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

Math Olympiad | A Nice Algebra Problem

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[danmimis]

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]

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

[walterwen]

A Nice Algebra Problem: ³√979146657 =?
979146657 = – 20853343 = 10⁹ – 20853000 – 343
= 10⁹ – + 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