Question Number 76580 by Rio Michael last updated on 28/Dec/19
$$\mathrm{how}\:\mathrm{many}\:\mathrm{divisors}\:\mathrm{does}\:\mathrm{180045}\:\mathrm{has}? \\ $$
Commented by mr W last updated on 28/Dec/19
$$\frac{\mathrm{180045}}{\mathrm{5}}=\mathrm{36009} \\ $$$$\frac{\mathrm{36009}}{\mathrm{3}}=\mathrm{12003} \\ $$$$\frac{\mathrm{12003}}{\mathrm{3}}=\mathrm{4001}={prime} \\ $$$$\Rightarrow\mathrm{180045}=\mathrm{3}^{\mathrm{2}} ×\mathrm{5}×\mathrm{4001} \\ $$$$ \\ $$$${number}\:{of}\:{divisors}\:{from}\:\mathrm{180045}: \\ $$$$\left(\mathrm{2}+\mathrm{1}\right)\left(\mathrm{1}+\mathrm{1}\right)\left(\mathrm{1}+\mathrm{1}\right)=\mathrm{3}×\mathrm{2}×\mathrm{2}=\mathrm{12} \\ $$
Commented by Rio Michael last updated on 28/Dec/19
$$\mathrm{sir}\:\mathrm{why}\:\mathrm{are}\:\mathrm{you}\:\:\mathrm{doing}\:\mathrm{an}\:\mathrm{addition}\:\mathrm{plus}\:\mathrm{1}? \\ $$
Commented by Kunal12588 last updated on 28/Dec/19
$${if}\:{N}={p}_{\mathrm{1}} ^{{a}_{\mathrm{1}} } .{p}_{\mathrm{2}} ^{{a}_{\mathrm{2}} } .{p}_{\mathrm{3}} ^{{a}_{\mathrm{3}} } \ldots{p}_{{n}} ^{{a}_{{n}} } ;\:\:{p}_{\mathrm{1},\mathrm{2},\mathrm{3}…} {are}\:{prime} \\ $$$${then},\:{d}\left({N}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}\left({a}_{{i}} +\mathrm{1}\right) \\ $$$${d}\left({N}\right)\:{is}\:{a}\:{function}\:{which}\:{tells}\:{no}.\:{of}\:{divisers} \\ $$
Commented by JDamian last updated on 28/Dec/19
$${Kunal}:\:{you}\:{should}\:{use}\:\boldsymbol{\mathrm{N}}\:{for}\:{the}\:{number} \\ $$$${and}\:\boldsymbol{\mathrm{n}}\:{for}\:{the}\:{number}\:{of}\:{different}\:{primes}. \\ $$$${Otherwise}\:{it}\:{is}\:{confusing}. \\ $$
Commented by Kunal12588 last updated on 28/Dec/19
$${ohh},\:{thank}\:{you}\:{very}\:{much}\:{sir} \\ $$