# Let $a$ and $b$ be positive integers such that $a+b=57$ and LCM $[a,b]=680$. Find $a$ and $b$.

Let $a$ and $b$ be positive integers such that $a+b=57$ and LCM $[a,b]=680$. Find $a$ and $b$.

Step by step explanation please! I know $lcm*gcd=ab$ but I'm not sure how to incorporate this or if i need to incorporate this to solve the problem.

Thanks!

• have a look at my answer – Shobhit May 2 '14 at 20:28

Note that $a$ and $b$ are relatively prime. For any common divisor of $a$ and $b$ must divide $a+b$ and the lcm of $a$ and $b$. But $57$ and $680$ are relatively prime.

It follows that $ab=680$. Now we can substitute $57-a$ for $b$ in $ab=680$ and get a quadratic equation. More simply, $$(a-b)^2=(a+b)^2-4ab= 57^2-4(680)=529.$$ That gives $a-b=\pm 23$, and now solving for $a$ and $b$ is straightforward.

• ok, I understand that 680 and 57 are relatively prime but what equation are we plugging 57-a into? – Lil May 2 '14 at 20:21
• The equation $ab=680$. – André Nicolas May 2 '14 at 20:23
• when i have 57a-a^2=680 what do i do next? – Lil May 2 '14 at 20:26
• I have given an alternate path. But you can write $a^2-57a+680=0$, and either use the Quadratic Formula, or note the factorization $(a-17)(a-40)$. – André Nicolas May 2 '14 at 20:30
• ok, thank you. So last thing..you can always rewrite the lcm of two numbers as the product? – Lil May 2 '14 at 20:34

Let $a=pd$ and $b=qd$, where $d$ is the H.C.F of $a,b$ and so $p$ is prime to $q$.

Also $680.d = pd.qd$ or $680=pqd$.

Given $d(p+q)=57$.

Divide to get $\frac{p+q}{pq}=\frac{57}{680}$ or $(p+q)680=pq(57)$

From here $57$ is prime $680$, therefore $57$ divides $(p+q)$ also $(p+q)$ is prime to $pq$ , therefore $(p+q)$ divides $57$, hence we get $p+q=57$ and $pq=680$ solving we get $p=17, q=40$ or $p=40, q=17$ hence $d=1$.

Hence $(a,b)=(40,17),(17,40)$

Hint $\,\gcd(a,b)=1$ so we seek a factorization of $\,17\cdot 8\cdot 5\,$ into coprime factors $\,a,b\,$ with sum $57.$ It can only be $\ (a,b) = (17,\,8\cdot5)\,$ or its swap, since $\, 57 < 17\cdot 8, 17\cdot 5.$