$$ 16k \equiv 1 \pmod{5} \implies k \equiv 1 \pmod{5} $$

February 22, 2026

$16k \equiv 1 \pmod{5} \implies k \equiv 1 \pmod{5}$

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Solution: Let $t = \sin x + \csc x$. Note $\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{1}{\sin x \cos x}$. Let $y = \sin x \cos x = \frac{1}{2}\sin 2x$, so $\tan x + \cot x = \frac{2}{\sin 2x}$. The expression becomes $\left(\frac{2}{\sin 2x}\right)^2 + (\sin x + \csc x)^2$. Let $z = \sin x + \csc x \geq 2$ by AM-GM. Substitute $u = \sin 2x$, then minimize $\frac{4}{u^2} + \left(\frac{1}{\sin x} + \sin x\right)^2$. After calculus or substitution, the minimum occurs at
\boxed{9}Question: What is the smallest positive integer that is a multiple of 16 and one more than a multiple of 5?
Solution: Let the integer be $ n $. We have $ n \equiv 0 \pmod{16} $ and $ n \equiv 1 \pmod{5} $. Let $ n = 16k $. Substituting into the second congruence:
$$ n = 16(5m + 1) = 80m + 16 $$
The smallest positive $ n $ occurs when $ m = 0 $, giving $ n = 16 $. Verifying: $ 16 \div 16 = 1 $ and $ 16 - 1 = 15 $, which is divisible by 5.
Question: What is the sum of all the squares of the divisors of $ 12 $?

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