This problem is from a 1915 algebra textbook that my

Katherine Walls

Katherine Walls

Answered question

2021-12-12

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.

Answer & Explanation

Lindsey Gamble

Lindsey Gamble

Beginner2021-12-13Added 38 answers

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
(a2),(a1),a,(a+1)
(a2)×a=(a1)×(a+1)161
a22a=(a21)161
2a=162
a=1622=81
a2=812=79
a1=811=80
a=81
a+1=81+1=82
These four consecutive numbers are 79, 80, 81 and 82.

Lindsey Gamble

Lindsey Gamble

Beginner2021-12-14Added 38 answers

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+3161
4x2x=1613
2x=158
x=79

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