Math Olympiad | A Nice Factorial Problem | VIJAY Maths

[@ganeshdas3174]

On factorisation of 840 it gives four continued factors. They are 4 × 5 × 6 × 7= 840 (x+1) (x + 2) (x+3) (x +4) =840 (x+1) (x+2) (x+3) (x+4) = (3+1) (3+2) (3+3) (3+4) implies x = 3

[@geralynpinto5971]

Brilliant. There's so much I can learn from you.

[@DecodeLearningUPSC]

umm i got answer in 2 mins by factorising rhs. am i doing something wrong?

[@skkarimullah9988]

Nah that's just a faster solution well done

[@PrithwirajSen-nj6qq]

We may take a brief logical approach please. The LHS is a number (840) expressed in ratio of two Factorials. The factorial as the Numerator is greater than the factorial as the Denominator Now 6!= 720 As the N/D is greater than 720 we may say that numerator will be a factorial greater than 6! So just take the Numerator =7! = (x +4)!=5040 Here x =3 So Denominator will be x! =3! Now 7!/3! =5040/6=840 Hence x =3 is the required answer.

[@raghvendrasingh1289]

👍 (x+4)(x+3)(x+2)(x+1) = 840 let x = y - 5/2 (y +3/2)(y+1/2)(y-1/2)(y - 3/2) = 840 (y^2 - 9/4)(y^2 - 1/4) = 840 y^4 - 5y^2/2+9/16 = 840 (y^2 - 5/4)^2 = 841 y^2 = 29+5/4 y^2 = 121/4 y = 11/2 x = y - 5/2 = 3

[@vijaymaths5483]

Nice approach, I like how you used the substitution to simplify the problem!

[@andrepiotrowski5668]

x^4 < 840 => x < sqrt(sqrt(840)) < sqrt(29) < 6 and 840 is divisible by 5. So only x=1, x=2, x=3 or x=4 need to be checked. (x=5: x is less than 6 as requested, but 6 * 7 * 8 * 9 is not divisible by 5) x=4: 5 * 6 * 7 * 8 = 1680 - no x=3: 4 * 5 * 6 7 = 840 - yes (here we can stop, because for lower x the result will be less than 840) x=2: 3 * 4 * 5 * 6 = 360 - no x=1: 2 * 3 * 4 * 5 = 120 - no (x=0: 1 * 2 * 3 * 4 = 24 - no - not divisible by 5)

[@adribber]

there is no condition for x, so i think about the real solution is 3 & - 8. (x + 4)!/x! = 840 x!(x + 1)(x + 2)(x + 3)(x + 4)/x! = 840 (x + 1)(x + 2)(x + 3)(x + 4) = 840 (x + 1)(x + 2)(x + 3)(x + 4) = 4*5*6*7 & (x + 1)(x + 2)(x + 3)(x + 4) = (- 7)*(- 6)*(- 5)*(- 4) x + 1 = 4 & x + 1 = - 7 ∴ x = 3 , x = - 8 ∴ eq. must have a factor as '3' & '- 8' (x + 1)(x + 4)(x + 2)(x + 3) = 840 (x² + 5x + 4)(x² + 5x + 6) = 840 (x² + 5x + 4){(x² + 5x + 4) + 2} = 840 (x² + 5x + 4)² + 2(x² + 5x + 4) = 840 x⁴ + 25x² + 16 + 2(5x³ + 20x + 4x²) + 2(x² + 5x + 4) = 840 x⁴ + 10x³ + 35x² + 50x - 816 = 0 x⁴ - 81 + 10x³ - 270 + 35x² - 315 + 50x - 150 = 0 (x⁴ - 3⁴) + 10(x³ - 3³) + 35(x² - 3²) + 50(x - 3) = 0 (x² + 3²)(x + 3)(x - 3) + 10(x - 3)(x² + 3x + 3²) + 35(x + 3)(x - 3) + 50(x - 3) = 0 (x - 3){(x² + 9)(x + 3) + 10(x² + 3x + 9) + 35(x + 3) + 50} = 0 (x² + 9)(x + 3) + 10(x² + 3x + 9) + 35(x + 3) + 50 = 0 x³ + 3x² + 9x + 27 + 10x² + 30x + 90 + 35x + 105 + 50 = 0 x³ + 13x² + 74x + 272 = 0 (x³ + 512) + (13x² + 208x + 832) - (134x + 1072) = 0 (x³ + 8³) + 13(x² + 16x + 8²) - 134(x + 8) = 0 (x + 8)(x² - 8x + 8²) + 13(x + 8)² - 134(x + 8) = 0 (x + 8){(x² - 8x + 8²) + 13(x + 8) - 134} = 0 (x + 8)(x² - 8x + 64 + 13x + 104 - 134) = 0 (x + 8)(x² + 5x + 34) = 0 ∴ (x - 3)(x + 8)(x² + 5x + 34) = 0 x² + 5x + 34 = 0 D = 5² - 4*1* 34 < 0 rejected ∴ (x - 3)(x + 8) = 0 ∴ x = 3 , x = - 8

[@PrithwirajSen-nj6qq]

If we take x = -8 Then the numerator will be factorial of - ve 4 And Denominator will be factorial of - ve 8 But factorial always be of a positive number and zero. Hence x = -8 may not be taken as an answer. Please see.

[@adribber]

@@PrithwirajSen-nj6qq Thanks for your comment. It made me aware of something I didn't know.

[@PrithwirajSen-nj6qq]

@@adribber OK, enjoy Math. Here there are two solutions worked out by me. May see. Think one of these two will make u happy. Happy New Year to all of you.

[@AmédéeGerest]

Maybe you could simplify the process? By taking a = x^2 +5x +4 a(a+2) = 840 By decomposing 840 you get 28.30 Then a = 28 And then you just solve x^2 + 5x - 24 = 0 And get - 8 and +3 Is it correct?