China | Can you solve this? | A Nice Math Olympiad Algebra Problem
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- Опубликовано: 13 сен 2024
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Even if you didn't recognize 120 as 5!, you could easily reason as follows: We have four consecutive integers whose product is 120. Thus, they must be in the neighborhood of the fourth root of 120. Now 3^4 = 81 < 120 and 4^4=256 > 120, so these consecutive integers must be near 3 or 4. So if we just try 2*3*4*5, we've got it. So x = 1. Similarly, (-2)*(-3)*(-4)*(-5) = 120. So x = -6 as well.
Why assuming x is an integer?
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@@HenriLaporte-kv6qq It doesn't have to be, but these type of problems often do have integer solutions. If I couldn't find it quickly, I might have guessed there aren't integer solutions. Try the low hanging fruit first, I guess. My bigger point is that often you can make a simple educated guess before pulling out the big guns. Being a 4th order polynomial, turns out the complex roots don't have integer components. In fact the complex part isn't even rational. x = 1/2 (- 5 +/- sqrt(39)i)
120 =2*3*4*5. x+1=2 => x=1. 120=(-2)*(-3)*(-4)*(-5), => x+1=-5; x=-6. По схеме Горнера приходим к уравнению x^2 + 5x + 16=0 с корнями (-5+/-V39*i)/2.
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Fact the first, we need 4 consecutive whole numbers that multiply to equal 120.
Fact the second, 120 prime factorization is 2, 2, 2, 3, 5
2 * 3 * 4 * 5 fits both of these.
Fact the third, an even set of negative multiplied together result in a positive number.
so -2, -3, -4, -5 also satisfy.
So x = 1 or x = -6
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I was thinking along similar lines though you put this much better than I could have!
another approach
let t = x + 2.5 then (t - 1.5)(t - 0.5)(t + 0.5)(t + 1.5) = 120
=> (t^2 - 0.5^2)(t^2 - 1.5^2) = 120 => 16t^4 - 40t^2 - 1911 = 0
=> t^2 = 49/4 => t = ±7/2 => x = t -2.5 = -6 or 1 (real solutions only)
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Yes, yes, moreover, both methods are relevant in the generalized version (x+a)(x+b)(x+c)(x+d)=e, where a+d=b+c ...
Умножим (х+1) на (х+4) имеем х²+5х+4. Умножим (х+2) на (х+3) имеем х²+5х+6. Обозначим z=х²+5х+4, тогда уравнение примет вид: z(z+2)=120, очевидно z1=10 z1=-12 далее последовательно решаем 1)х²+5х+4=10 и 2)х²+5х+4=-12. Получаем х1=-6, х2=1, в случае 2) нет действительных корней.
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Math Olympiad: (x + 1)(x + 2)(x + 3)(x + 4) = 120; x =?
First method:
(x + 1)(x + 2)(x + 3)(x + 4) = 120 = (2)(3)(4)(5) = (1 + 1)(1 + 2)(1 + 3)(1 + 4)
= (- 6 + 1)(- 6 + 2)(- 6 + 3)(- 6 + 4) = (- 5)(- 4)(- 3)(- 2); x = 1 or x = - 6
Missing two complex value roots
Second method:
[(x + 2)(x + 3)][(x + 1)(x + 4)] = (x² + 5x + 6)(x² + 5x + 4) = 0
(x² + 5x + 6)(x² + 5x + 6 - 2) = (x² + 5x + 6)² - 2(x² + 5x + 6) = 120
(x² + 5x + 6)² - 2(x² + 5x + 6) - 120 = (x² + 5x + 6 - 12)(x² + 5x + 6 + 10) = 0
(x² + 5x - 6)(x² + 5x + 16) = (x - 1)(x + 6)(x² + 5x + 16) = 0
x - 1 = 0, x = 1; x + 6 = 0, x = - 6 or x² + 5x + 16 = 0, x = (- 5 ± i√39)/2
Answer check:
(x + 1)(x + 2)(x + 3)(x + 4) = 120; Confirmed as shown in First method
x = (- 5 ± i√39)/2:
[(- 5 ± i√39)/2 + 1][(- 5 ± i√39)/2 + 2][(- 5 ± i√39)/2 + 3][(- 5 ± i√39)/2 + 4]
= (1/16)(- 3 ± i√39)(1 ± i√39)(1 ± i√39)(3 ± i√39) = (1/16)(- 39 - 9)(- 39 - 1)
= (1/16)(- 48)(- 40) = 120; Confirmed
Final answer:
x = 1, x = - 6, Two complex value roots; x = (- 5 + i√39)/2 or x = (- 5 - i√39)/2
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120 is 5!
