BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
Featured Course
Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.
We're adding nine odd numbers here- the sum will be odd. The sum will also be a multiple of 3, since each term in the sum is divisible by 3. So the sum is an odd multiple of 3, and the remainder will be 3 when you divide by 6.nikhilagrawal wrote:What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6?
I have to disagree that the remainder is 3 when you divide an odd multiple of 3 by 6.Ian Stewart wrote:So the sum is an odd multiple of 3, and the remainder will be 3 when you divide by 6.nikhilagrawal wrote:What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6?
OK, the remainder should be 1, not 5. When an odd number is divided by 2, the remainder's always 1, my mistake. But the answer is still None of these, since 1 isn't an option either.pepeprepa wrote:Just one remark about Saule.
I think I see where your error is.
You say "Eliminate 3 from both numerator and denominator, and you'll be left with 2 in the denominator."
I think you cannot do that when you want to find a remainder.
Counter-exemple:
15 divided by 6 gives you a remainder of 3
15/6=(3*5)/(3*2)=5/2
5 divide by 2 gives you a remainder of 1
Am I right about this fact?
New here Create free account
