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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

Martin Sleziak
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super saiyan
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  • See this link your question is similar to this. http://math.stackexchange.com/questions/1143716/find-two-numbers-whose-sum-is-20-and-lcm-is-24/1143731 – Kanwaljit Singh Jan 01 '17 at 12:17

5 Answers5

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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 $a=51,34$.

8hantanu
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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.

Oscar Lanzi
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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 $\frac{\text{lcm}}{\text{gcd}} =120$

Community
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lab bhattacharjee
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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.

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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 $a_1+b_1=5, a_1\cdot b_1 =\frac{102}{17}=6$

Roman83
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