Basic Mathematics & Logarithm
Logarithmic Inequalities and Bounds
Grade Class 11
Question:
<p>Let \(\log_5 N = I_1 + f_1\) and \(\log_3 N = I_2 + f_2\), where \(I_1, I_2\) are integers and \(f_1, f_2 \in [0,1)\). If \(I_1 = 3\) and \(I_2 = 4\), then the number of possible integral values of \(N\) is less than or equal to</p>
\(119\)
\(116\)
\(117\)
\(118\)
Step-by-Step Solution
Key Concept: Again turn the integer-part data into interval bounds and count the admissible integers carefully.
Notice that $I_1=3$ gives $125 \le N < 625$. A clever move here is to combine this with $I_2=4$, which gives $81 \le N < 243$. Their intersection is $125 \le N < 243$. So the number of integral values is $243-125=118$. Therefore the count is less than or equal to $119$ and $118$, but not to $117$ or $116$.
Correct Answer: A, D