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# Help!

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The sum of the positive divisors of a positive integer of the form 2^i*3^j is equal to 600. What is i+j?

MIRB16  May 5, 2018
#1
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2^3 x 3^3 =216

Divisors of 216 =1 + 2 + 3 + 4 + 6 + 8 + 9 + 12 + 18 + 24 + 27 + 36 + 54 + 72 +108 + 216 = 600

i + j = 3 + 3 = 6

Guest May 5, 2018
#2
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There is a formula for the sum of the divisors, but you have to know the number whose divisors are to be summed up. For your particular problem, it goes like this:

[(2^(i + 1) - 1) / (2 - 1)] * [(3^(j + 1) - 1)] / (3 - 1) = 600

I don't know how to formally solve for 2 variables in one equation without resorting to trial and error, which quickly leads to i = 3 and j = 3, which give: 2^3 x 3 ^3 = 216. Then you would find the divisors of 216, as done above, which sum up to 600.

Guest May 5, 2018
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This is impressive, Mr. BB. There’s no blarney, bullshit, or slop in your post. Looks like GETSMART is doing his job.

GingerAle  May 6, 2018
#4
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Solving two variables in a single equation is usually a trial and error process. However, in this case, you can easily narrow the process to a few guesses using logic.

The (i) and (j) values are integers; there are no fractional or decimal components in them.

Because the sum of integers is very low, it reasonable to assume the (i,,j) exponents are low—less than 10.

To solve, start by assuming a central value (between 1 and 10) for (j), and then use logarithms to isolate for (i).

Because (2) is one of the factors, its resolution will be its multiples 2,4,8,16,32,64,128,256, ... ect. Seeing one of these indicates you’ve found the solution set.

$$(( 2^i-1)*\dfrac{(3^j - 1)}{(2)} = 600 \leftarrow \tiny \text{ The (2) and (1) are common to all equations.}\\ ( 2^i) = \dfrac{1200}{(3^j - 1)} +1\leftarrow \tiny \text{ Leave it in this form to streamline the solution search.}\\ \text {Example: assume (j) = 6 and solve for (i)}\\ ( 2^i) = \dfrac{1200}{(3^6 - 1)} +1\\ ( 2^i) = \dfrac{150}{91} +1\\ Log_2( 2^i) = Log_2 \left( \dfrac {150}{91} +1\right) \small \text{ Note the fraction; This is not the solution.}\\ \text {Example: assume (j) = 5 and solve for (i)}\\ ( 2^i) = \dfrac{1200}{(5^5 - 1)} +1\\ ( 2^i) = \dfrac{600}{121} +1\\ Log_2( 2^i) = Log_2 \left( \dfrac {600}{121} +1\right) \small \text{ Again, note the fraction; This is not the solution.}\\ \text {Example: assume (j) = 4 and solve for (i)}\\ ( 2^i) = \dfrac{1200}{(5^4 - 1)} +1\\ ( 2^i) = 15 +1\\ Log_2( 2^i) = Log_2 \left( 16\right) \small \text{ Note the integer 16, corresponding to i=4; This is the solution.}\\$$

The solution is j=4 and i=4.  Subtract (1) from each and use these as the exponents for the primes 2, and 3 to find the number (N) for which 600 is the sum of its divisors.

There are similar techniques for (N) that have three or more unique prime factors.

GA

Edit: Corrected to indicate base (2) log.

GingerAle  May 6, 2018
edited by GingerAle  May 7, 2018