The 1 does nothing.
So, you need four terms. 5,4,3,2. x = 1
Then negative also.
-5,-4,-3,-2. x = -6
2 imaginary roots also.
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Этот ролик можно было бы сократить вдвое, если бы автор не занимался переписыванием выражений при каждом удобном случае.
Anyone with knowledge of elementary maths can guess the x=1 solution in 10 seconds.
But the equation is a quartic so has four solutions: 1, -6, -5/2 +/- i root(39) / 2
Only the complex solutions get anywhere near a "Maths Olympiad" type problem!
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120 | 2
60 | 2
30 | 2
15 | 3
5 | 5
The secuence needs to be consecutive. So;
2x3x2x2x5 which is equal to 2x3x4x5, not hard to guess.
So x=1 or x=-6
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x=t-1 substitute
t(t+1)(t+2)(t+3)=120 rewritten
(t²+3t)(t²+3t+2)=120 multiply middle terms and outer terms
(t²+3t)(t²+3t+1)+t²+3t=120 split 2=1+1, distribute to the constant
(t²+3t)(t²+3t+1)+t²+3t+1=121 add 1 to both sides
(t²+3t+1)²=121 factor the trinomial out and combine like terms.
(t²+3t+1)²-11²=0 subtract 121 from both sides
(t²+3t-10)(t²+3t+12)=0 factor difference of squares
(t-2)(t+5)(t²+3t+12)=0 factor left quadratic.
t= 2 or -5 or (-3+i√39)/2 or (-3-i√39)/2 complete the square on right quadratic, solve for t with 0 product property
x=t-1, so
x=-6 or 1 or (-5+i√39)/2 or (-5-i√39)/2
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Or -5*-4*-3*-2 = 120
x = -6
Plus two imaginary roots.
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Way too complex a solution process. If four factors multiplied = 120, none can be much bigger than 5. So, 120/5 = 24. 24 = 2x3x4. So, We have 2x3x4x5 = 120. (X=1) = 2. X = 1.
1+2+3+4=10，1*2*3*4=24（-120=-96=16*-6因16-6=10，所以因式分解（x^2+5x-6）（x^2+5x+16）
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X=1
when see 120, I recognise 5 factorial : 2x3x4x5 = 120. So, I find a 1st solution : x = 1 (because (1+1)x(1+2)x(1+3)x(1+4)=120
Of course, -2x-3x-4x-5 is also equal to 120, so, I find a second solution : x = -6 (because (-6+1)x(-6+2)x(-6+3)x(-6+4)= 120
After, if you divide the initial polynomial by (x-1) and by (x + 6) : the result is a quadratic equation, very easy to resolve
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1
Interesting. Integerwise it's sort of 4 consecutive numbers multiplied, while having 120 dividing 2, 3, 4, 5, 6 but no 7 - simple trial and error, x=1, before recognizing, there must be 4 solutions 😅 So at least one could construct the same solution 2*3*4*5, but as (-5)(-4)(-3)(-2) by choosing x= -6. The funny expand & polynomial division, which is chilling, should gibe a quadratical for the rest.
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This is proper way but too long for me. I prefer to use a.b.c.d=120, since we known 120 can be divide 4 times: 6.20 is 2.3.4.5 and -5.-4.-3.-2, the rest is easy.
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120 = 5 (24) = 5(6)(4) = (2)(3)(4)(5) = (-2)(-3)(-4)(-5)
x + 1 = 2, x = 1
x + 1 = -5, x = -6
To find the remaining two solutions
(x + 1)(x + 2)(x + 3)(x + 4) = (x2 + 3x + 2)(x + 3)(x + 4) = (x^3 + 6x^2 + 11x + 6)(x + 4) = x^4 + 10x^3 + 35x^2 + 50x + 24 = 120
x^4 + 10x^3 + 35x^2 + 50x - 96 = 0
(x - 1)(x^3 + 11x^2 + 46x - 96) = 0
(x - 1)(x + 6)(x^2 + 5x + 16) = 0
x = 1, -6, (-5 +/- i✓39)/2
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Ans. x=1. Now 2×3×4×5=120. So, (1+1)(1+2)((1+3)(1+4).
x={1,-2,-3 and -6}ans
Как то всегда помнил, что 3*4=12. 3 и 4 идут подряд, а 120\12=10. 10 это 2* 5.ряд сложился 2*3*4*5. Значит х это 2-1=1
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It’s about the longest possible way to solve this
There was definitely no need for the root formula
As you can solve it like this
X^2+5x-6=0
(X+6)(x-1)=0
X+6=0
X=-6
X-1=0
X=1
Much faster, much easier, much less room for Making mistakes.
