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$a=bq+r$, where \[0\text{ }\le \text{ }r\text{ }<\text{ }b ........................(1)\]

$a$ is the dividend. $b$ is the divisor. $q$ is the quotient and $r$ is the remainder.

Apply this lemma and represent the general form of all those numbers according to the condition given in the question for each of the three numbers individually. For the question, we will assume the positive integer to be $x$ and then we will apply Euclid’s Lemma for the first condition and get the equation as $x=5p+3$. Similarly, we will do it for other conditions too. Then, using the fact that the divisor will be the LCM of 5,7,8 we will compute the divisor and proceed further. After computing the divisor, we have to use the Lemma again to get a single common representation that will satisfy all the given conditions.

Let us know about Euclid’s Division Lemma first before solving the question

According to Euclid's Division Lemma, if we have two positive integers $a$ and $b$, then there would be whole numbers $q$ and $r$ that satisfy the equation:

$a=bq+r$, where \[0\text{ }\le \text{ }r\text{ }<\text{ }b........................(1)\]

$a$ is the dividend. $b$ is the divisor. $q$ is the quotient and $r$ is the remainder.

Extending this concept in the given question,

Let us suppose that when a positive integer $x$ is divided by 5,7 and 8, the remainder is 3,2,5 respectively.

Then using (1) we can represent $x$ as,

$\therefore x=5p+3$ then $x$ could be 3,8, 13,23,...93,...

Or, $x=7q+2$ then $x$ could be 2,9,16,...93,...

Or, $x=8r+5$ then $x$ could be 5,13,...93,...

(Here $p,q$ and $r$ could be any positive integer)

Now, according to the question we have to find a single number which can satisfy all the given conditions.

Let’s say that $x$ satisfies all these conditions then the divisor will be the least common multiple (LCM) of 5,7 and 8.

$\therefore Divisor=LCM(5,7,8)$

As two of the numbers are prime so, LCM will be the product of these three numbers.

$\Rightarrow Divisor=5\times 7\times 8=280$

And finally, the remainder for $x$ will be the first common multiple of all these three integers, which is 93.

Using (1), we can write $x$ as,

$\therefore x=280t+93$ where $t$ can be any positive integer

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