2
$\begingroup$

I am stuck on this question,

Find two natural numbers, whose sum is $85$ and LCM is $102$.

I just broke $102$ as $17*2*3$ and saw that $85=17*2+17*3$. So numbers are $34$ and $51$.

But I need a mathematical way of solving all such kind of problems, as I just did this problem in hit and trial fashion.

Thanks

$\endgroup$
1
4
$\begingroup$

HINT:

Let the two numbers be $a,b$

and $(a,b)=d>0$ and $\dfrac aA=\dfrac aB=d\implies(A,B)=1$

$102=ABd$ and $85=(A+B)d$

As $85=5\cdot17,d$ must be one of $1,5,17$

but $d$ must divide $102=2\cdot3\cdot17$ as well.

See also: Find all pairs of positive integers that add up to $667$ and their $\frac{\text{lcm}}{\text{gcd}} =120$

$\endgroup$
4
$\begingroup$

Here is a little twisted but different approach. Let $a$ and $b$ be the natural numbers.

Given $a+b=85\Rightarrow b=85-a$

Since $LCM(a,b)=102$

then $ab=102k\Rightarrow a(85-a) = 102k$ where k is some natural number.

Solving the above quadratic we get $$a=\frac{85\ ^+_- \sqrt{85^2-408k}}{2}$$ Since the term under the square-root should be an odd perfect square, we can easily find $k$ to be $17$.

Hence $a=51,34$.

$\endgroup$
4
$\begingroup$

Here is what I would considervan even simpler solution.

Since the sum is divisible by $17$ which is prime, either (a) both unknown numbers are divisivle by $17$ or (b) neither is. The lcm is divisble by $17$ so (a) must apply. Then the lcm is divisible by $3$ , another prime factor, so at least one of the unknowns must also be divisible by $3$. Thus, an unknown must be divisible by $3\times 17=51$ and less than $85$, so must be $51$ itself. Leaving $34$ for the other one.

$\endgroup$
3
$\begingroup$

Maybe this can help:

Say the required two numbers are $A$ and $B$. We are given that: $A+B=85=5\times 17...(1)$ and $lcm(A,B)=102=2\times 3\times 17$. As $gcd(A,B)$ divides both $A$ and $B$, it has to divide $A+B$ also so, hence that makes $gcd(A,B)=17$.

Now using the identity : $A*B=gcd(A,B)\times lcm(A,B)$, we have: $AB=17\times 102...(2)$ Solving $(1)$ and $(2)$ gives us the required answer. Hope it helps.

$\endgroup$
3
$\begingroup$

Hint:

Let $a+b=85=17\cdot5$ and $[a,b]=102=17\cdot 2\cdot3$.

Denote $\gcd(a;b)=d$. Then $[a,b]=\frac{ab}{d}=a_1b_1d_1$

Then $a=a_1d, b=b_1d$, where $\gcd(a_1;b_1)=1$.

Then $d(a_1+b_1)=5\cdot17$

1) Let $d=1$, then $a_1+b_1=85, a_1\cdot b_1 =102$

2) Let $d=5$ then $a_1+b_1=17, a_1\cdot b_1 =\frac{102}5$

3) Let $d=17$ then $a_1+b_1=5, a_1\cdot b_1 =\frac{102}{17}=6$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.