Levans writes a positive fraction in which the numerator and denominator are integers, and the numerator is 1 greater than the denominator. He then writes several more fractions. To make each new fraction, he increases both the numerator and the denominator of the previous fraction by 1. He then multiplies all his fractions together. He has 3 fractions, and their product equals 10. What is the value of the first fraction he wrote?

Guest Nov 12, 2022

#1**0 **

*Levans writes a positive fraction in which the numerator and denominator are integers, and the numerator is 1 greater than the denominator. He then writes several more fractions. To make each new fraction, he increases both the numerator and the denominator of the previous fraction by 1. He then multiplies all his fractions together. He has 3 fractions, and their product equals 10. What is the value of the first fraction he wrote?*

I tried solving this, using x as the denominator in the first fraction, and going from there.

I got the final answer as x = 1/3 which isn't the desired answer. Now I'm going to try brute force.

I'll put them in sort of a table to see if there's a pattern to be discerned.

(2•3•4) = 24 the denominator would be (1•2•3) = 6 Resultant is 24/6 = 4**.**00

(3•4•5) = 60 the denominator would be (2•3•4) = 24 Resultant is 60/24 = 2**.**50

(4•5•6) = 120 the denominator would be (3•4•5) = 60 Resultant is 120/60 = 2**.**00

(5•6•7) = 210 the denominator would be (4•5•6) = 120 Resultant is 210/120 = 1**.**75

(6•7•8) = 336 the denominator would be (5•6•7) = 210 Resultant is 336/210 = 1**.**60

The above trend is going the wrong direction so I tried graphing it.

(x+1) (x+1+1) (x+1+1+1)

I took this formula to Desmos y = –––– • ––––––– • –––––––––––

(x) (x+1) (x+1+1)

The only place when y = 10 , x is less than 1.

The problem has no solution which is an integer

unless I've made a mistake in my initial conception.

_{.}

Guest Nov 13, 2022