the product of four consecutive positive integers is 840 find the number
Answers
Answer:
Step-by-step explanation:Let the consecutive numbers are ( x - 2 ) , ( x - 1 ) , x , ( x + 1 ).
According to the question :
Product of the same consecutive numbers = 840
( x - 2 )( x - 1 )( x )( x + 1 ) = 840
x( x - 2 )( x - 1 )( x + 1 ) = 840
( x² - 2x )( x² - 1 ) = 840
x²( x² - 1 ) - 2x( x² - 1 ) = 840
x⁴ - x² - 2x³ + 2x = 840
x⁴ - 2x³ - x² + 2x + 840 = 0
x⁴ + 5x³ - 7x² - 35x² + 34x² + 170x - 168x - 840 = 0
( x + 5 )( x - 6 )( x² - x + 28 ) = 0
x = - 5 or 6
As numbers are positive, x = 6
Therefore required positive number are : 4 , 5 , 6 , 7.
Method 2 :
As the product is 840,
840 = 2 × 2 × 2 × 3 × 5 × 7
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Numbers which can be consecutive : 3 , 5 , 7 but 4 and 6 are missing, in order to make them consecutive we want 4 and 6. Extra numbers : 2³
2 × 2 can be written as 4 .
Now, Number which can be consecutive 3 , 4 , 5 , 7 . Still we want 6 and we have 2 as extra but on factorising 6 we get 6 can be written as 2 × 3. So multiplying extra ( 2 ) and 3 , we get 6
840 = 3 × 4 × 5 × 2 × 7
840 = 4 × 5 × 2 × 3 × 7
840 = 4 × 5 × 6 × 7
Therefore, required positive numbers are 4 , 5 , 6 , 7.