This problem is from a 1915 algebra textbook that my grandmother in law had when she took algebra in highschool. There are four consecutive integers such that product of the first and third integers is 161 less than the product of the second and fourth integers.

Question

Lindsey Gamble · Accepted Answer

Given there are four consecutive numbers such that product of the first and third integers is 161 less than the product of the second and fourth integers. Let the four consecutive integers be (a−2),(a−1),a,(a+1) (a−2)×a=(a−1)×(a+1)−161 ⇒a2−2a=(a2−1)−161 ⇒−2a=−162 ⇒a=1622=81 a−2=81−2=79 a−1=81−1=80 a=81 a+1=81+1=82 These four consecutive numbers are 79, 80, 81 and 82.

Lindsey Gamble · Accepted Answer

Let the first integer be x.  Next three consecutive integers will be  x+1,x+2,x+3  The four consecutive integer are  x(1st)  x+1(2nd)  x+2(3nd)  x+3(4nd)  Product of first and 3rd integer is 161 less than product of 2nd and 4th integer.  x(x+2)=(x+1)(x+3)−161  x2+2x=x2+4x+3−161  4x−2x=161−3  2x=158  x=79

This problem is from a 1915 algebra textbook that my

Answered question

Answer & Explanation

New Questions in Algebra I