Competitive math players -- How do they solve so fast?

[member 624364]

Hi, I have watched some videos of Raytheon Mathcounts Competitions and the likes on Youtube recently. I was awe struck by the shear speed that some of the players were able to solve the problems.

Some, so fast that I don't believe a human brain could actually properly solve if given to randomly with no prior experience related to that specific problem while only equipped with the foundational knowledge to solve such problems.

The players at various points blurted out correct answers within seconds before the question could even be fully read. I believe this must be like Rubiks cube competitive solvers where they have practiced so many possible situations and have memorized them into almost like the brains version of muscle memory with the solution or precise way to solve them ready at a moments notice.

The only way I can think of it is like students memorising squares, cubes and multiplicative tables that they can give off at a moments notice, except they are instead memorising outpoints and methods to more advanced problems.

Am I approximately correct in my hypothesis? I just don't know any other way they could solve so quickly. I have never engaged in competitive mathematics at any point in my life so forgive me if the answer is pretty obvious/trivial.

 

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Practice. Usually there is not much to calculate if you recognize how to do it.
You certainly want to know some basic results by heart, but you don't need many. Square numbers below 100, cubes of 2, 3, 4, 5 - things like that.

Spoilers:

Spoiler

• For the first question it is sufficient to know that 50% is 1/2. Then technically you divide 3 by 1/2, but that is clearly 3*2=6, a trivial multiplication.
• The n! answer came really quick, but again you don't have to actually calculate anything. n!/(n-1)! will give the factor n for the numerator and n!/(n+1)! will give the factor n+1 for the denominator. You just have to make sure you don't mess up the +1 (in other words: 8 is not the right answer). It helps if you encountered a similar expression before, but they do pop up once in a while in mathematics.
• Juice: Volume scales with the height cubed, 2/3 of the height is left, so 8/27 of the volume is left. You should certainly know these cubes. Here comes the first actual calculation: 1 - 8/27 = 19/27 because 27-8=19.
• 13/(p²-3): This one is a bit trickier. 13 is a prime, so the denominator can only be 1 or 13. A denominator of 1 means p²=4 or p=2, and a denominator of 13 means p²=16 or p=4. Sum: 2+4=6.
• The cube root of 97336: There are tricks to calculate this very fast, but if you don't know them: 50^3 = 125000 which is a bit too large, but 40^3 = 64000 is too low. The result has to be somewhere around 45. The cube is even so the base number has to be even as well. If you cube 44 then the number ends with 4 (as 4^3=64 ends in 4), if you cube 46 the number ends in 6 as powers of 6 always do, if you cube 8 you get whatever but not 6, and 48 looks too large anyway. The answer has to be 46. You can find this faster than it is to type all this.
• (x^2+2x+1)/(x^2+6x+9)=36/49: This needs some time and it doesn't surprise me the early answer was wrong. 36 and 49 are both squares, but they don't equal the numerator and denominator. You can recognize them as (x+1)^2 and (x+3)^2, then you can take the square root on both sides: (x+1)/(x+3)=6/7. The difference between numerator and denominator on the left side is twice as large, so let's expand the right side by 2 to match it: (x+1)/(x+3)=12/14 - okay that works, x=11.

Edit: Some more. All these can be solved faster than writing down the solution path.