#15 Divisor Counting

\(84=2^2\cdot3\cdot7\). Hence \(\tau(84)=(2+1)(1+1)(1+1)=12\).

\(144=2^4\cdot3^2\). Thus \(\tau(144)=(4+1)(2+1)=15\).

The exponent of \(2\) can be chosen in \(6\) ways, and the exponent of \(3\) in \(4\) ways. Total: \(6\cdot4=24\) divisors.

Every positive divisor of \(p^a\) has the form \(p^k\), where \(k=0,1,\ldots,a\). There are \(a+1\) choices for \(k\). Hence there are \(a+1\) divisors.

An odd divisor contains no factor \(2\), so it can only be \(1\) or \(3\). Answer: \(2\) odd divisors.

\(\tau(360)=(3+1)(2+1)(1+1)=4\cdot3\cdot2=24\).

\(1000=2^3\cdot5^3\). Therefore \(\tau(1000)=(3+1)(3+1)=16\).

\(720=2^4\cdot3^2\cdot5\). The odd part is \(3^2\cdot5\), so the number of odd divisors is \((2+1)(1+1)=6\).

\(225=15^2\) is a square. A square has an odd number of positive divisors, because the divisor \(\sqrt{225}=15\) is unpaired. Hence \(\tau(225)\) is odd.

Candidates: \(2^7=128\), \(2^3\cdot3=24\), \(2\cdot3\cdot5=30\). The smallest is \(24\). Answer: \(24\).

Candidates: \(2^9=512\) and \(2^4\cdot3=48\). The smallest number is \(48\).

If \(\tau(n)=3\), then \(n=p^2\) for a prime \(p\). We need \(p^2<100\), so \(p<10\). The primes are \(2,3,5,7\). The numbers are \(4,9,25,49\).

If \(d\mid n\), then \(\frac{n}{d}\mid n\). Divisors come in pairs except when \(d=\frac{n}{d}\), i.e. \(d^2=n\). Thus an unpaired divisor exists exactly when \(n\) is a square. Hence the number of divisors is odd exactly for squares.

The exponent patterns are \(14\), or \(4\) and \(2\). Candidates: \(2^{14}\) and \(2^4\cdot3^2=144\). The smallest number is \(144\).

Important candidates are \(2^7\cdot3^2=1152\), \(2^5\cdot3^3=864\), \(2^5\cdot3\cdot5=480\), \(2^3\cdot3^2\cdot5=360\), \(2^2\cdot3\cdot5\cdot7=420\). The smallest is \(360\), and the remaining patterns with larger exponents are larger. Answer: \(360\).

\(360=2^3\cdot3^2\cdot5\). Then \(360^2=2^6\cdot3^4\cdot5^2\), so \(\tau(360^2)=7\cdot5\cdot3=105\).

\(36=2^2\cdot3^2\), so \(\tau(36)=9\). The product of divisors is \(36^{9/2}=6\cdot36^4\). This is the exact value.

If \(\tau(n)=4\), the possible exponent patterns are \(3\), or \(1\) and \(1\). Therefore \(n=p^3\) for a prime \(p\), or \(n=pq\), where \(p,q\) are distinct primes.

Let \(n=2^a m\), where \(m\) is odd. The odd divisors of \(n\) are exactly the divisors of \(m\). We need \(\tau(m)=2\), so \(m\) is an odd prime. Therefore \(n=2^a p\), where \(a\ge0\), and \(p\) is an odd prime.

\(10^6=2^6\cdot5^6\), so \(\tau(10^6)=(6+1)(6+1)=49\).

Candidates are \(2^8\cdot3^3=6912\), \(2^5\cdot3^5=7776\), \(2^5\cdot3^2\cdot5=1440\), \(2^3\cdot3^2\cdot5^2=1800\), \(2^2\cdot3^2\cdot5\cdot7=1260\). The smallest is \(1260\). Other patterns with a larger single exponent are even larger. Answer: \(1260\).

For \(12\) divisors, the exponent patterns are as before. Odd candidates include \(3^{11}\), \(3^5\cdot5\), \(3^3\cdot5^2\), \(3^2\cdot5\cdot7=315\). The smallest is \(315\). Answer: \(315\).

Let \(n=2^a m\), with \(m\) odd. Then \(\tau(n)=(a+1)\tau(m)\), while \(\tau(2n)=(a+2)\tau(m)\). The condition gives \(a+2=2(a+1)\), hence \(a=0\). Thus \(n\) must be odd. All odd \(n\) work.

To minimise the number, it is enough to consider \(2^{a_1}3^{a_2}5^{a_3}\cdots\), where \(a_1\ge a_2\ge a_3\ge\cdots\). We need \((a_1+1)(a_2+1)\cdots=60\). The main candidates are \(2^9\cdot3^2\cdot5=23040\), \(2^5\cdot3^4\cdot5=12960\), \(2^4\cdot3^3\cdot5^2=10800\), \(2^4\cdot3^2\cdot5\cdot7=5040\). Patterns with one or two very large exponents are even larger. The smallest candidate is \(5040\), and \(\tau(5040)=\tau(2^4\cdot3^2\cdot5\cdot7)=5\cdot3\cdot2\cdot2=60\). Answer: \(5040\).