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.

[hillbillydeluxe]

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 “.

[panjirizki]

Did I just watch math problem as entertainment?

[benperry]

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.

[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.

[tomknoll]

It's a nice, very logical explanation, but I solved this in about 30 seconds in my head. I just thought of two numbers that, when multiplied together would get something less than 54 and which, when added would roughly equal an amount that, when then added to the product of their multiplication might get close to 54. I first tried 6 and 7: 42+13 = 55. So then I tried 5 and 8: 40+13 = 53. So then I tried 4 and 9: 36+13 = 49. Finally, I tried 4 and 10 which gave the correct answer.

[konyecstrengthlife]

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.

[david_allen]

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.

[Robertita]

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.

[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.

[larrycornell]

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.

[bhoopendramasram]

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

[JuanDelaCruz-ilwy]

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

[ZRogers]

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.

[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.

[radhakrishnanpp]

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

[magicman-svjm]

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.

[ralfp]

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.

[pietergeerkens]

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.