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

Жыл бұрын

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

@suraj7938

Жыл бұрын

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

@thebeastslayer6329

Жыл бұрын

I searched google and it says 0 is a poitive integer

@mraprender9074

Жыл бұрын

@@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

Жыл бұрын

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

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

Жыл бұрын

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

@alexmorrison2360

Жыл бұрын

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

@thechumpsbeendumped.7797

Жыл бұрын

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

@hillbillydeluxe27

Жыл бұрын

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

@tarvoc746

Жыл бұрын

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

Did I just watch math problem as entertainment?

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.

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

Жыл бұрын

that's the way I did it

@dre3951

Жыл бұрын

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

@ralfp8844

Жыл бұрын

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

Жыл бұрын

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

@Lolwutdesu9000

Жыл бұрын

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.

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

10 ай бұрын

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

@Jeph629

10 ай бұрын

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

@PlatuzCubing

9 ай бұрын

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

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.

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

Жыл бұрын

Excellent

@ashokkhullar6650

Жыл бұрын

good thinking.

@19Szabolcs91

Жыл бұрын

I like this solution, as well.

@Joosher56

Жыл бұрын

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

@MaharshiRay

Жыл бұрын

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

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.

@user-fo5dl6cf7r

Жыл бұрын

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

@henrikstenlund5385

Жыл бұрын

@@user-fo5dl6cf7r 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

Жыл бұрын

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

@prithvidhyani1991

10 ай бұрын

@@user-fo5dl6cf7r your second solution is invalid because the question clearly asks for positive integers...

I remember studying this years and years ago and high school. Thank you for the refresher!!

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

Жыл бұрын

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

Жыл бұрын

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

@vaibhav-kr1cx

Жыл бұрын

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

Жыл бұрын

exactly same thought process flow bro

@meowcat5596

Жыл бұрын

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

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 :-)

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

@LKLogic

Жыл бұрын

🙏🏻🥰⭐️

@danielgautreau161

Жыл бұрын

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

Жыл бұрын

its a nice algebra problem

@deadbush2812

Жыл бұрын

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

@davidhart5426

Жыл бұрын

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 ...

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

Жыл бұрын

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

Жыл бұрын

@@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

10 ай бұрын

Yes! And congrats on the ascii graphics!

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!!

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

@ANunes06

Жыл бұрын

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

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

Жыл бұрын

Far easier to follow and far more concise.

@raviprabhala

Жыл бұрын

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

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.

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.

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

Жыл бұрын

The math is not good.

@shortaybrown

Жыл бұрын

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

Ah, this was A level maths in 1971 with Mr Martin at Whitchurch High School. Nice guy, used to drive a Ford Pop. I never found a use for integrating 1 + Tan^2 (x) though.

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.

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.

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

Жыл бұрын

You done it cool🤟

@GurpreetSinghMadaan

Жыл бұрын

Saw your solution after I had posted, almost same reasoning

@karriem5666

Жыл бұрын

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

@user-rf9me7xm1w

Жыл бұрын

That’s what I did, much quicker.

@leif1075

Жыл бұрын

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?

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

Жыл бұрын

your way is *much* more complicated

@pietergeerkens6324

Жыл бұрын

@@chad0x But more general.

@trietphanminh6334

Жыл бұрын

this looks easier for asian🤣

@gvc76

Жыл бұрын

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

@reconquistahinduism346

Жыл бұрын

@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.

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

Жыл бұрын

@@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.

I just substituted random numbers starting from 9 and 3, and concluded via trial and error that the answer must be x = 10 and y = 4, or x = 4 and y = 10. I'm lucky it was not too difficult of an equation - unless I find I'm horribly wrong.

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

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

Жыл бұрын

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

Жыл бұрын

You got it immediately.

@EgoTrip42

Жыл бұрын

so much easier to understand

@FuriousMaximum

Жыл бұрын

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.

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.

Thanks, I have not done algebra in a long time. It is good to remember.

