One linked question #2

Number Theory Level pending

Let's consider P to be the product of the all positive divisors integers of 1000, then what is the value of

$\sqrt[24]{P}$ + $\sqrt[12]{P}$ + $\sqrt[8]{P}$ + $\sqrt[6]{P}$ + $P^0$ ?

The answer is 11111.

1 solution

Guillermo Templado
Oct 18, 2015

1000 = $(2^3)\times (5^3)$ . Then, the number of positive divisors integers of 1000 is, therefore, $4 \times 4$ = 16
$\Rightarrow$ P = $\sqrt (1000^{16})$ = $\sqrt (10^{48})$ = $10^{24}$ $\Rightarrow$ $\sqrt[24]{P}$ + $\sqrt[12]{P}$ + $\sqrt[8]{P}$ + $\sqrt[6]{P}$ + $P^0$ = 10 +100 + 1000 + 10000 + 1 = 11111

Sorry, but shouldn't it be 1000^8 instead (for your 3rd step), because 16 factors mean 8 ways to make 1000? Each single factor isn't equal to one thousand.

Raghav Arora - 5 years, 7 months ago

The positive divisors integers of 1000 from the lowest term to the biggest term are 1,2,4,5,8,10,20,25,40,50,100,125,200,250,500 and 1000, is it ok?Now , we are going to multiply them. P=(1 x 1000) x (2 x 500) x (4 x 250) x (5 x 200) x (8 x125) x (10 x 100) x (20 x 50) x (25 x 40) = 1000 ^ 8 =10 ^24. In the 3rd step I wrote the sqrt of 1000 ^16... Does this explain your doubt?

Guillermo Templado - 5 years, 7 months ago

Ohh, sorry, made a silly mistake. My bad!

Raghav Arora - 5 years, 7 months ago

@Raghav Arora – Don't worry. It happens to everybody, cheers

Guillermo Templado - 5 years, 7 months ago

One linked question #2

The answer is 11111.

1 solution

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