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Olympiad workbook
Exam
Olympiad workbook
Subject
IOQM
Chapter
Permutation and Combination

Olympiad workbook · IOQM · Permutation and Combination

Let n=2 ^ 19 3^ 12 . Let M denote the number of positive divisors...

Let n=219312n=2{ }^{19} 3^{12}. Let MM denote the number of positive divisors of n2n^2 which are less than nn but would not divide nn. What is the number formed by taking the last two digits of MM (in the same order)?

Correct answer

28

Step-by-step solution

n2=238⋅324n^2=2^{38} \cdot 3^{24} ∴\therefore Required number of divisors =(38+1)(24+1)+12−20×13=228∴28\begin{aligned} & =\frac{(38+1)(24+1)+1}{2}-20 \times 13 \\ & =228 \\ \therefore & 28 \end{aligned}
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