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

@carstenlarsen8144

Жыл бұрын

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

@reiniernn9071

Жыл бұрын

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

Жыл бұрын

go n practice math if u wonder at this

@rickdesper

Жыл бұрын

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

Жыл бұрын

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

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

@vimmivimmi3173

Жыл бұрын

Same here

Nicely solved

Nicely done! A pleasure to watch.

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

Жыл бұрын

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

@WikiBidoz

Жыл бұрын

@@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

Жыл бұрын

@@WikiBidoz 👍

@rickdesper

Жыл бұрын

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

Жыл бұрын

@@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.

After factoring x you could arrange it so that y = 54-x / x+1 And then let y = f(x) = 54-x / x+1 since y = f(x) f(x) = 54-x / x+1 , 0 Now we essentially just want a positive integer pair of x and y which we can get if the fraction is a positive whole number, so we can use algebraic long division to get the quotient -1 and remainder 55 ∴ f(x) ≡ -1 + 55/(x +1) focusing on the right fraction part we can see 55 is in the numerator so in order for this fraction to give a whole number x+1 must be a factor of 55, which are 1,55 and 5 so we get x = 0, x = 54, x = 4 respectively but our domain restriction was 0 ∴x + y = 4 + 10 = 14

@patinho5589

Жыл бұрын

I don’t follow your logic for the step where you say x must be less the than 54.

@georgeplatsakis7277

Жыл бұрын

@@patinho5589 if x= 54 the y=f(x) becomes zero.But y must be a positive integer.if xmore than 54 then y=f(x) becomes negative ( again rejected) .

@patinho5589

Жыл бұрын

@@georgeplatsakis7277 ahh yes thank you. Your f(x) must be a positive integer because it’s y. I forgot that. Thanks.

@georgeplatsakis7277

Жыл бұрын

Your approach is elegant ,much more" mathematical" than the approach in the video.Congrats!

@f4rensabri

Жыл бұрын

Back in my time in school, we solved this using the f(x) function as well, so I understand this method better than the video method. 👍

Awesome video and explanation 👍

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

@LKLogic

Жыл бұрын

Thank you so much 🙂

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

Жыл бұрын

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

@theprofessor5625

Жыл бұрын

@@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

Жыл бұрын

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

@amishmittal2954

Жыл бұрын

​@@MizzSparkle90lmao not at all

@vbanksd

Жыл бұрын

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

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

@devenderkumar2597

Жыл бұрын

Both variables are still unknown but we do know their sum

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

that was a very insightful way to re-balance the equation, by adding the 1 to both sides so, you could factor out the y+1, thanks!

@user-fo5dl6cf7r

Жыл бұрын

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

@markgreen2170

Жыл бұрын

@@user-fo5dl6cf7r and you did not pay attention to details, it is clearly stated x and y are positive numbers greater than ZERO ...but, thanks for the complement, for more complicated problems, I would jump to chatGPT and ask for a quick list comprehension routine in python, that included the appropriate filters:)

@banhimitrakundu943

Жыл бұрын

@@user-fo5dl6cf7r are you dumb ?? x and y are positive integers

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

Жыл бұрын

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

Жыл бұрын

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

Жыл бұрын

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

@kathyg8535

Жыл бұрын

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

@thearcheologist6679

Жыл бұрын

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.

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

Жыл бұрын

where is this school?

@SurajSingh-wi9vd

Жыл бұрын

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

@sirmixalot3332

Жыл бұрын

Kansas City, Missouri

@MasterYoist

Жыл бұрын

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

Жыл бұрын

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 ?

Bixi dhiig oo badbaadi NAFTAADA # Ramadan Mubarak

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

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

Жыл бұрын

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?

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

Жыл бұрын

very fast & very good solution.

