Canadian Junior Math Olympiad Problem | Algebra | 2 Methods

Canadian Junior Math Olympiad Problem | Algebra | Solved by 2 Ways

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[crkr]

We can arrange like this
a+2ab+b=38
2a+4ab+2b+1=77
(2a+1)(2b+1)=77 11×7
a=5 0r 3
b=5 or 3
a=3, b=5 satisfies two equations
It takes justv25 seconds

[GillesF]

Great video but a question is there: at 2:38 and from << b² - b - 20 = 0 >> you use a not obvious trick instead of (see below) ...

b² - b - 20 = 0

delta = (-1)² - 4*1*-20 = 1 + 80 = 81

√(delta) = √81 = 9

root #1: b = (-(-1) + 9)2*1 = 10/2 = 5

root #2: b = (-(-1) - 9)2*1 = 10/2 = -4

...

(same remark in the second method).

🙂

[Osirion]

Here's a cool method assuming a and b are both positive integers :
First, transform the first 2 equations into
a(1+b)=18
b(1+a)=20
Now let x and y be factors of 18, then xy=18 and therefore, (x+1)(y-1)=20
By looking at the factors of 18 and 20, we need to find a couple that satisfies the equations above
For 18 we have : (1*18), (2*9), (3*6)
For 20 we have : (1*20), (2*10), (4*5)
If we look closely, (3, 6) and (4, 5) is the couple we're looking for as they differ by +1 and -1
Meaning that if x=3 and y=6, then we get that (x+1)(y-1)=4*5
So it must satisfy both equations, therefore we have an answer
Again this only works if a and b are positive integers, but it's still a cool method that requires a bit of logical thinking instead of just pure algebra