Find three consecutive numbers whose sum is 2001
Hbhalla:
666+667+668 makes the desired number.
Answers
Answered by
5
Let the 3 numbers be n-1,n,n+1
n-1+n+n+1=2001
3n=2001
.:n=2001/3
n=667
.: The 3 numbers are 666,667 and 668
n-1+n+n+1=2001
3n=2001
.:n=2001/3
n=667
.: The 3 numbers are 666,667 and 668
Answered by
6
Let the numbers be :(x),(x+1),(x+2) As total sum is 2001 (x)+(x+1)+(x+2)=2001 3x+3=2001 3x=2001-3 3x=1998 x=666 1st number=x=666 2nd number=x+1=667 3rd number=x+2=668 Hope it helps.
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