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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  .
a,b&caredistinctprimesandx,y,z{0,1,2,}.Whatisthenumberofdivisors,commontothenumbersaxbycz,axbzcy,aybxcz,aybzcx,azbxcy&azbycz.
Commented by mrW2 last updated on 17/Jan/18
According to Q27888, it is [1+min(x,y,z)]^3 .
AccordingtoQ27888,itis[1+min(x,y,z)]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
YessirIthoughtsoandcanbeextendedto[1+min(x1,x2,..,xn)]nCommondivisors:a10,1,..,min(x1,x2,..,xn)a20,1,..,min(x1,x2,..,xn)an0,1,..,min(x1,x2,..,xn)Numberofcommondivisors=[1+min(x1,x2,..,xn)]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
Thatsabsolutelycorrectsir.Ithinkyoumeantinlastline[1+min(x1,x2,..,xn)]n
Commented by Rasheed.Sindhi last updated on 17/Jan/18
Yes Sir commit mistake mistakenly.
YesSircommitmistakemistakenly.

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