SOLUTION: If the product of 3 consecutive numbers is 21924 what is the smallest of these numbers?
Algebra.Com
Question 1013364: If the product of 3 consecutive numbers is 21924 what is the smallest of these numbers?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Look at the cube root of 21924 almost 28
26*27*28 does not have a 4 for the final digit, but 27*28*29 does.
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Alternatively,
numbers are x, x+1, x+2
(x^2+x)(x+2)=x^3+3x^2x=21924
x^3+3x^2+2x-21924=0
x=27
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