A number theory problem by A Former Brilliant Member

Number Theory Level 3

Let $a, b$ and $c$ be positive distinct integers such that $a+b+c=36$ and $abc=1680$ . Find $3bc\left(1-\dfrac{b}{36}-\dfrac{c}{36}\right)$ .

2 solutions

A Former Brilliant Member
Jul 25, 2017

$a+b+c=36$ $\color{#D61F06}\large \implies$ $a=36-b-c$

$abc=1680$ $\color{#D61F06}\large \implies$ $a=\dfrac{1680}{bc}$

$a=a$

$36-b-c=\dfrac{1680}{bc}$

$bc-\dfrac{b^2c}{36}-\dfrac{bc^2}{36}=\dfrac{140}{3}$

$bc\left(1-\dfrac{b}{36}-\dfrac{c}{36}\right)=\dfrac{140}{3}$

$3bc\left(1-\dfrac{b}{36}-\dfrac{c}{36}\right)=\dfrac{140}{3}(3)=$ $\large \color{plum}\boxed{140}$

This is incomplete. You need to prove that there exists positive integers, a,b,c satisfying the 3 conditions.

Pi Han Goh - 3 years, 10 months ago

Mohan Nayak
Aug 10, 2017

Prime factorize 1680=>2^(4) *(3) *(5) *(7) ,partition the factors such that their sum a+b+c=36. Guess a, b, c as 10,12,14