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June 30th, 2014, 10:05 PM   #1
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Find out the least number

Help me to solve math homework problem.

What is the least number, which when divided by 12, 15, 20 and 54 leaves remainder of 8 in each case?

Please tell me the procedure to solve this problem.

Last edited by skipjack; July 1st, 2014 at 01:54 AM.
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July 1st, 2014, 01:56 AM   #2
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Obviously, the least natural number with that property is 8.
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July 1st, 2014, 02:51 AM   #3
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If the trivial solution 8 is not allowed or not good then the answer is $\displaystyle LCM (12,15,20,54) + 8$ Where LCM is the lowest common multiple.

This could be calculated like this :
$\displaystyle \frac {12*15*20*54 } { GCD(12,15,20,54) } + 8$ Where GCD is the greatest common divisor. The greatest common divisor is obviously 1. And thus the final answer is

$\displaystyle (12*15*20*54) + 8 \; = \; 194408$

Last edited by gelatine1; July 1st, 2014 at 03:43 AM.
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July 1st, 2014, 01:30 PM   #4
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$\displaystyle \rm \ L \ C\ M\ (12,\ 15,\ 20,\ 54) = 540$
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July 1st, 2014, 03:47 PM   #5
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Whoops then I assume I extended the formula $\displaystyle LCM(a,b) = \frac {a*b}{GCD(a,b)}$ in a wrong way...

The result could have been calculated as follows:

$\displaystyle LCM(12,15,20,54)+8 = LCM \left( \frac {12*15}{GCD(12,15)} , \frac {20*54}{GCD(20,54)} \right) + 8$

$\displaystyle = LCM (60,540) + 8 = \frac {60*540}{GCD(60,540)} + 8 = 548$

Which should be correct this time. My apologies for the wrong answer before.
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July 1st, 2014, 04:38 PM   #6
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Quote:
Originally Posted by gelatine1 View Post
My apologies for the wrong answer before.
Because it's Canada Day, Father Denis forgives you;
but you must recite a full Holy Rosary
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July 2nd, 2014, 03:08 AM   #7
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Find the LCM of 12,15,20,54,

The LCM will 6,15,10,27

Thus we will find the ans.

The ans will be 548

For more details you can take help online tutor.
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July 2nd, 2014, 04:34 AM   #8
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Quote:
Originally Posted by aaliyah View Post
Find the LCM of 12,15,20,54,
The LCM will 6,15,10,27
Thus we will find the ans.
The ans will be 548
For more details you can take help online tutor.
You've lost me; anyway:
LCM(12,15,20,54) = 540
LCM(6,15,10,27) = 270

Which online tutor are you advertising for?
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July 2nd, 2014, 04:35 AM   #9
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$\displaystyle \Large{\color{blue}{\ \qquad 12=2^2\cdot3
\\\;\\
\ \qquad 15=3\cdot5
\\\;\\
\ \qquad 20=2^2\cdot5
\\\;\\
\ \qquad 54=2\cdot3^3
\\
\_\_\_\_\_\_\_\_\_\_\_\_\_
\\\;\\
LCM = 2^2\cdot3^3\cdot5 = 4\cdot27\cdot5 = 20\cdot27 = 540 .}}


$
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July 2nd, 2014, 04:39 AM   #10
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Quote:
Originally Posted by aurel5 View Post
$\displaystyle \Large{\color{blue}{\ \qquad 12=2^2\cdot3
\\\;\\
\ \qquad 15=3\cdot5
\\\;\\
\ \qquad 20=2^2\cdot5
\\\;\\
\ \qquad 54=2\cdot3^3
\\
\_\_\_\_\_\_\_\_\_\_\_\_\_
\\\;\\
LCM = 2^2\cdot3^3\cdot5 = 4\cdot27\cdot5 = 20\cdot27 = 540 .}}


$
In case the numbers are very big it will be faster to calculate the gcd (using the euclidean algorithm) instead of factorising the numbers.
Thanks from CRGreathouse
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