@avinashbalakrishna1927

Жыл бұрын

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

@martinbennett2228

Жыл бұрын

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

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

you can deduce they must both be even numbers since 2 does not have a solution 4 is the next candidate and then it becomes easy

From the very beginning is should be noted that the equation is symmetric relative to swapping x and y. The reasoning makes no sense starting at the point where 4 variants are suggested (1×55, 55×1, 11×5, 5×11). At this point, the answer is already found.

@rickdesper

Жыл бұрын

She's providing an exhaustive proof that the solution given is the only possible one.

@Micro-Moo

Жыл бұрын

@@rickdesper My point is: part of this proof is redundant due to the reason I've mentioned. The use of symmetry would make it more elegant and avoid redundancy.

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

Жыл бұрын

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)

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

Brilliant! Love it!

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

Жыл бұрын

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... 🤔😔

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

Жыл бұрын

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

@aaabbb-py5xd

Жыл бұрын

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

@sirmixalot3332

Жыл бұрын

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

@mattdebyl8321

Жыл бұрын

​@@aaabbb-py5xdno, it's not.

@aaabbb-py5xd

Жыл бұрын

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

Very helpful ❤

@LKLogic

Жыл бұрын

I'm so glad!❤️

i never saw or thought about this result representation with x + y before. nice to know

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

Жыл бұрын

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

Жыл бұрын

@@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

Жыл бұрын

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

Жыл бұрын

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

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

Жыл бұрын

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.

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

you can also change the form to an equation of (y+x)^2 and (y+x), then you can use the quadratic formula.

I let y = x + z and solved the quadratic for x. To make x a positive integer, z has to be 6. Then x = 4 and y = 10, with x + y = 14. But LKLogic's trick is better.

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

Жыл бұрын

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

@carstenlarsen8144

Жыл бұрын

i fpound 10 and 4 the same way for a start

@cron93

Жыл бұрын

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

@khalilfuller4939

Жыл бұрын

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

@kularathnehasindu1575

Жыл бұрын

Maths is not about guessing tho

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

Thank you and God bless you. I'm subscribing now.

@LKLogic

Жыл бұрын

Thanks for subbing!❤️

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

Жыл бұрын

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

@simens8646

Жыл бұрын

@@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.

Some sort of ‘brute force’ attack seems to work, viz. directly seeing x = (54-y)/(y+1) must be a (positive) integer. ie. k=y+1 must divide 54-y. But as y = -1 (mod k), this is the same as 54-(-1) = 55 = 0 (mod k), or that k divides 55 (that is necessary and sufficient). ie. k=1,5,11,55 are the only possibilities. But k=1 gives y=0, and k=55 gives x=0 so those are ruled out. k=5, 11 (ie.y=4,10) gives x=10,4 (50/5, 44/11 resp.) of course the solns _had_ to be symmetric.. so we get x+y=14. This is essentially the same as what you did, but without waiting for a trick to work. 😉

@DimkaTsv

Жыл бұрын

Our math teacher would've likely banned such solution. Because it is not solution at this point, it is "educated guess", thus not accepted as viable result. On Olympiad you would've also likely to get points deducted. I thought to reform equation by defining "y" through "x" and then solving equation with only "x" as variable. But as my math skills for simplification dulled to almost nothingness (yay for many years of no practice, i guess), i couldn't guarantee proper transcription.

@nnaammuuss

Жыл бұрын

@@DimkaTsv ‘brute force’ is the exact opposite of guessing. It means, you take no nondeterministic steps at all, bulldoze your way through the symbols, and you still produce a valid solution. In this case, a valid solution which is also the order of length of the proposed solution. If your teacher _would_ not accept (and that is a very big if), it is sad, might likely be caused by the fact the she/he is unable to parse through a chain of arguments, and tries to match a given solution with the one she/he's been taught. That really amounts to complete mathematical illiteracy-a lack of understanding of _what mathematics is!_ , let alone learning some particular bit of it. We mathematicians (I'm an algebraic geometer) often lament (well, actually laugh) about how schools teach nothing. If your olympiad bit is correct (a bigger if), the same remark. As it happens, I _was_ in the olympiads (a long time ago, now) and did encounter such resistance to reasoning, but only at the lower (regional) levels. I hope they've grown out of such drawbacks, if they had it, by now. 👍

