error. I started by assuming Y = 7 and guessing several possible X answers. Eventually, I arrived at Y = 10 and with that, X = 4. The full answer is 4 + (4x10) + 10 = 54.

[@sajidkhot8414]

Very nicely explained.... Thank you so much, Ma'am.... Best wishes... Sajid

[@pavlovvi]

x + xy + y = 54; -> (x + 1)*(y + 1) = 55; Since x & y are integers , the only 4 solutions are possible: 1. x + 1 = 5; y + 1 = 11; 2. x + 1 = 11; y + 1 = 5; 3. x + 1 = -5; y + 1 = -11; 4. x + 1 = -11; y + 1 = -5; Hence only 4 possible solutions for x & y are: 1. x = 4; y = 10; 2. x = 10; y = 4; 3. x = -6; y = -12; 4. x = -12; y = -6; Hence, for (x + y) there are 2 solutions: -18 and +14

[@roderickyoung1243]

I wasn't so clever. Guess and check by inspection. (x=1) : 1 + y + y = 54 ; 2y = 53; 53 not divisible by 2 2 + 3y = 54; 3y = 52; 52 not divisible by 3 3 + 4y = 54; 4y = 51; nope 4 + 5y = 54; 5y = 50; y=10, x=4; x+y=14 assume from the structure of the problem that if there are other solutions, the sum is the same.

[@RajatKumar-ld6gl]

I tried different method x+xy+y = 54 x(1+y) +y = 54 x(1+y) = 54-y substitute y= z-1 x(1+z-1) = 54-z+1 xz= 55-z x = (55/z ) - 1 as x and y are positive integer so z will be integer too and z >=2 (as z>=1) And if x is integer then 55/z should be integer so z can be either 5 or 11 if z= 5 then x=10, y= 4 Or if z = 11 then x = 4, y = 10 In both the conditions x+y = 14

[@abdullahimohamed5432]

I usually don't understand but this video helped me out thanks 🙏 go on and help others thank you very much

[@shobhitupadhyay8047]

Yeah coming same answer but through different logic , firstly get the factorial x(1+y)+y=54 , then here by seeing through equation , we can only get the equation , 10*5+4=54 , thus by comparing x and y we got we got x=10 and y=4 . Then, simply we got X+Y=14......

[@briseboy]

As there are quite a few comments suggesting alternate methods of solution, i suggest that those who have difficulty understanding at 2x Normal speed, slow down their video to 1.5 Normal. Since algebra is a separate school course from geometry, it remains useful to become adept at both by choosing to solve through indirect AND direct proofs. In reference to the ensuing comments, Narcissism & Braggadocio are NOT included in useful alternate solutions.

[@ukaszkonieczny2573]

The result 14 is only special solution. The question is (x+y). Correct rounded general solution is (x+y)=13. Special where both x and y are integers is (x+y)= {-58,-18,14,54}.

[@yakovspivak962]

Two answers: X = 10, Y = 4 X = 54, Y = 0 Plus, of course, symmetrical X, Y values.

[@qujie8212]

The way I did it: For this: 54-y = x•(y+1) Since you know that the numbers cant be that high really you then just try out y=1, y=2, y=3 etc (each one takes max 10-15s) and that way you've solved it within probably less than 2mins

[@satyanaraharimallisetty]

The idea💡that Adding +1 or subtraction of -1 in LHS and RHS and balancing ⚖️ the equation was first done by ancient Indian lawyers while giving judgement in odd cases. You can see in Thenali Ramakrishna Life story.

[@walrtbstudios5430]

Subtract xy from both sides: x + y = 54 - xy. As both numbers must be even and have a product a little less than 54, simple trial and error immediately led me to 4 and 10. Took me less than 30s. I know trial and error isn’t allowed in arithmetic (show your working!), but Occam’s Razor people..

[@123idolfan]

At 0:20 I knew that x+y=14. The "positive integer" info was obviously too helpful.

[@LKLogic]

Thank you dear