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Find-21-22-23-24-1-

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Question Number 212219 by hardmath last updated on 06/Oct/24
Find: (√(21∙22∙23∙24 + 1)) = ?
$$\mathrm{Find}: \\ $$$$\sqrt{\mathrm{21}\centerdot\mathrm{22}\centerdot\mathrm{23}\centerdot\mathrm{24}\:+\:\mathrm{1}}\:=\:? \\ $$
Answered by Rasheed.Sindhi last updated on 06/Oct/24
(√((22.5−1.5)(22.5−.5)(22.5+.5)(22.5+1.5)+1)) (√((22.5^2 −.5^2 )(22.5^2 −1.5^2 )+1)) (√(506∙504+1)) (√((505+1)(505−1)+1)) (√((505^2 −1^2 )+1)) (√(505^2 −1+1)) (√(505^2 )) 505
$$\sqrt{\left(\mathrm{22}.\mathrm{5}−\mathrm{1}.\mathrm{5}\right)\left(\mathrm{22}.\mathrm{5}−.\mathrm{5}\right)\left(\mathrm{22}.\mathrm{5}+.\mathrm{5}\right)\left(\mathrm{22}.\mathrm{5}+\mathrm{1}.\mathrm{5}\right)+\mathrm{1}}\: \\ $$$$\sqrt{\left(\mathrm{22}.\mathrm{5}^{\mathrm{2}} −.\mathrm{5}^{\mathrm{2}} \right)\left(\mathrm{22}.\mathrm{5}^{\mathrm{2}} −\mathrm{1}.\mathrm{5}^{\mathrm{2}} \right)+\mathrm{1}} \\ $$$$\sqrt{\mathrm{506}\centerdot\mathrm{504}+\mathrm{1}} \\ $$$$\sqrt{\left(\mathrm{505}+\mathrm{1}\right)\left(\mathrm{505}−\mathrm{1}\right)+\mathrm{1}} \\ $$$$\sqrt{\left(\mathrm{505}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} \right)+\mathrm{1}} \\ $$$$\sqrt{\mathrm{505}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}} \\ $$$$\sqrt{\mathrm{505}^{\mathrm{2}} } \\ $$$$\mathrm{505} \\ $$
Commented by racer last updated on 13/Oct/24
hh
Answered by Frix last updated on 06/Oct/24
x(x+1)(x+2)(x+3)+1=(x^2 +3x+1)^2 x=21 ⇒ answer is 505
$${x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)+\mathrm{1}=\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${x}=\mathrm{21}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{505} \\ $$
Answered by Rasheed.Sindhi last updated on 07/Oct/24
(√(21∙22∙23∙24 + 1)) =(√(((42)/2)∙((44)/2)∙((46)/2)∙((48)/2)+1)) =(√(((42.44.46.48)/2^4 )+1)) =(√((((45−3)(45−1)(45+1)(45+3))/(16))+1)) =(√((((45^2 −1^2 )(45^2 −3^2 ))/(16))+1)) =(√(((2024∙2016)/(16))+1)) =(√((((2020+4)(2020−4))/(16))+1)) =(√(((2020^2 −4^2 )/4^2 )+1)) =(√(((2020^2 )/4^2 )−1+1)) =(√((2020)/4)) =505
$$\:\sqrt{\mathrm{21}\centerdot\mathrm{22}\centerdot\mathrm{23}\centerdot\mathrm{24}\:+\:\mathrm{1}} \\ $$$$=\sqrt{\frac{\mathrm{42}}{\mathrm{2}}\centerdot\frac{\mathrm{44}}{\mathrm{2}}\centerdot\frac{\mathrm{46}}{\mathrm{2}}\centerdot\frac{\mathrm{48}}{\mathrm{2}}+\mathrm{1}} \\ $$$$=\sqrt{\frac{\mathrm{42}.\mathrm{44}.\mathrm{46}.\mathrm{48}}{\mathrm{2}^{\mathrm{4}} }+\mathrm{1}}\: \\ $$$$=\sqrt{\frac{\left(\mathrm{45}−\mathrm{3}\right)\left(\mathrm{45}−\mathrm{1}\right)\left(\mathrm{45}+\mathrm{1}\right)\left(\mathrm{45}+\mathrm{3}\right)}{\mathrm{16}}+\mathrm{1}}\: \\ $$$$=\sqrt{\frac{\left(\mathrm{45}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} \right)\left(\mathrm{45}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)}{\mathrm{16}}+\mathrm{1}} \\ $$$$=\sqrt{\frac{\mathrm{2024}\centerdot\mathrm{2016}}{\mathrm{16}}+\mathrm{1}}\: \\ $$$$=\sqrt{\frac{\left(\mathrm{2020}+\mathrm{4}\right)\left(\mathrm{2020}−\mathrm{4}\right)}{\mathrm{16}}+\mathrm{1}} \\ $$$$=\sqrt{\frac{\mathrm{2020}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}} }+\mathrm{1}} \\ $$$$=\sqrt{\frac{\mathrm{2020}^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}} }−\mathrm{1}+\mathrm{1}}\: \\ $$$$=\sqrt{\frac{\mathrm{2020}}{\mathrm{4}}} \\ $$$$=\mathrm{505} \\ $$
Commented by hardmath last updated on 08/Oct/24
thankyou dear professors
$$\mathrm{thankyou}\:\mathrm{dear}\:\mathrm{professors} \\ $$

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