@DimkaTsv

Жыл бұрын

@@nnaammuuss she won't accept brute force because getting answer through it doesn't necessary mean it is FULL answer. Also because she, as math and geometry teacher supposed to teach you how to simplify and solve things, and not bruteforce them. This particular equation goes borderline. So it probably could've been accepted, but likely it would be better to solve it with more defining way. And by deduct points on olympad it means that you get 4/5 from example. Meaning solution is still valid, just have flaw that could cause severe mistake.

@nnaammuuss

Жыл бұрын

@@DimkaTsv if so, that is very sad, as you're describing someone as your teacher who's in complete darkness about the business of mathematics, and still gets paid as a teacher. and... um, flaw? 😊 which? you do realize that you haven't been able to point out a single flaw, but are using flowery words like “full” or appealing to authority (and that authority, a school teacher, dear me! who might not _like_ a particular unfamiliar solution)? If there's a flaw, it's not a solution. mathematics is not where you jump to a conclusion based on wishful thinking and stuff, and then bargain or vote on it. About a hundred years now, we have reduced it all to a complete mechanical system, a machine can check a proof and whether it's correct or not is not a matter of opinion. if you haven't understood the difference between ‘proof’, and ‘excuse’, you haven't learnt mathematics at all. And, that is the sad failing of most our school systems-it doesn't even leave students with even a rudimentary sketch of its fundamental nature, and when they join research one's obliged to train them from scratch (and undo the harmful effects of the superstitions taught to them). Very very sad in deed.

@DimkaTsv

Жыл бұрын

@@nnaammuuss my teacher beautifully explained me in school how using brute force without enough data can lead to not correct results. For example missing bunch of other possible solutions. For this specific example it is kind of fine to do, because of HEAVY restrictions for answer (without them there are 8 different answers). Remove them, and good bye brute forced solution. Nobody forbade us from using brute force, but if anything you get answer and not solution, so it won't count as proper SOLUTION. And teacher needs to see solution to understand if her explanation on topic was understood by students. And that same exact teacher was best i ever had in school and wish i had her since my 5-th grade, and not just 10-11. Because she managed to teach me more things that other teachers did. And yet, sorry as i am unable to provide correct result at this point, because i hadn't practiced math in years, so i lost my touch completely. "Machines that can check proof"... Those machines that for common user can fuck up simple floating point math? Like 0.1+0.2 can give you 0.3000(0)4.. sometimes... (if developer hadn't fixed such behaviour). And other machines are being used to bruteforce some theorems to find if they have exceptions, and despite fact that they don't find anything, those theorems won't ever be actually resolved, because who knows what will happen with some number far towards infinity? That is why brute force is often not an acceptable as solution, just "partially correct" answer (up to point where you really find every answer). It also too resource intensive to brute force something and not resolve it. Also... Schools don't teach kids to become math scientists. It is not their job, as 90+% of students won't ever go into math science sphere. They teach BASICS, and how to apply them. Not to say most kids are barely able to grasp what they are being taught. Teachers in school teach kids how to use their brains and existing ways to convert, simplify and solve, precisely because not everything is possible to solve straight on with brute force, and moreover for much more equations brute force likely just won't show every possible answer, even though will show some if them. Sadly, i am not that proficient in math, because i am unable to see proper way to convert and reduce equations. Often i start in right direction, but in the end i just oversimplify equation to negate itself into 0.

Although this is a sure fire way to solve, it’s extremely difficult to solve in short periods of time. I believe there are two other ways to finish this equation, that both come to the same conclusion. Guess and check also takes into account human error, so issues occur more often than not

When you know x any y are postive integers better is put up some positive integers in the first place without solving equation. As the number is 54 and it is also small so try taking x and y values which when multiplies results as a number little less than 54. So x =10 and y =4

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

Жыл бұрын

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

11 ай бұрын

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

@AcePhotoSverige

11 ай бұрын

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

@markzed66

10 ай бұрын

I solved it, but there was pondering involved.

