IMO 1959 - Problem 2
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- čas přidán 8. 09. 2024
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At 4:45, when you replace the square root with the mod, I took a different path.
Since there are 2 possible square roots (one positive, and one negative), this becomes 2 equations (only the LHS shown):
2x+2(x-1), and
2x-2(x-1)
That still gives us the same 2 results, but without the same constraints on x.
A^2 = 2, or
x=(A^2 + 2)/4
If we try out A=1, and x=3/4 - with the right choice of +ve square root, or -ve square roots, the solution holds good.
Interesting approach. Thanks for sharing
back in 1960 i could be a imo participant lol, today it is much tougher
Nice it's a simple problem liked it
Nice 👍
Cool
Thanks for checking 👍