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# Halp Plez

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The product of the proper positive integer factors of \$n\$ can be written as \$n^{(ax+b)/c}\$, where \$x\$ is the number of positive divisors \$n\$ has, \$c\$ is a positive integer, and the greatest common factor of the three integers \$a\$, \$b\$, and \$c\$ is \$1\$. What is \$a+b+c\$?

Dec 25, 2020

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By the product formula, the product of the proper divisors of n is n^((2n + 1)/2), so a + b + c = 2 + 1 + 2 = 5.

Dec 25, 2020
#2
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a=1, b = -2, c=2
A few examples:
20 = (1, 2, 4, 5, 10, 20) > 6 (divisors)
20^[a6 + (-2)] / 2 =2 =20^2 =400=[10*5*4*2*1]

64 = (1, 2, 4, 8, 16, 32, 64) > 7 (divisors)
64^[a7 + (-2)] / 2 =2.5=64^2.5=32,768=[32*16*8*4*2*1]

24 = (1, 2, 3, 4, 6, 8, 12, 24) > 8 (divisors)
24^[a8 + (-2)] / 2 =3 =24^3=[12*8*6*4*3*2*1]

36 = (1, 2, 3, 4, 6, 9, 12, 18, 36) > 9 (divisors)
36=^[a9 + (-2)] / 2 =3.5=36^3.5=[18*12*9*6*4*3*2*1]

48 = (1, 2, 3, 4, 6, 8, 12, 16, 24, 48) > 10 (divisors)
48^[a10 + (-2)] / 2 =4 =48^4=[24*16*12*8*6*4*3*2*1]

60 = (1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60) > 12 (divisors)
60^[a12 + (-2)] / 2 =5=60^5=[30*20*15*12*10*6*5*4*3*2*1]

So: a + b + c =1 + (-2) + 2 = 1

Dec 26, 2020