A factorial \(n!\) is the product \(1 \cdot 2 \cdot 3 \cdots n\), with the convention that \(0! = 1\). For reference, \(10! = 3{,}628{,}800\). Factorials connect directly to combinations through the binomial coefficient \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\); for example, \(\binom{10}{3} = \frac{10!}{3! \cdot 7!} = 120\).
To count trailing zeros in \(n!\), count factors of 5, since factors of 2 are always more abundant: \(\lfloor n/5 \rfloor + \lfloor n/25 \rfloor + \lfloor n/125 \rfloor + \cdots\). For 25! this gives \(\lfloor 25/5 \rfloor + \lfloor 25/25 \rfloor = 5 + 1 = 6\) trailing zeros. The same method finds the highest prime power dividing a factorial: the largest power of 2 in 100! is \(\lfloor 100/2 \rfloor + \lfloor 100/4 \rfloor + \lfloor 100/8 \rfloor + \lfloor 100/16 \rfloor + \lfloor 100/32 \rfloor + \lfloor 100/64 \rfloor = 50 + 25 + 12 + 6 + 3 + 1 = 97\), and the highest power of 3 in 50! is \(\lfloor 50/3 \rfloor + \lfloor 50/9 \rfloor + \lfloor 50/27 \rfloor = 16 + 5 + 1 = 22\). To estimate the number of digits in \(n!\), use \(\lfloor \log_{10}(n!) \rfloor + 1\), giving 65 digits for 50!.
Perfect squares and cubes form a familiar landscape worth memorizing: squares are \(n^2\) for \(n \geq 0\), starting 0, 1, 4, 9, 16, 25, … and cubes are \(n^3\), starting 0, 1, 8, 27, 64, 125, 216, …. Between 50 and 200 the perfect squares are 64, 81, 100, 121, 144, 169, and 196, seven in total. A square always ends in 0, 1, 4, 5, 6, or 9, so no square can end in 2, 3, 7, or 8; the last two digits of \(n^2\) also follow a fixed 22-element pattern mod 100. The factorization \(x^2 - y^2 = (x-y)(x+y)\) is the master tool for difference-of-squares problems, while \(x^2 + y^2\) does not factor over the integers (it does over the complex numbers as \((x+iy)(x-iy)\)). Pythagorean triples make \(a^2 + b^2\) a perfect square: the smallest is (3, 4, 5), and every primitive triple takes the form \(a = m^2 - n^2\), \(b = 2mn\), \(c = m^2 + n^2\) with coprime \(m > n\) of opposite parity. The irrationality of \(\sqrt{2}\) follows a classic proof by contradiction: if \(\sqrt{2} = p/q\) in lowest terms, then \(2q^2 = p^2\), forcing \(p\) even, then \(q\) even, contradicting lowest terms.
Repeating decimals convert to fractions through a single algebraic move. Let \(x\) equal the repeating block, then \(10^k x - x\) (where \(k\) is the block length) clears the repeating part. So \(0.\overline{1} = 1/9\), \(0.\overline{3} = 1/3\), \(0.1\overline{6} = 1/6\) (since \(0.1\overline{6} = 0.2 - 0.0\overline{3} = 1/5 - 1/30 = 1/6\)), \(0.\overline{142857} = 1/7\), and \(0.\overline{123} = 123/999 = 41/333\). A terminating decimal like 0.625 simply reduces: \(625/1000 = 5/8\).