Math Olympiad Radical Simplification | Nice Cube Root | Can You Simplify this ?

Math Olympiad Radical Simplification | Nice Cube Root
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[@pkpain6306]

Excellent.

[@superacademy247]

Many thanks!

[@alexandergroysman2357]

Taking cube root has an easy way to evaluate. Last digit is 7, meaning result has to end with 3(since 3*3*3 is 27 ends with 7). Now drop last 3 digits - you get 979146. Number closest to it in cube is 99(since 100 is 100000). So answer would be 993.

[@barneyhall2753]

One problem, nothing in the question indicates that the answer is an integer.

[@user-qr7dw4hk6x]

Меньше минуты: на конце 7, значит, оканчивается на 3 . Почти миллиард, значит очень близко к тысяче, откуда 993. С комсомольским приветом оли пиаде.

[@mustafasevimli9011]

Aklımdan geçeni bul şeklinde bir soru

[@rainerzufall42]

Test!

[@rainerzufall42]

TH-cam is a mess! It is incontinent..

[@rainerzufall42]

Number theory (asuming, we are looking for an integer) tells us, that only ***3 ^ 3 gives us ***7.
My approach: x = 1000 - 10 * y - 7 = 993, 983, 973, ...

[@rainerzufall42]

= 979.146.657 - 29.581.470 y + 297.900 y^2 - 1.000 y^3.
I could stop here, because we know, what we've already suspected: y = 0.

[@rainerzufall42]

But we owe us a test: y = 1:
983^3 = 979.444.557 - 29.582.470 = 949.862.087. Using the calculator => check!

[@rainerzufall42]

The original solution is 993.

[@MrGeorge1896]

We know the cube root we are looking for is a natural number.
1000³ = 1,000,000,000 is greater than 979,146,657 but 900³ = 729,000,000 is smaller, so the cube root we are looking for is between 901 and 999.
As the last digit is a 7 and b³ ≡ 7 (mod 10) is only true for b = 3 (3*3*3 = 27) we know the last digit must be a three, so the cube root is 903 + 10a.
In order to find a we have to solve (900 + 10a + 3)³ ≡ 57 (mod 100) via the binomial formula. All terms with 900 cancel out as 900 ≡ 0 (mod 100) so we get 3 * 100a² * 3 + 3 * 10a * 3² + 3³ ≡ 57 (mod 100). 100a² is divisible by 100 so it cancels out, too.
270a + 27 ≡ 57 (mod 100)
270a ≡ 30 (mod 100)
27a ≡ 3 (mod 10)
7a ≡ 3 (mod 10)
So a = 9 is the only possible solution: 7 * 9 = 63 ≡ 3 (mod 10)
So the only possible natural number which gives 979,146,657 when cubes is 993.

[@superacademy247]

Thanks for bringing in wealth of experience in modular Math.🙏🙏🙏

[@rainerzufall42]

Also a nice approch, with even more number theory, and with adding instead of subtracting. Signs are ugly, but I've suspected the solution near 1000, not near 900. Thus 1000 - 10y - 7 instead of 900 + 10a + 3. Essentially the same! For complicate problems, I would have probably chosen this approach, but I didn't fear the calculation, and I got y = 0, which made it pretty easy, so I've calculated 983^3 as well...