MathProblemSolver101

@Melody

Note $153 = 3^2 \cdot 17$

Rearrange the orig. equation: $x^3 = 5 - 153y^3$

You try all the possible combinations of possible modulos.

At one point or another, you end up with $\pmod 9$

So, if we try to simplify mod 9...

$x^3 \equiv 5 \pmod 9$

Consider this.

You have $x = n-1, n, n+1, n = 3k, k \in \mathbb{Z}$

If you have $x=n,$ obviously $x^3 \equiv 0 \pmod 9$

But…

If you have $x=n-1,$ you have $n^3 - 3n^2 + 3n - 1 \equiv -1 \pmod 9$

If you have $x = n+1,$ you have $n^3 + 3n^2 - 3n + 1 \equiv 1 \pmod 9$

So $x^3 \equiv \{-1, 0, 1\} \pmod 9$

But we have $x^3 \equiv 5 \pmod 9$ which is a contradiction. Thus there are 0 solutions.

x + x-13 + 1/3 x = 85

7/3 x = 98

1/3x = 14

x = 42

x-13 = 29

1/3 x = 14

(42,29,14)

I think PIE would be faster, though this is very rigorous without a doubt.

$\frac{\lfloor \frac{36}{4} \rfloor + \lfloor \frac{36}{6} \rfloor - \lfloor \frac{36}{24}\rfloor}{36}$

$\frac{14}{36}$

$\frac{7}{18}$

In two hours, one airplane has traveled 840 km and the other 750 km.

By the cos rule, we have $1500^2 = 840^2 + 750^2 - 2(750)(840) \cos \gamma$

$981900 = - 2(750)(840) \cos \gamma$

$\cos \gamma = 0.77928571428$

$\gamma = \cos ^{-1} 0.77928571428$

$\gamma = 38.739424597856$

$\gamma = \boxed{39^{\circ}}$

$8 = 2^3$

$2^4, 2^6, 2^8$

$2^4 \cdot {3^2, 5^2, 7^2}$

$2^6 \cdot {3^2}$

$3+3+1=\boxed{7}$

$k=3$

1+7+13+...+259

floor(259/6)=43

44(1)+43(6)=44+43*6=44(mod 6) = 2 (mod 6)

remainder is 2.

Guest, just because an answer was incorrect doesn't mean you can't see the flaw or the generic way of solving.