Math Olympiad Question | Nice Algebra Equation Solving | You should be able to solve this!

Math Olympiad Question | Nice Algebra Equation Solving | You should be able to solve this!#olympiad
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[@genevaspring8974]

Practical solution I found. (x+6)!/(x+2)! = (x+6)(x+5)(x+4)(x+3). you work with factorials, the answer is expected to be integer. Decompose. 1680 = 2*2*2*2*3*5*7. Reorganise as a product of 4 following integers, with obviously 7 in them (you wont have one of the numbers begin 14 or 21), you find quicly 1680 = 5*6*7*8. x+3=5, so x=2.

[@AbhenandanJain]

Right 👍👍

[@ulf-nicklassdegenhardt-mei3121]

Well... This translates to: Find the 4 consecutive numbers, whose product is 1680, and substract 2 of the smallest to get X... Easy...

[@ulf-nicklassdegenhardt-mei3121]

After watching the video: Yes, you can do it the complicated way, but... Just a little bit thinking outside formulas would have been easier...

[@halneufmille]

For those interested, I the four answers are: x=2, x=-11, x=(-9+√(-159))/2, x=(-9-√(-159))/2. I found the two real solutions using Gauss method.

[@deathwing3087]

Other solution : Easy to see that we have to multiply 4 successive integers, so we expect the result to be more or less a power of 4 of the integers. It is then easy to test some integers. 6^4 = 36*36 = 1296, we are almost there ! Then we take the 4 numbers as slightly above a 6*6*6*6, let's try 5*6*7*8. It works :)

[@nikitaluzhbin8982]

b = -(x1 + x2) c = x1*x2 Vieta theorem