Let n=2 ¹⁹ 3¹² . Let M denote the number of positive divisors of n^2 which are less than n but would not divide n . What is the number formed by taking the last two digits of M (in the same order)?

Question

Quantrex Academy · Accepted Answer

Correct answer: 28

n^2=2³⁸ 3²⁴ Required number of divisors aligned & = (38+1)(24+1)+1 2 -20 13 & =228 & 28 aligned

Let n=2 ¹⁹ 3¹² . Let M denote the number of positive divisors of n^2 which are less than n but would not divide n . What is the number formed by taking the last two digits of M (in the same order)?

Correct answer

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