Folding fun

Each fold will reduce its length by $1/2$ . So, the answer will be the largest number $n$ such that $2^n < 90$ . So, $n = \boxed{64}$

Each fold will reduce the string's length by $\dfrac{1}{2}$ , therefore the length after each fold forms a geometric progression with $r= \dfrac{1}{2}$

The first fold, $a = \dfrac{90}{2} = 45$ mm (Note: 1 cm = 10 mm)

Now, we want to find the maximum value of $n$ such that $T_n > 1$ mm

$45\left(\dfrac{1}{2}\right)^{n-1} > 1\\ \left(\dfrac{1}{2}\right)^{n-1} > \dfrac{1}{45}\\ \log_{10} \left(\left(\dfrac{1}{2}\right)^{n-1}\right) > \log_{10} \left( \dfrac{1}{45} \right)\\ (n-1)\log_{10} \left(\dfrac{1}{2}\right) > \log_{10} \left( \dfrac{1}{45} \right)\\ (n-1) < \dfrac{\log_{10} \left( \dfrac{1}{45} \right)}{\log_{10} \left(\dfrac{1}{2}\right) }\\ (n-1) <5.49\\ n < 6.49$

Therefore, the maximum number of folds is $\boxed{6}$

Note: The symbol is changed to " $<$ " because we divide $\log_{10} \left(\dfrac{1}{2} \right)$ , which is a negative number