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Not really. sqrt is obtained by integer casting; what you suggested might end up not accounting for some of the divisors.

Say n = 6.
Then sqrt = (int) Math.sqrt(6) = 2
So your loop would run only once with this statement:
sum += 1+6

After that, i < sqrt so we exit the loop.
Since 2*2 != 6, you don't add 2 to the sum.

So your suggested modification would result in sum = 7 for n = 6, which is the wrong sum.

You have probably caught this but for anyone who have not, always pay close attention to the difference between (< vs <= ) when you do integer casting!

## Day 19: Interfaces

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it would be slightly more efficient, though, to do

`for(... i < sqrt ...)`

and`if (sqrt*sqrt==n) sum += sqrt;`

in the end.Not really. sqrt is obtained by integer casting; what you suggested might end up not accounting for some of the divisors.

Say n = 6. Then sqrt = (int) Math.sqrt(6) = 2 So your loop would run only once with this statement: sum += 1+6

After that, i < sqrt so we exit the loop. Since 2*2 != 6, you don't add 2 to the sum.

So your suggested modification would result in sum = 7 for n = 6, which is the wrong sum.

You have probably caught this but for anyone who have not, always pay close attention to the difference between (< vs <= ) when you do integer casting!