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

Easy Math Editor

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Comments

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.

Sandeep Bhardwaj - 6 years, 2 months ago

Ok! Thank you very much sir!

tanveen dhingra - 6 years, 2 months ago

Does HCF and LCM being fractions have any meaning?

Janardhanan Sivaramakrishnan - 6 years, 2 months ago

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

A Former Brilliant Member - 6 years, 2 months ago

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

I'm getting contradictory conclusions.

Raghav Vaidyanathan - 6 years, 2 months ago

@Raghav Vaidyanathan well I got answer in decimals.

A Former Brilliant Member - 6 years, 2 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?

tanveen dhingra - 6 years, 2 months ago

Never think like this. Everyone makes mistakes

A Former Brilliant Member - 6 years, 2 months ago

Just ask her/him to re check it.

A Former Brilliant Member - 6 years, 2 months ago

@A Former Brilliant Member – 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

tanveen dhingra - 6 years, 2 months ago

@Tanveen Dhingra – 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.

A Former Brilliant Member - 6 years, 2 months ago

@A Former Brilliant Member – Yes. I guess they are co-prime.

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

tanveen dhingra - 6 years, 2 months ago

@Tanveen Dhingra – Well could I send it tomorrow. I will recheck and confirm then send it.

A Former Brilliant Member - 6 years, 2 months ago

@A Former Brilliant Member – Ok! Thank you

tanveen dhingra - 6 years, 2 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.

@tanveen dhingra @Kalash Verma

Raghav Vaidyanathan - 6 years, 2 months ago

Thank you @Sandeep Bhardwaj sir Is this fine?

tanveen dhingra - 6 years, 2 months ago

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

A Former Brilliant Member - 6 years, 2 months ago

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

Raghav Vaidyanathan - 6 years, 2 months ago

@Raghav Vaidyanathan – 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 .

A Former Brilliant Member - 6 years, 2 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?

Siddhartha Srivastava - 6 years, 2 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.

Raghav Vaidyanathan - 6 years, 2 months ago

dhruv kaushik - 6 years, 2 months ago

@Sandeep Bhardwaj @Calvin Lin

tanveen dhingra - 6 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$