Find the smallest natural number $n$ with exactly $\\tau(n)=\\var{t[j]}$ divisors.
\n \n \n \nSmallest=[[0]]
\n \n \n \nWhat is the second smallest?
\n \n \n \nSecond smallest=[[1]]
\n \n \n", "steps": [{"type": "information", "prompt": "\n \n \nRecall that if $n=p_1^{\\beta_1}\\times p_2^{\\beta_2}\\times \\cdots \\times p_m^{\\beta_m}$ is the prime factorization of $n$, then the number of divisors is given by:
\n \n \n \n$\\tau(n)=(\\beta_1+1)\\times (\\beta_2+1) \\times\\cdots \\times (\\beta_m+1)$
\n \n \n \nHence if $\\tau(n)= \\var{ta}$, we factorize $\\var{ta}$ to look for possible values of the $\\beta_j$
\n \n \n \nIn this example there are two possible factorizations.
\n \n \n", "showCorrectAnswer": true, "scripts": {}, "marks": 0}], "showCorrectAnswer": true, "marks": 0}], "statement": "", "tags": ["checked2015", "MAS3214", "number of divisors", "number theory", "tau(n)"], "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "", "licence": "Creative Commons Attribution 4.0 International", "description": "Given $m \\in \\mathbb{N}$, find the smallest natural number $n \\in \\mathbb{N}$ with $\\tau(n)=m$ divisors.
"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "\n \n \nRecall that if $n=p_1^{\\beta_1}\\times p_2^{\\beta_2}\\times \\cdots \\times p_m^{\\beta_m}$ is the prime factorization of $n$, then the number of divisors is given by:
\n \n \n \n$\\tau(n)=(\\beta_1+1)\\times (\\beta_2+1) \\times\\cdots \\times (\\beta_m+1)$
\n \n \n \nHence if $\\tau(n)= \\var{ta}$, we factorize $\\var{ta}$ to look for possible values of the $\\beta_j$
\n \n \n \nNow there are two possible factorizations i.e.
\n \n \n \nFactorization 1. $\\var{ta}= \\var{ta}$
\n \n \n \nFactorization 2. $\\var{ta}= \\var{fact[0]+1}\\times\\var{fact[1]+1}$
\n \n \n \nWe have only one prime factor and $n = p_1^{\\beta_1} $ for some prime $p$.
\n \n \n \nIn this case we have $\\tau(n)= 1+\\beta_1 =\\var{ta} \\Rightarrow \\beta_1 = \\var{ta-1}$
\n \n \n \nHence possible values for $n$ are $2^{\\var{ta-1}}$ and $3^{\\var{ta-1}}$.
\n \n \n \nNo need to look at any more powers of primes as we want the smallest 2 and all other prime powers are greater than these two.
\n \n \n \nBut note that these may not be the smallest values of $n$ as we have now to look at case 2.
\n \n \n \nIn this case we have two distinct prime factors as
\n \n \n \n\\[n=p_1^{\\beta_1}p_2^{\\beta_2} \\Rightarrow \\tau(n)=(1+\\beta_1)(1+\\beta_2)= \\var{fact[0]+1}\\times\\var{fact[1]+1} \\Rightarrow \\beta_1=\\var{fact[0]},\\;\\;\\beta_2=\\var{fact[1]}\\]
\n \n \n \nSo now we look at the possibilities.
\n \n \n \nThere is no point in looking at any primes greater than 5 as combinations of powers of the primes 2,3 and 5 will give the two smallest values of $n$.
\n \n \n \nAlso the pair of primes 3, 5 cannot give one of the smallest two values as $3^{\\beta_1}5^{\\beta_2}$ is greater than both $2^{\\beta_1}5^{\\beta_2}$ and $2^{\\beta_1}3^{\\beta_2}$
\n \n \n \nLook at the following table:
\n \n \n \n| Powers of Prime Numbers | ||
|---|---|---|
| Prime Number Pairs | $(\\var{fact[0]},\\var{fact[1]})$ | $(\\var{fact[1]},\\var{fact[0]})$ |
| $(2,3)$ | $2^{\\var{fact[0]}}\\times 3^{\\var{fact[1]}}=\\var{2^fact[0]* 3^fact[1]}$ | $2^{\\var{fact[1]}}\\times 3^{\\var{fact[0]}}=\\var{2^fact[1]* 3^fact[0]}$ |
| $(2,5)$ | $2^{\\var{fact[0]}}\\times 5^{\\var{fact[1]}}=\\var{2^fact[0]* 5^fact[1]}$ | $2^{\\var{fact[1]}}\\times 5^{\\var{fact[0]}}=\\var{2^fact[1]* 5^fact[0]}$ |
We see from this table for Factorization 2 and from Factorization 1 that the two smallest values for $n$ are $\\var{poss[0]}$ and $\\var{poss[1]}$.
\n \n \n", "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}]}], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}