The product of four consecutive natural numbers is 360. find the numbers.
Answer :
We know natural numbers " The whole numbers from 1 upwards: 1, 2, 3, ... and no negative numbers and no fractions. "
Let Four consecutive natural numbers = x , ( x + 1 ) , ( x + 2 ) , ( x + 3 ) , Here x , can't be negative as we know natural numbers doesn't have negative numbers .
From question we get
x ( x + 1 ) ( x + 2 ) ( x + 3 ) = 360
x ( x + 3 ) ( x + 1 ) ( x + 2 ) = 360
( x2 + 3x ) ( x2 + 3x + 2 ) = 360
[ ( x2 + 3x + 1 ) - 1 ] [ ( x2 + 3x + 1 ) + 1 ] = 360
[ ( x2 + 3x + 1 )2 - 12 ] = 360
( x2 + 3x + 1 )2 - 1 = 360
( x2 + 3x + 1 )2 = 361
( x2 + 3x + 1 ) = 19
x2 + 3x - 18 = 0
x2 + 6x - 3x- 18 = 0
x ( x + 6 ) - 3 ( x + 6 ) = 0
( x - 3 ) ( x + 6 ) = 0
So,
x = 3 and - 6 , We neglect x = - 6 , As we assumed x as a natural number .
So,
x = 3 ,
x + 1 = 3 + 1 = 4 ,
x + 2 = 3 + 2 = 5
And
x + 3 = 3 + 3 = 6 ,
And
Four consecutive natural numbers whose product is 360 = 3 , 4 , 5 and 6 ( Ans )
We know natural numbers " The whole numbers from 1 upwards: 1, 2, 3, ... and no negative numbers and no fractions. "
Let Four consecutive natural numbers = x , ( x + 1 ) , ( x + 2 ) , ( x + 3 ) , Here x , can't be negative as we know natural numbers doesn't have negative numbers .
From question we get
x ( x + 1 ) ( x + 2 ) ( x + 3 ) = 360
x ( x + 3 ) ( x + 1 ) ( x + 2 ) = 360
( x2 + 3x ) ( x2 + 3x + 2 ) = 360
[ ( x2 + 3x + 1 ) - 1 ] [ ( x2 + 3x + 1 ) + 1 ] = 360
[ ( x2 + 3x + 1 )2 - 12 ] = 360
( x2 + 3x + 1 )2 - 1 = 360
( x2 + 3x + 1 )2 = 361
( x2 + 3x + 1 ) = 19
x2 + 3x - 18 = 0
x2 + 6x - 3x- 18 = 0
x ( x + 6 ) - 3 ( x + 6 ) = 0
( x - 3 ) ( x + 6 ) = 0
So,
x = 3 and - 6 , We neglect x = - 6 , As we assumed x as a natural number .
So,
x = 3 ,
x + 1 = 3 + 1 = 4 ,
x + 2 = 3 + 2 = 5
And
x + 3 = 3 + 3 = 6 ,
And
Four consecutive natural numbers whose product is 360 = 3 , 4 , 5 and 6 ( Ans )