<p>If the integers <i>m</i> and <i>n</i> are chosen at random from 1 to 100, then the probability that a number of the form <i>7<sup>n</sup> + 7<sup>m</sup></i> is divisible by 5 equals</p>
<p>(a) <span>\(\frac{1}{4}\)</span></p>
<p>(b) <span>\(\frac{1}{2}\)</span></p>
<p>(c) <span>\(\frac{1}{8}\)</span></p>
<p>(d) <span>\(\frac{1}{3}\)</span></p>
Step-by-Step Solution
Key Concept: Use the cyclic pattern of last digits of powers of 7 (period 4) and count the favorable cases where m and n satisfy the divisibility condition.
<p><strong>Solution:</strong> Let <i>I = 7<sup>n</sup> + 7<sup>m</sup></i>. We observe that <i>7<sup>1</sup>, 7<sup>2</sup>, 7<sup>3</sup></i> and <i>7<sup>4</sup></i> end in 7, 9, 3 and 1, respectively. Thus, <i>7<sup>i</sup></i> ends in 7, 9, 3 or 1 according as <i>i</i> is of the form <i>4k + 1, 4k + 2, 4k − 1</i> or <i>4k</i>, respectively.</p><p>If <i>S</i> is the sample space, then <i>n(S) = (100)<sup>2</sup></i>.</p><p><i>7<sup>m</sup> + 7<sup>n</sup></i> is divisible by 5, if</p><p>(i) <i>m</i> is of the form <i>4k + 1</i> and <i>n</i> is of the form <i>4k − 1</i>, or</p><p>(ii) <i>m</i> is of the form <i>4k + 2</i> and <i>n</i> is of the form <i>4k</i>, or</p><p>(iii) <i>m</i> is of the form <i>4k − 1</i> and <i>n</i> is of the form <i>4k + 1</i>, or</p><p>(iv) <i>m</i> is of the form <i>4k</i> and <i>n</i> is of the form <i>4k + 2</i>.</p><p>Thus, number of favourable ordered pairs <i>(m, n) = 4 × 25 × 25</i>.</p><p>Required probability <span>\(= \frac{4 \times 25 \times 25}{(100)^2} = \frac{1}{4}\)</span></p><p>∴ Answer is (b) <span>\(\frac{1}{2}\)</span></p>
Correct Answer: b