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1.
(x + 1)(x + 2)(x + 3)(x + 4) = 120
120 = 5! = 2¹3¹4¹5¹
x + 1 = 2 => x = 1
x + 4 = -2 => x = -6
(x² + 5x + 4)(x² + 5x + 6) = 120
x² + 5x = u
(u + 4)(u + 6) = 120
u = 6 ∨ u = -16
x² + 5x = 6
x(x + 5) = 6
x = 1 ∨ x = -6
x² + 5x = -16
x² + 5x + 16 = 0 => complex solution
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X = 1
(1+1)(1+2)(1+3)(1+4)=2x3x4x5=
120.
a(a+b)(a+2b)(a+3b)+b^4 = (a^2 + 3ab +b^2)^2
In this question, a = x+1, b = 1, so that (x+1)(x+2)(x+3)(x+4)+1=(x^2 + 5x + 5)^2 = 11^2
x^2+5x+5=±11
x= 1, -6, (-5±√39i)/2
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Most obvious quick answer. 120 = 5!, which => x = 1.
x=1
got it thru trial n error😀
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I knew X=1 just looking at it, took me maybe 5 seconds to go through the whole operation
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Prime roots of 120 ( got by dividing by successive prine numbers ie 3 then 3 then 5 then 7 etc.) = 5 x3x2^3 =5x3x2x2^2= 2x3x4x5 =(x+1)(x+2)((x+3)(x+4)
=> x=1
X=1. It took me only 10 seconds. Just don't ask me to show my work.
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🎉🤭guess why the authorities didn’t put this one in the multiple choiceS😂?
Sometimes, it's quicker to use some logic. Must be a whole number, can't be too big.... let's try x=1. Voila, it works. Problem solved.
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（x+1)(x+2)(x+3)(x+4)= 2*3*4*5 = 120
x = 1
That one took me about 15 seconds but, of course, it misses the other roots.
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5!=120. 3 Sekunden Zeit
Too easy……
2*3*4*5=120
х=1
2*3*4*5 = 120 so x = 1
Sustituyó x= 0 y da 24. Sustituyo x = 1 y da 120
Ya está 😂
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+-√121.
(x + 1) (x + 2) (x + 3) (x + 4) = 5!
(x + 1) (x + 2) (x + 3) (x + 4) = 5 · 4 · 3 · 2 · 1
(x + 1) (x + 2) (x + 3) (x + 4) = (1 + 1) (1 + 2) (1 + 3) (1 + 4)
x = 1.
Ok, that's too easy. 🤣
Again:
(x + 1) (x + 2) (x + 3) (x + 4) = 120
(x² + 3x + 2) (x² + 7x + 12) = 120
x⁴ + 10x³ + 35x² + 50x + 24 = 120
x⁴ + 10x³ + 35x² + 50x − 96 = 0
(x − 1) (x³ + 11x² + 46x + 96) = 0
(x − 1) (x + 6) (x² + 5x + 16) = 0
x₁ = 1 ∨ x₂ = −6
x₃,₄ = −5/2 ± √(25/4 − 16)
= −5/2 ± √(−39/4)
= −5/2 ± i√39 / 2
x₃ = (−5 − i√39) / 2 ∨ x₄ = (−5 + i√39) / 2
𝕃ₓ = {−6, 1, (−5−i√39)/2, (−5+i√39)/2}
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If you can’t see (x=1, -6 )these problems aren’t for you.
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Один из ответоа Х равен 1
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а где проверка для Х3 и Х4?
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Кто вас учил такому решению?!. Это полная безграмотность в математике.!
x equals one and you can spot at once that 5! is 120, so one is obviously the solution... ...5 seconds-challenge, right?
Le p'tit Daniel
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X=1，x= －6，30seconds find solution
Half solution.
1
x = 1
1
1
x =1
1