By how many of the following numbers is $2^{12} - 1$ divisible? $2, 3, 5, 7, 10, 11, 13, 14$

### Concept & Formula This problem tests algebraic factorization, specifically using the difference of two squares to break down large exponents into prime factors. $$ x^2 - y^2 = (x - y)(x + y) $$ ### Step-by-Step Solution * **Given:** The expression $2^{12} - 1$ and a list of potential divisors: $2, 3, 5, 7, 10, 11, 13, 14$. * **Rewrite as difference of squares:** We can express $2^{12}$ as $(2^6)^2$. $$(2^6)^2 - 1^2$$ * **Factorize:** Apply the formula $(x - y)(x + y)$. $$(2^6 - 1)(2^6 + 1)$$ * **Evaluate the brackets:** We know $2^6 = 64$. $$(64 - 1)(64 + 1)$$ $$63 \times 65$$ * **Find prime factors:** Break down $63$ and $65$ into their prime factorization. * $63 = 9 \times 7 = 3^2 \times 7$ * $65 = 5 \times 13$ So, the complete factorization of $2^{12} - 1$ is $3^2 \times 5 \times 7 \times 13$. * **Compare with the given list:** * The individual prime factors present are $3, 5, 7$, and $13$. * Let's check the composites in the list: $10$ requires a factor of $2$ (none present), $11$ is not present, $14$ requires a factor of $2$ (none present). * Therefore, the numbers from the list that divide the expression are exactly four: **$3, 5, 7, 13$**. ### Exam Strategy & Shortcut Memorize powers of 2 up to $2^{12}$. $2^{10} = 1024$, so $2^{12} = 4096$. The question asks for factors of $4096 - 1 = 4095$. Any number ending in 5 is odd, eliminating all even divisors immediately ($2, 10, 14$). $4095$ ends in 5, so $5$ is a factor. Sum of digits: $4+0+9+5=18$ (divisible by $9$, thus by $3$). Divide $4095 \div 45 = 91$. $91$ is famously $7 \times 13$. Thus, the factors from the list are $3, 5, 7, 13$ (a total of 4 numbers). ### Common Pitfall A common mistake is forgetting to factor further after reaching $63 \times 65$ and trying to manually divide $4095$ by every single number on the list, which wastes valuable time and risks arithmetic errors. ### Final Answer Therefore, the correct answer is **4**.