sudhir3127 wrote:Find the sum of all possible divisiors of 720.
I know a formula to solve the problem,. just exploring different ways of solving this problem.will share my thoughts after some discussion.
I don't think many people will find this problem 'easy', as the title of this thread suggests! Still, it's a useful problem to think about, so I'll put my response in spoiler tags, in the hope that people will attempt the problem before reading a solution.
[spoiler]
If you know the formula, of course the problem is not difficult. Most people won't know the formula, though. From the prime factorization of 720, one can count the number of divisors (thirty), and then could make a good estimate by adding the largest divisors of 720- you might add 720+360+240+180+144+120 = 1764, and then guess that the 'average size' of the remaining 24 divisors is roughly 30, which would give you an estimate of 1764 + 720 = 2484, which is not far off from the correct answer. Still, we can get the exact answer without a formula:
Prime factorize:
720 = (2^4)*(3^2)*(5)
-The divisors of 3^2 * 5 are simply: {1, 3, 5, 3^2, 3*5, (3^2)*5}, or, if you prefer, {1, 3, 5, 9, 15, 45}.
-Notice that we get all of the other divisors of 720 by multiplying the six numbers in the list {1, 3, 5, 9, 15, 45} by 2^1, by 2^2, by 2^3 and by 2^4 (so there are 30 divisors in total).
The sum of the divisors of 45, i.e. of this set: {1, 3, 5, 9, 15, 45}, is:
1 + 3 + 5 + 3^2 + 3*5 + (3^2)*5 = 78
Call this sum S. The sum of the divisors of 720 is thus
S + (2^1)*S + (2^2)*S + (2^3)*S + (2^4)*S
= S*(1 + 2 + 4 + 8 + 16)
= S*31
= 2418
[/spoiler]
I'd be surprised (though not completely shocked) to see this on a GMAT- I think it would be considered a bit too difficult for the test, and possibly too time-consuming. No test-taker would be expected to know the formula.
