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# Problem

+1
20
2
+289

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

Jan 13, 2024

#2
+36926
+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
+289
+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
+36926
+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 .......

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