<p>m, n are positive integers such that g.c.d. (m, n) = 1 and mn = 25!. The number of rational numbers <span class="math-tex">\(\frac{m}{n}\)</span> < 1 is</p>
<p style="display:inline">64</p>
<p style="display:inline">256</p>
<p style="display:inline">108</p>
<p style="display:inline">512</p>
Step-by-Step Solution
Key Concept: Since m and n are coprime, each distinct prime power in the prime factorization of 25! must be assigned as an inseparable block to either m or n, and the condition m < n halves the total 2^k possibilities.
<html><body><p>25! = 2<sup>22</sup> <span class="math-tex">\(\times\)</span> 3<sup>10</sup> <span class="math-tex">\(\times\)</span> 5<sup>6</sup> <span class="math-tex">\(\times\)</span> 7<sup>3</sup> <span class="math-tex">\(\times\)</span> 11<sup>2</sup> <span class="math-tex">\(\times\)</span> 13 <span class="math-tex">\(\times\)</span> 17 <span class="math-tex">\(\times\)</span> 19 <span class="math-tex">\(\times\)</span> 23<br/>
<img alt="" data-imgur-src="jH0guPk.png" src="https://media-mycbseguide.s3.amazonaws.com/images/imgur/1623761569-sgf7hb.jpg" style="width: 150px; height: 91px;"/><br/>
Since 25! = mn and g.c.d. of (m, n) = 1<br/>
Therefore, in box of m (or n), powers of 2, 3, ..., 23 are to be filled such that no power of 2, 3, ..., 23 is left to go to the box of n (or m).<br/>
So, basically these 9 bases 2, 3, 5, ..., 19, 23 have to be placed in the boxes of m and n (along with their powers.)<br/>
<span class="math-tex">\(\Rightarrow\)</span> Every base has two options of boxes.<br/>
<span class="math-tex">\(\Rightarrow\)</span> Number of representations of mn = 25! is 2<sup>9</sup><br/>
<span class="math-tex">\(\Rightarrow\)</span> Required number such that <span class="math-tex">\(\frac{m}{n}\)</span> < 1<br/>
= <span class="math-tex">\(\frac{2^{9}}{2}\)</span> = 2<sup>8</sup> = 256</p></body></html>
Correct Answer: B