Let n = 219312. Let M denote the number of positive divisors of n2 which are less than n but would n

Question

Let n = 219312. Let M denote the number of positive divisors of n2 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)?

Accepted Answer

{"content":[{"content":[{"marks":[{"type":"bold"}],"text":"Step 1:","type":"text"},{"type":"text","text":" n2 = 238 · 324"}],"type":"paragraph"},{"content":[{"marks":[{"type":"bold"}],"text":"Step 2:","type":"text"},{"type":"text","text":" =>     Required number of divisors"}],"type":"paragraph"},{"content":[{"marks":[{"type":"bold"}],"text":"Step 3:","type":"text"},{"text":" ( 38 + 1)( 24 + 1) + 1","type":"text"}],"type":"paragraph"},{"content":[{"marks":[{"type":"bold"}],"text":"Step 4:","type":"text"},{"text":" =                            − 20 \times 13","type":"text"}],"type":"paragraph"}],"type":"doc"}

Guide / Hint

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