What is the smallest positive integer, other than $1$, that is both a perfect cube and a perfect fourth power?

If $n$ is a perfect cube, then all exponents in its prime factorization are divisible by $3$. If $n$ is a perfect fourth power, then all exponents in its prime factorization are divisible by $4$. The only way both of these statements can be true is for all the exponents to be divisible by $\mathop{\text{lcm}}[3,4]=12$, so such an $n$ must be a perfect twelfth power. Since we aren't using $1^{12}=1,$ the next smallest is $2^{12}=4096.$

@Eq Nat answer (@OfNat.ofNat Nat (nat_lit 4096) (instOfNatNat (nat_lit 4096)))

theorem mathd_numbertheory_296
  (n : ℕ)
  (h₀ : 2 ≤ n)
  (h₁ : ∃ x, x^3 = n)
  (h₂ : ∃ t, t^4 = n) :
  4096 ≤ n := sorry


A line $\ell$ passes through the points $B(7,-1)$ and $C(-1,7)$.  The equation of this line can be written in the form $y=mx+b$; compute $m+b$.

The line through points $B$ and $C$ has slope $\dfrac{-1-7}{7-(-1)}=-1$.  Since $(7,-1)$ lies on the line, the line has equation $$y-(-1)=-1(x-7),$$or $y = -x + 6$.  Thus $m=-1$, $b=6$, and $m+b=-1+6=5$.

@Eq Real answer
  (@OfNat.ofNat Real (nat_lit 5)
    (@instOfNatAtLeastTwo Real (nat_lit 5) Real.instNatCast
      (@instNatAtLeastTwo (@OfNat.ofNat Nat (nat_lit 3) (instOfNatNat (nat_lit 3))))))

theorem mathd_algebra_142
  (m b : ℝ)
  (h₀ : m * 7 + b = -1)
  (h₁ : m * (-1) + b = 7) :
  m + b = 5 := sorry


Five plus $500\%$ of $10$ is the same as $110\%$ of what number?

@Eq Real answer
  (@OfNat.ofNat Real (nat_lit 50)
    (@instOfNatAtLeastTwo Real (nat_lit 50) Real.instNatCast
      (@instNatAtLeastTwo (@OfNat.ofNat Nat (nat_lit 48) (instOfNatNat (nat_lit 48))))))


Given that $(1+\sin t)(1+\cos t)=5/4$ and
:$(1-\sin t)(1-\cos t)=\frac mn-\sqrt{k},$
where $k, m,$ and $n_{}$ are [[positive integer]]s with $m_{}$ and $n_{}$ [[relatively prime]], find $k+m+n.$

From the givens,
$2\sin t \cos t + 2 \sin t + 2 \cos t = \frac{1}{2}$, and adding $\sin^2 t + \cos^2t = 1$ to both sides gives $(\sin t + \cos t)^2 + 2(\sin t + \cos t) = \frac{3}{2}$.  Completing the square on the left in the variable $(\sin t + \cos t)$ gives $\sin t + \cos t = -1 \pm \sqrt{\frac{5}{2}}$.  Since $|\sin t + \cos t| \leq \sqrt 2 < 1 + \sqrt{\frac{5}{2}}$, we have $\sin t + \cos t = \sqrt{\frac{5}{2}} - 1$.  Subtracting twice this from our original equation gives $(\sin t - 1)(\cos t - 1) = \sin t \cos t - \sin t - \cos t + 1 = \frac{13}{4} - \sqrt{10}$, so the answer is $13 + 4 + 10 = 027$.

@Eq Nat answer (@OfNat.ofNat Nat (nat_lit 27) (instOfNatNat (nat_lit 27)))

h₃

(1 - Real.sin t) * (1 - Real.cos t) = m/n - Real.sqrt k

theorem aime_1995_p7
  (k m n : ℕ)
  (t : ℝ)
  (h₀ : 0 < k ∧ 0 < m ∧ 0 < n)
  (h₁ : Nat.gcd m n = 1)
  (h₂ : (1 + Real.sin t) * (1 + Real.cos t) = 5/4)
  (h₃ : (1 - Real.sin t) * (1- Real.cos t) = m/n - Real.sqrt k):
  k + m + n = 27 := sorry


When a number is divided by 5, the remainder is 3. What is the remainder when twice the number is divided by 5?

If our number is $n$, then $n\equiv 3\pmod5$.  This tells us that  \[2n=n+n\equiv 3+3\equiv1\pmod5.\] The remainder is $1$ when the number is divided by 5.

@Eq Nat answer (@OfNat.ofNat Nat (nat_lit 1) (instOfNatNat (nat_lit 1)))

theorem mathd_numbertheory_185
  (n : ℕ)
  (h₀ : n % 5 = 3) :
  (2 * n) % 5 = 1 := sorry


Assuming $x\ne0$, simplify $\frac{12}{x \cdot x} \cdot \frac{x^4}{14x}\cdot \frac{35}{3x}$.

We have  \begin{align*}
\frac{12}{x \cdot x} \cdot \frac{x^4}{14x}\cdot \frac{35}{3x} &=
\frac{12 \cdot x^4 \cdot 35}{x^2\cdot 14x \cdot 3x}\\& = \frac{(4 \cdot 3) \cdot (5 \cdot 7) \cdot x^4}{(3 \cdot 2 \cdot 7)(x^2 \cdot x \cdot x)}\\
&= \frac{2\cdot 2 \cdot 3 \cdot 5 \cdot 7}{2 \cdot 3 \cdot 7}\cdot\frac{x^4}{x^{4}}\\
&= 2 \cdot 5 = 10.
\end{align*}

@Eq Real answer
  (@OfNat.ofNat Real (nat_lit 10)
    (@instOfNatAtLeastTwo Real (nat_lit 10) Real.instNatCast
      (@instNatAtLeastTwo (@OfNat.ofNat Nat (nat_lit 8) (instOfNatNat (nat_lit 8))))))


If $n$ is a multiple of three, what is the remainder when $(n + 4) + (n + 6) + (n + 8)$ is divided by $9$?

We see that $(n + 4) + (n + 6) + (n + 8) = 3n + 18.$ We can see that this must be a multiple of $9,$ since $18$ is a multiple of $9$ and $3n$ is as well, since we are given that $n$ is a multiple of $3.$ Therefore, our answer is $0.$

@Eq Nat answer (@OfNat.ofNat Nat (nat_lit 0) (instOfNatNat (nat_lit 0)))

theorem mathd_numbertheory_582
  (n : ℕ)
  (h₀ : 0 < n)
  (h₁ : 3∣n) :
  ((n + 4) + (n + 6) + (n + 8)) % 9 = 0 := sorry


Summing all three equations yields that $5a + 5b + 5c = -3 + 9 + 19 = 25$. Thus, $a + b + c = 5$. Subtracting this from each of the given equations, we obtain that $2a = -8, 2b = 4, 2c = 14$. Thus, $a = -4, b = 2, c =7$, and their product is $abc = -4 \times 2 \times 7 = -56$.

@Eq Real answer
  (@Neg.neg Real Real.instNeg
    (@OfNat.ofNat Real (nat_lit 56)
      (@instOfNatAtLeastTwo Real (nat_lit 56) Real.instNatCast
        (@instNatAtLeastTwo (@OfNat.ofNat Nat (nat_lit 54) (instOfNatNat (nat_lit 54)))))))

theorem mathd_algebra_338
  (a b c : ℝ)
  (h₀ : 3 * a + b + c = -3)
  (h₁ : a + 3 * b + c = 9)
  (h₂ : a + b + 3 * c = 19) :
  a * b * c = -56 := sorry