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# Find the greatest number of $6$ digits number exactly divisible by $24,15$ and$36$ .

Last updated date: 29th May 2024
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Hint-This question can be solved by simply finding the LCM of given numbers and then divide the greatest $6$ digit number by this LCM and then subtract the remainder from the greatest $6$ digit number.

Now we have to find the greatest number of $6$ digits number exactly divisible by $24,15$ and$36$
And we know that the greatest number of $6$ digits $= 999999$
Now, firstly we find the LCM of $24,15$ and $36$
$24 = 2 \times 2 \times 2 \times 3 = {2^3} \times 3 \\ 15 = 3 \times 5 \\ 36 = 2 \times 2 \times 3 \times 3 = {2^2} \times {3^2} \\$
The LCM must have all of these factors.
LCM$\left( {24,15,36} \right) = {2^3} \times {3^2} \times 5 = 360$
Now, we will divide the greatest number of $6$ digits by this LCM.
We will do it so as to find the remainder, then we subtract the remainder from the greatest number of $6$ digits, and then we get the greatest number of $6$ digits exactly divisible by $24,15$and $36$.
Now,
$999999 = 360 \times 2777 + 279$
Quotient$= 2777$
Remainder$= 279$
Now, the required number is
$= 999999 - 279 \\ = 999720 \\$
$\dfrac{{999720}}{{24}} = 41,655 \\ \dfrac{{999720}}{{15}} = 66,648 \\ \dfrac{{999720}}{{36}} = 27,770 \\$
Hence, $999720$ is the greatest number of $6$ digits exactly divisible by $24,15$ and$36$.

Note- Whenever we face such types of questions the key concept is that we should find the LCM of given numbers and then divide the greatest $n$ digit number by this LCM and then subtract the remainder from the greatest $n$ digit number.