A Nice Algebra Problem | Exponent Root Simplification👇👇👇 | 99% did NOT figure out!

Math Olympiad | A Nice Algebra Problem
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[@danmimis4576]

1000^3 = 10^9 which is slightly bigger than the given Number N that has 9 digits. If we add the N's digits it's a multiple of 3 (actually of 27) so the Result R is a number 990 < 3k < 1000. It cannot be 996 cuz N is odd, it cannot be 999 cuz 9^3 ends in 9, so IT HAS TO be 993.

[@tubetigeerr]

the soluation here is making the problem even complicated, harder!

[@tubetigeerr]

XXXXX< 1000,000,,000. the last digi must be XX3, so the answers must be : 993 or 983, or 973, or 963, or 953,...... 913. try 990x990x990(easy to calculate by hand) we find it the closest one. so the answer is 993.

[@jeffw1267]

Really, you don't even have to calculate by hand. If 1000^2 is 1,000,000, then 990 is 10 away from 1000. The 990 is multiplied by itself, so the 1,000 at the start of the number would be expected to drop from 1000 to 990 to 980, because of the ten difference between 1000 and 990 counted twice. If I take 990^2 and multiply by 990 to get 990^3, I'd expect to get about 970 as the first digits of the number, because I'd lose another ten just like when I squared the number.
So 990^3 is about 970,000,000. The starting number is more than that, but less than 1000^3. The only cube root in that range that will give a final digit of 7 is 993 (because 3^3 is 27, with a final digit of 7). Every other final digit, when cubed, will give a unique final digit that is not 7. So 993 must be the answer.
It took me about five seconds to find the cube root in my head. Mental calculators of the past would do such calculations and make them seem more difficult than they really were.
I knew that 99^2=9801 without having to calculate it, and that's why I was able to find the answer so fast. Anybody can do it if they know a little about the properties of numbers.

[@tubetigeerr]

see the 7, i know that the last digi must be 3. because 3x3x3 =xxx7.

[@walterwen2975]

A Nice Algebra Problem: ³√979146657 =?
979146657 = 1000000000 - 20853343 = 10⁹ - 20853000 - 343
= 10⁹ - 21000000 + 147000 - 343 = (10³)³ - 3(10³)²(7) + 3(10³)(7²) - 7³ = (10³ - 7)³
³√979146657 = ³√[(10³ - 7)³] = 10³ - 7 = 1000 - 7 = 993