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nick1816
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a, b and c are integers such that 0<a<b<c<15. Also, a+1/b=1/GCD(a,b)+ 24 Oct 2019, 18:21
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34% (02:16) correct 66% (02:14) wrong based on 50 sessions

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a, b and c are integers such that \(0<a<b<c<15\).

Also, \(a+\frac{1}{b}= \frac{1}{GCD(a,b)}+\frac{1}{GCD(b,c)}+\frac{1}{GCD(a,c)}\)

Find a+b+c ?

{ GCD- greatest common divisor}

A. 10
B. 14
C. 18
d. 22
E. 26
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B
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a, b and c are integers such that 0<a<b<c<15. Also, a+1/b=1/GCD(a,b)+ 24 Oct 2019, 22:02
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nick1816
a, b and c are integers such that \(0<a<b<c<15\).

Also, \(a+\frac{1}{b}= \frac{1}{GCD(a,b)}+\frac{1}{GCD(b,c)}+\frac{1}{GCD(a,c)}\)

Find a+b+c ?

{ GCD- greatest common divisor}

A. 10
B. 14
C. 18
d. 22
E. 26
Show more

You have to analyze the two sides and find the values for each variable..

(I) Let us work for a..
If a=1, then LHS =\(a+\frac{1}{b}=1+\frac{1}{b}<2.......RHS= \frac{1}{GCD(a,b)}+\frac{1}{GCD(b,c)}+\frac{1}{GCD(a,c)}=1+\frac{1}{GCD(b,c)}+1>2\)..Not possible
a=1, means GCD(a,any number) will be 1

Now each fraction on RHS can be maximum 1, so RHS will be <3, so a can be 2 if not 1, as the LHS will remain between 2 and 3
a=2..

(II) Let us check for b and c
Now \(a+\frac{1}{b}= \frac{1}{GCD(a,b)}+\frac{1}{GCD(b,c)}+\frac{1}{GCD(a,c)}\).....\(2+\frac{1}{b}= \frac{1}{1}+\frac{1}{GCD(b,c)}+\frac{1}{1}=2+\frac{1}{GCD(b,c)}\)....So, \(\frac{1}{b}=\frac{1}{GCD(b,c)}\) OR b=GCD(b,c), which is possible only when b is a factor of c.
Now, b and c have to be odd...To make GCD(a,b) and GCD(a,c) as 1, a and b AND a and c should be co-prime..
If b=3, c can be 3*3 and next 3*5=15, which does not fit in
If b=5, c can not be <15..
Hence b=3, c=9

\(a+b+c=2+3+9=14\)

B
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a, b and c are integers such that 0<a<b<c<15. Also, a+1/b=1/GCD(a,b)+ 01 Nov 2019, 11:12
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chetan2u
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a, b and c are integers such that \(0<a<b<c<15\).

Also, \(a+\frac{1}{b}= \frac{1}{GCD(a,b)}+\frac{1}{GCD(b,c)}+\frac{1}{GCD(a,c)}\)

Find a+b+c ?

{ GCD- greatest common divisor}

A. 10
B. 14
C. 18
d. 22
E. 26

You have to analyze the two sides and find the values for each variable..

(I) Let us work for a..
If a=1, then LHS =\(a+\frac{1}{b}=1+\frac{1}{b}<2.......RHS= \frac{1}{GCD(a,b)}+\frac{1}{GCD(b,c)}+\frac{1}{GCD(a,c)}=1+\frac{1}{GCD(b,c)}+1>2\)..Not possible
a=1, means GCD(a,any number) will be 1

Now each fraction on RHS can be maximum 1, so RHS will be <3, so a can be 2 if not 1, as the LHS will remain between 2 and 3
a=2..

(II) Let us check for b and c
Now \(a+\frac{1}{b}= \frac{1}{GCD(a,b)}+\frac{1}{GCD(b,c)}+\frac{1}{GCD(a,c)}\).....\(2+\frac{1}{b}= \frac{1}{1}+\frac{1}{GCD(b,c)}+\frac{1}{1}=2+\frac{1}{GCD(b,c)}\)....So, \(\frac{1}{b}=\frac{1}{GCD(b,c)}\) OR b=GCD(b,c), which is possible only when b is a factor of c.
Now, b and c have to be odd...To make GCD(a,b) and GCD(a,c) as 1, a and b AND a and c should be co-prime..
If b=3, c can be 3*3 and next 3*5=15, which does not fit in
If b=5, c can not be <15..
Hence b=3, c=9

\(a+b+c=2+3+9=14\)

B
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What is LHS and RHS?
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a, b and c are integers such that 0<a<b<c<15. Also, a+1/b=1/GCD(a,b)+ 02 Nov 2019, 05:48
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AUTricco27
chetan2u
nick1816
a, b and c are integers such that \(0<a<b<c<15\).

Also, \(a+\frac{1}{b}= \frac{1}{GCD(a,b)}+\frac{1}{GCD(b,c)}+\frac{1}{GCD(a,c)}\)

Find a+b+c ?

{ GCD- greatest common divisor}

A. 10
B. 14
C. 18
d. 22
E. 26

You have to analyze the two sides and find the values for each variable..

(I) Let us work for a..
If a=1, then LHS =\(a+\frac{1}{b}=1+\frac{1}{b}<2.......RHS= \frac{1}{GCD(a,b)}+\frac{1}{GCD(b,c)}+\frac{1}{GCD(a,c)}=1+\frac{1}{GCD(b,c)}+1>2\)..Not possible
a=1, means GCD(a,any number) will be 1

Now each fraction on RHS can be maximum 1, so RHS will be <3, so a can be 2 if not 1, as the LHS will remain between 2 and 3
a=2..

(II) Let us check for b and c
Now \(a+\frac{1}{b}= \frac{1}{GCD(a,b)}+\frac{1}{GCD(b,c)}+\frac{1}{GCD(a,c)}\).....\(2+\frac{1}{b}= \frac{1}{1}+\frac{1}{GCD(b,c)}+\frac{1}{1}=2+\frac{1}{GCD(b,c)}\)....So, \(\frac{1}{b}=\frac{1}{GCD(b,c)}\) OR b=GCD(b,c), which is possible only when b is a factor of c.
Now, b and c have to be odd...To make GCD(a,b) and GCD(a,c) as 1, a and b AND a and c should be co-prime..
If b=3, c can be 3*3 and next 3*5=15, which does not fit in
If b=5, c can not be <15..
Hence b=3, c=9

\(a+b+c=2+3+9=14\)

B

What is LHS and RHS?
Show more

LHS = Left Hand Side
RHS = right Hand Side

in the equation a-7 = b+1,
a-7 is LHS and
b+1 is RHS.

:)
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a, b and c are integers such that 0<a<b<c<15. Also, a+1/b=1/GCD(a,b)+ 02 Nov 2019, 05:50
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Can anyone give a more detailed solution for this? I fail to understand a lot of concepts used by Chetan2u.

egmat #egmat
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a, b and c are integers such that 0<a<b<c<15. Also, a+1/b=1/GCD(a,b)+ 17 Mar 2024, 12:52
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