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The product of four consecutive positive integers is 1 less than 461². What is the least of these four numbers?

 Jan 13, 2024

Best Answer 

 #2
avatar+36916 
+1

Let the numbers be     n-1     n      n+1    and   n+2

   The numbers are ALMOST 

 n^4 = 461^2 - 1 = 212520      <======take 4th root of both sides

n ~~  21 

                   so try  20 21 22 23    =  212520     Bingo !   On first try .......

 Jan 13, 2024
edited by ElectricPavlov  Jan 13, 2024
 #1
avatar+288 
+1

I think to do n(n+1)(n+2)(n+3)=461^2 -1, then do

 

(n^2 +n)(n+2)(n+3)=212520

 

(n^3+3n^2+2n)(n+3)=212520

 

n^4+3n^3+3n^3+9n^2+2n^2+6n=212520

 

n^4 + 6n^3 + 11n^2 + 6n = 212520

 

What now? Or is there another way to do this?

 

Pls help

 Jan 13, 2024
 #2
avatar+36916 
+1
Best Answer

Let the numbers be     n-1     n      n+1    and   n+2

   The numbers are ALMOST 

 n^4 = 461^2 - 1 = 212520      <======take 4th root of both sides

n ~~  21 

                   so try  20 21 22 23    =  212520     Bingo !   On first try .......

ElectricPavlov Jan 13, 2024
edited by ElectricPavlov  Jan 13, 2024

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