I know $60=(a+1)(b+1)$ Which means $ab+a+b+1=60$ Now $n^2$ is $P^2a \times Q^2b$ So number of factors $=(2a+1)(2b+1)$ But I am not getting the answer. Any hint or correction? Please help
If a +ve integer n= P^a × Q^b (where P,Q are distinct primes and a,b +ve integers)has 60 +ve factors, the largest number of factors of n^2?
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number-theory
1 Answers
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Either the problem is missing information, or there are multiple possible answers.
WLOG, I'll assume $n = p^aq^b$ with $a \ge b$.
If $n = p^{29}q^1$ (which has $30 \cdot 2 = 60$ factors), then $n^2 = p^{58}q^2$ has $59 \cdot 3 = 177$ factors.
If $n = p^{19}q^2$ (which has $20 \cdot 3 = 60$ factors), then $n^2 = p^{38}q^4$ has $39 \cdot 5 = 195$ factors.
If $n = p^{14}q^3$ (which has $15 \cdot 4 = 60$ factors), then $n^2 = p^{28}q^6$ has $29 \cdot 7 = 203$ factors.
If $n = p^{11}q^4$ (which has $12 \cdot 5 = 60$ factors), then $n^2 = p^{22}q^8$ has $23 \cdot 9 = 207$ factors.
If $n = p^9q^5$ (which has $10 \cdot 6 = 60$ factors), then $n^2 = p^{18}q^{10}$ has $19 \cdot 11 = 209$ factors.