Watch out your page! Can't see the bottom part with the solution. Very nice anyway, I love this channel 💙

@pnachtwey

Жыл бұрын

The answer is obvious anyway. The two answers must be even. My first guess was 6x8 since 6x8 is less than 54, but that didn't work. 6x10 doesn't work and 8x4 didn't work. I hate "gimmick' math problems like this where the solution is not general and the limitations make it easier to rapidly work it out in your head. so 4x10 so 14 is the sum. This took only a few seconds because of the constraints.

Thanks and thumbs up..

You can do polynomial division (54-y) by (1+y). That's an alternative solution

You need to fix the thumbnail, as the condition of x and y as positive integers do not appear in it. People tend to try to solve from the thumbnail before watching the video, and when you do not get all the conditions you are kind of disappointed because trying to solve something entirely different.

@johncox7169

Жыл бұрын

Every single one of these math video I have seen do that, so I always start the video to see what the hidden change is before working the problem. Is it annoying? yes. But I know it is gonna happen, so it really isn't that big of a deal anymore.

very clever to add 1 to both sides 👍

@carstenlarsen8144

Жыл бұрын

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

Жыл бұрын

thats the process mennn🤣🤣

I think you should take one topic for example factor make video with explanation

what a crazy math!!🙏🙏🙏

That's fantastic ❤️👏

@LKLogic

Жыл бұрын

Thank you! 😊

@kapa4163

Жыл бұрын

I don’t understand

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

Жыл бұрын

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

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.

adding that " 1 " to each part of the equation was too genius. i didn't even thought about that.

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.

very simple: y = (54-x)/(x+1) = (55-x-1)/(x+1) = 55/(x+1) - 1 => 55 is divisible by (x+1) => 55 = 5*11 or 1*55, so non-negative integer X = 54, 10, 4 or 0, but x and y should be positive => only one solution 4 + 10 = 10 + 4 = 14

@LKLogic

Жыл бұрын

Visit My another channel @Lklogy and if possible share with your friends too🙏🏻🥰

@thijsg717

Жыл бұрын

This is false. x=54 and y=0 will give you 54. Their sum equals 54. x=6.5 and y=6 1/3 will give you 54. Their sum equals 12.88888...3 The real solution should be this y = 55(x+1)-1+x

@thijsg717

Жыл бұрын

Ah welp🙃 I missed the integer part. My bad.

Don't forget to cancel the random tangent of 7 quantified by reversing the back quotient needed to verify parallel series

Interesting the number of approaches to reach the same solutions.

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

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

Жыл бұрын

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

Жыл бұрын

@@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

Жыл бұрын

@@larsprins3200 Thx!

i realy like you all giving more ways to solve keep it up :D

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?

That made me realize why I choose to be a medicine practitioner🙂🙂

@sherylbegby

Жыл бұрын

Right. I always suspected medicine was really easy. So I picked up a biology textbook

10 and 4 by inspection. x + y = 14

@jmack619

Жыл бұрын

What is inspection.. to you?

@jim2376

Жыл бұрын

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

@JohnSmith-nx7zj

Жыл бұрын

But how do you know those are the only solutions?

@jim2376

Жыл бұрын

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

Nice explanation ma'am

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..

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

Жыл бұрын

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

Жыл бұрын

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

Жыл бұрын

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

Жыл бұрын

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

@smgdfcmfah

Жыл бұрын

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

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

Жыл бұрын

Yes this is what i did

@48Ballen

Жыл бұрын

where is the X**2 term?

@Vardaris

Жыл бұрын

@@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

Жыл бұрын

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

@khalilfuller4939

Жыл бұрын

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

Great❤nice video🤩👍👍

This is indeed a gem of channel

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

Жыл бұрын

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