Question and Answers Forum

All Questions      Topic List

Number Theory Questions

Previous in All Question      Next in All Question      

Previous in Number Theory      Next in Number Theory      

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 bymrW2 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 byRasheed.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 bymrW2 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 byRasheed.Sindhi last updated on 17/Jan/18

Yes Sir commit mistake mistakenly.

$$\mathrm{Yes}\:\mathrm{Sir}\:\mathrm{commit}\:\mathrm{mistake}\:\mathrm{mistakenly}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com