Sum of three digits=product of three digits.if the numbers are consecutive natural numbers the numbers are...
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Hi there!
Here's the answer:
•°•°•°•°<><><<><>><><>°•°•°•°•°
Given
3 Digits are consecutive Natural No.
Let the No.s be x, x+1, x+2
Sum of digits = x + (x+1) + (x+2)
= 3x+3
= 3(x+1)
Product of digits = x(x+1)(x+2)
As per the given data,
3(x+1) = x(x+1)(x+2)
=> x(x+2) = 3
=> x² + 2x - 3 = 0
=> x² + 3x - x - 3 = 0
=> x(x+3) - 1(x+3) = 0
=> (x-1)(x+3) = 0
•°• x = 1 or x = -3
But x£N ; so x = 1
•°• The No.s are 1 , 2 , 3
Sum = 1 + 2 + 3 = 6
Product = 1 × 2 × 3 = 6
•°•°•°•°<><><<><>><><>°•°•°•°•°
Hope it helps
Here's the answer:
•°•°•°•°<><><<><>><><>°•°•°•°•°
Given
3 Digits are consecutive Natural No.
Let the No.s be x, x+1, x+2
Sum of digits = x + (x+1) + (x+2)
= 3x+3
= 3(x+1)
Product of digits = x(x+1)(x+2)
As per the given data,
3(x+1) = x(x+1)(x+2)
=> x(x+2) = 3
=> x² + 2x - 3 = 0
=> x² + 3x - x - 3 = 0
=> x(x+3) - 1(x+3) = 0
=> (x-1)(x+3) = 0
•°• x = 1 or x = -3
But x£N ; so x = 1
•°• The No.s are 1 , 2 , 3
Sum = 1 + 2 + 3 = 6
Product = 1 × 2 × 3 = 6
•°•°•°•°<><><<><>><><>°•°•°•°•°
Hope it helps
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