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Let-N-be-a-positive-integer-with-prime-factorisation-N-p-1-m-1-p-2-m-2-p-3-m-3-p-n-1-m-n-1-p-n-m-n-where-n-m-i-Z-and-p-r-is-prime-How-many-proper-factors-does-N-have-Inv




Question Number 1942 by Yozzi last updated on 25/Oct/15
Let N be a positive integer with prime  factorisation     N=p_1 ^m_1  p_2 ^m_2  p_3 ^m_3  ×...×p_(n−1) ^m_(n−1)  p_n ^m_n    where  n,m_i ∈Z^+  and p_r  is prime.  How many proper factors does N have?  Investigate cases where n=1,n=2, n=3  and n=4. What is the smallest positive  integer with 12 proper factors?  What is the smallest positive integer  with at least 12 proper factors?  (A proper factor of a positive number N  is positive nteger M such that M≠1 and M≠N.)
$${Let}\:{N}\:{be}\:{a}\:{positive}\:{integer}\:{with}\:{prime} \\ $$$${factorisation}\: \\ $$$$\:\:{N}={p}_{\mathrm{1}} ^{{m}_{\mathrm{1}} } {p}_{\mathrm{2}} ^{{m}_{\mathrm{2}} } {p}_{\mathrm{3}} ^{{m}_{\mathrm{3}} } ×…×{p}_{{n}−\mathrm{1}} ^{{m}_{{n}−\mathrm{1}} } {p}_{{n}} ^{{m}_{{n}} } \\ $$$${where}\:\:{n},{m}_{{i}} \in\mathbb{Z}^{+} \:{and}\:{p}_{{r}} \:{is}\:{prime}. \\ $$$${How}\:{many}\:{proper}\:{factors}\:{does}\:{N}\:{have}? \\ $$$${Investigate}\:{cases}\:{where}\:{n}=\mathrm{1},{n}=\mathrm{2},\:{n}=\mathrm{3} \\ $$$${and}\:{n}=\mathrm{4}.\:{What}\:{is}\:{the}\:{smallest}\:{positive} \\ $$$${integer}\:{with}\:\mathrm{12}\:{proper}\:{factors}? \\ $$$${What}\:{is}\:{the}\:{smallest}\:{positive}\:{integer} \\ $$$${with}\:{at}\:{least}\:\mathrm{12}\:{proper}\:{factors}? \\ $$$$\left({A}\:{proper}\:{factor}\:{of}\:{a}\:{positive}\:{number}\:{N}\right. \\ $$$$\left.{is}\:{positive}\:{nteger}\:{M}\:{such}\:{that}\:{M}\neq\mathrm{1}\:{and}\:{M}\neq{N}.\right) \\ $$
Commented by prakash jain last updated on 25/Oct/15
Total factors,K=Π_(i=1) ^n (m_i +1)  proper factors=K−2
$$\mathrm{Total}\:\mathrm{factors},\mathrm{K}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}\left({m}_{{i}} +\mathrm{1}\right) \\ $$$$\mathrm{proper}\:\mathrm{factors}=\mathrm{K}−\mathrm{2} \\ $$
Answered by prakash jain last updated on 25/Oct/15
12 proper factor ⇒14 total factors  5×3=15>14  m_1 =4, m_2 =2  Smallest integer=2^4 ×3^2 =16×9=144
$$\mathrm{12}\:\mathrm{proper}\:\mathrm{factor}\:\Rightarrow\mathrm{14}\:\mathrm{total}\:\mathrm{factors} \\ $$$$\mathrm{5}×\mathrm{3}=\mathrm{15}>\mathrm{14} \\ $$$${m}_{\mathrm{1}} =\mathrm{4},\:{m}_{\mathrm{2}} =\mathrm{2} \\ $$$$\mathrm{Smallest}\:\mathrm{integer}=\mathrm{2}^{\mathrm{4}} ×\mathrm{3}^{\mathrm{2}} =\mathrm{16}×\mathrm{9}=\mathrm{144} \\ $$

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