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# An Unsolvable Problem!

Hey guys!

My teacher gave us a problem.

Eventually later on even she could not solve it!

Can you help me!

Here is the question:

H.C.F of $$\frac{a}{b}$$ and $$\frac{c}{d}$$ is $$\frac{2}{105}$$

L.C.M of $$\frac{a}{b}$$ and $$\frac{c}{d}$$ is $$\frac{12}{5}$$

H.C.F. of $$\frac{a}{c}$$ and $$\frac{b}{d}$$ is $$\frac{1}{210}$$

L.C.M of $$\frac{a}{c}$$ and $$\frac{b}{d}$$ is 60.

Find the value of a + b + c + d

Note by Tanveen Dhingra
2 years, 8 months ago

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Simply use the property : $\boxed{L.C.M(A,B) \times H.C.F.(A,B)=A \times B}$

Go on for the calculation part and you'll get the answer.

If you don't get the values even then, it means required values of $$a,b,c$$ and $$d$$ don't simply exist.

- 2 years, 8 months ago

Ok! Thank you very much sir!

- 2 years, 8 months ago

Does HCF and LCM being fractions have any meaning?

- 2 years, 8 months ago

Yeah It has. For example LCM of 1/3 and and 7/12 is 7/3 @Janardhanan Sivaramakrishnan

- 2 years, 8 months ago

Have you considered the fact that this problem may be wrong??

- 2 years, 8 months ago

Even I thought so but since my teacher gave me this question I thought it ought to be correct. Are you sure that it is wrong?

- 2 years, 8 months ago

Just ask her/him to re check it.

- 2 years, 8 months ago

Okay! But she seemed pretty sure! And she was ready to give a solution. But I think it was wrong. So means this question is wrong

- 2 years, 8 months ago

Well! If there is some info given about a,b,c,d Like are they co-prime or anything I THINK IT CAN BE SOLVED .I solved it assuming that a,b,c,d are co-prime.

- 2 years, 8 months ago

Yes. I guess they are co-prime.

Maam said that they are in their simplest form. Could you send me a solution?

- 2 years, 8 months ago

Well could I send it tomorrow. I will recheck and confirm then send it.

- 2 years, 8 months ago

Ok! Thank you

- 2 years, 8 months ago

Never think like this. Everyone makes mistakes

- 2 years, 8 months ago

@Raghav Vaidyanathan well I got answer in decimals.

- 2 years, 8 months ago

Using the fact that $$H.C.F(x,y)*L.C.M(x,y)=x*y$$ where $$x,y$$ are both natural numbers or rational fractions, we can get the following from the given data:

$$\frac { ac }{ bd } =\frac { 8 }{ 7\times 25 } \\ \frac { ab }{ cd } =\frac { 2 }{ 7 }$$

From the above equations, we can get:

$$\frac { a }{ d } =\frac { 4 }{ 35 } \\ \frac { c }{ b } =\frac { 2 }{ 5 }$$

This means that we can take:

$$a=4p, d=35p, c=2q$$ and $$b=5q$$ where $$p$$ and $$q$$ are co-prime natural numbers.

$$\frac { a }{ b } =\frac { 4p }{ 5q } ,\frac { c }{ d } =\frac { 2q }{ 5p } \quad \quad ........(1)\\ \frac { a }{ c } =\frac { 2p }{ q } ,\frac { b }{ d } =\frac { q }{ 7p } \quad \quad ........(2)$$

The fractions mentioned in Eq$$(2)$$ have H.C.F as $$\frac { 1 }{ 2\times 3\times 5\times 7 }$$. What this means is that, in Eq$$(2)$$, one of the fractions MUST contain $$5$$ in the denominator(if neither contains $$5$$ in the denominator, then the H.C.F will not contain a $$5$$ in the denominator).

This means that either $$p$$ or $$q$$ is $$5$$. (Keep in mind that $$p$$ and $$q$$ are co-prime).

Now, if we observe Eq$$(1)$$, both fractions already have exactly one power of $$5$$ in the denominator(without considering value of $$p$$ or $$q$$). Observe that the H.C.F of the fractions in Eq$$(1)$$ also has exactly one power of $$5$$ in the denominator. To take H.C.F of two fractions, we take the highest power of each prime divisor from the denominator, and since the H.C.F only has one $$5$$ in the denominator, this means that both the fractions in Eq$$(1)$$ can contain maximum of only one $$5$$ in the denominator. But we have already seen that either $$p$$ or $$q$$ is $$5$$. So either way, one of the fractions WILL have a multiple $$25$$ in the denominator. But if $$25$$ was present in the denominator, then the H.C.F must also have a multiple of $$25$$ in its denominator. This is a contradiction. Hence, our equations are inconsistent and natural numbers $$a,b,c,d$$ do not exist for given conditions.

- 2 years, 8 months ago

Thank you @Sandeep Bhardwaj sir Is this fine?

- 2 years, 8 months ago

This means that we can take: $$a=4p, d=35p, c=2q$$ and $$b=5q$$ where $$p$$ and $$q$$ are co-prime natural numbers.

I think the problem is at this step. Why do $$p$$ and $$q$$ have to be co - prime?

- 2 years, 8 months ago

They don't have to be, even if we assume that $$p$$ and $$q$$ are rational numbers, we get the same result. The actual thing that makes a difference is $$p/q$$ which is also a rational number. The problem is when we find that $$p/q$$ must have and not have a $$5$$ in the denominator simultaneously, which is impossible.

- 2 years, 8 months ago

Yeah did the same way.That's why it was coming In decimals I think.

- 2 years, 8 months ago

I don't think a rational solution exists for given data.

- 2 years, 8 months ago

The reason I get an decimal is due to that never solved for values of A,B ,C,D instead used some equations to straight away find A,B,C,D. Could have done some mistake!!. @tanveen dhingra I do not think that it is worth to give a solution by me. My solution is 85 percent the same. Though his more clearer .

- 2 years, 8 months ago

hi

- 2 years, 8 months ago