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Question Number 27922 by Rasheed.Sindhi last updated on 17/Jan/18
a,b & c are distinct primes and x,y,z∈{0,1,2,...}. What is the number of divisors, common to the numbers a^x b^y c^z , a^x b^z c^y ,a^y b^x c^z ,a^y b^z c^x ,a^z b^x c^y & a^z b^y c^z .
$$\mathrm{a},\mathrm{b}\:\&\:\mathrm{c}\:\mathrm{are}\:\boldsymbol{\mathrm{distinct}}\:\boldsymbol{\mathrm{primes}}\:\mathrm{and} \\ $$$$\mathrm{x},\mathrm{y},\mathrm{z}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},…\right\}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\boldsymbol{\mathrm{divisors}}, \\ $$$$\boldsymbol{\mathrm{common}}\:\mathrm{to}\:\mathrm{the}\:\boldsymbol{\mathrm{numbers}}\:\boldsymbol{\mathrm{a}}^{\boldsymbol{\mathrm{x}}} \boldsymbol{\mathrm{b}}^{\boldsymbol{\mathrm{y}}} \boldsymbol{\mathrm{c}}^{\boldsymbol{\mathrm{z}}} , \\ $$$$\boldsymbol{\mathrm{a}}^{\boldsymbol{\mathrm{x}}} \boldsymbol{\mathrm{b}}^{\boldsymbol{\mathrm{z}}} \boldsymbol{\mathrm{c}}^{\boldsymbol{\mathrm{y}}} ,\boldsymbol{\mathrm{a}}^{\boldsymbol{\mathrm{y}}} \boldsymbol{\mathrm{b}}^{\boldsymbol{\mathrm{x}}} \boldsymbol{\mathrm{c}}^{\boldsymbol{\mathrm{z}}} ,\boldsymbol{\mathrm{a}}^{\boldsymbol{\mathrm{y}}} \boldsymbol{\mathrm{b}}^{\boldsymbol{\mathrm{z}}} \boldsymbol{\mathrm{c}}^{\boldsymbol{\mathrm{x}}} ,\boldsymbol{\mathrm{a}}^{\boldsymbol{\mathrm{z}}} \boldsymbol{\mathrm{b}}^{\boldsymbol{\mathrm{x}}} \boldsymbol{\mathrm{c}}^{\boldsymbol{\mathrm{y}}} \:\&\:\boldsymbol{\mathrm{a}}^{\boldsymbol{\mathrm{z}}} \boldsymbol{\mathrm{b}}^{\boldsymbol{\mathrm{y}}} \boldsymbol{\mathrm{c}}^{\boldsymbol{\mathrm{z}}} \:. \\ $$
Commented by mrW2 last updated on 17/Jan/18
According to Q27888, it is [1+min(x,y,z)]^3 .
$${According}\:{to}\:{Q}\mathrm{27888},\:{it}\:{is}\:\left[\mathrm{1}+{min}\left({x},{y},{z}\right)\right]^{\mathrm{3}} . \\ $$
Commented by Rasheed.Sindhi last updated on 17/Jan/18
Yes sir I thought so and can be extended to [1+min(x_1 ,x_2 ,..,x_n )]^n Common divisors: a_1 ^(0,1,..,min(x_1 ,x_2 ,..,x_n )) a_2 ^(0,1,..,min(x_1 ,x_2 ,..,x_n )) ...a_n ^(0,1,..,min(x_1 ,x_2 ,..,x_n )) Number of common divisors= [1+min(x_1 ,x_2 ,..,x_n )]^n
$$\mathrm{Yes}\:\mathrm{sir}\:\mathrm{I}\:\mathrm{thought}\:\mathrm{so}\:\mathrm{and}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{extended}\:\mathrm{to}\:\left[\mathrm{1}+\mathrm{min}\left(\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} ,..,\mathrm{x}_{\mathrm{n}} \right)\right]^{\mathrm{n}} \\ $$$$\mathrm{Common}\:\mathrm{divisors}: \\ $$$$\:\mathrm{a}_{\mathrm{1}} ^{\mathrm{0},\mathrm{1},..,\mathrm{min}\left(\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} ,..,\mathrm{x}_{\mathrm{n}} \right)} \mathrm{a}_{\mathrm{2}} ^{\mathrm{0},\mathrm{1},..,\mathrm{min}\left(\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} ,..,\mathrm{x}_{\mathrm{n}} \right)} …\mathrm{a}_{\mathrm{n}} ^{\mathrm{0},\mathrm{1},..,\mathrm{min}\left(\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} ,..,\mathrm{x}_{\mathrm{n}} \right)} \\ $$$$\mathrm{Number}\:\mathrm{of}\:\mathrm{common}\:\mathrm{divisors}= \\ $$$$\left[\mathrm{1}+\mathrm{min}\left(\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} ,..,\mathrm{x}_{\mathrm{n}} \right)\right]^{\mathrm{n}} \\ $$
Commented by mrW2 last updated on 17/Jan/18
That′s absolutely correct sir. I think you meant in last line [1+min(x_1 ,x_2 ,..,x_n )]^n
$${That}'{s}\:{absolutely}\:{correct}\:{sir}. \\ $$$${I}\:{think}\:{you}\:{meant}\:{in}\:{last}\:{line} \\ $$$$\left[\mathrm{1}+\mathrm{min}\left(\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} ,..,\mathrm{x}_{\mathrm{n}} \right)\right]^{\mathrm{n}} \\ $$
Commented by Rasheed.Sindhi last updated on 17/Jan/18
Yes Sir commit mistake mistakenly.
$$\mathrm{Yes}\:\mathrm{Sir}\:\mathrm{commit}\:\mathrm{mistake}\:\mathrm{mistakenly}. \\ $$

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