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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.

Sandeep Bhardwaj - 2 years, 8 months ago

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Ok! Thank you very much sir!

Tanveen Dhingra - 2 years, 8 months ago

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Does HCF and LCM being fractions have any meaning?

Janardhanan Sivaramakrishnan - 2 years, 8 months ago

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

Kalash Verma - 2 years, 8 months ago

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

I'm getting contradictory conclusions.

Raghav Vaidyanathan - 2 years, 8 months ago

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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 - 2 years, 8 months ago

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Just ask her/him to re check it.

Kalash Verma - 2 years, 8 months ago

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@Kalash Verma 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 - 2 years, 8 months ago

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@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.

Kalash Verma - 2 years, 8 months ago

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@Kalash Verma Yes. I guess they are co-prime.

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

Tanveen Dhingra - 2 years, 8 months ago

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

Kalash Verma - 2 years, 8 months ago

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@Kalash Verma Ok! Thank you

Tanveen Dhingra - 2 years, 8 months ago

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Never think like this. Everyone makes mistakes

Kalash Verma - 2 years, 8 months ago

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@Raghav Vaidyanathan well I got answer in decimals.

Kalash Verma - 2 years, 8 months ago

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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 - 2 years, 8 months ago

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Thank you @Sandeep Bhardwaj sir Is this fine?

Tanveen Dhingra - 2 years, 8 months ago

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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 - 2 years, 8 months ago

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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 - 2 years, 8 months ago

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

Kalash Verma - 2 years, 8 months ago

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

Raghav Vaidyanathan - 2 years, 8 months ago

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@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 .

Kalash Verma - 2 years, 8 months ago

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hi

Dhruv Kaushik - 2 years, 8 months ago

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