The number $(2^{48} - 1)$ is exactly divisible by two numbers between 60 and 70. The numbers are\n\n(A.A.O. Exam, 2010)

### Concept & Formula The problem tests the algebraic divisibility rule for the difference of powers. When a number is in the format $x^n - y^n$, it is always divisible by $(x - y)$ for any integer $n$, and divisible by $(x + y)$ if $n$ is an even integer. $$x^n - y^n = (x-y)(x^{n-1} + ... + y^{n-1})$$ ### Step-by-Step Solution * **Given:** The expression is $(2^{48} - 1)$. * **Calculation:** We need to find factors between 60 and 70. Let's rewrite $2^{48}$ so the base is close to 60-70. * We know that $2^6 = 64$. We can rewrite the power using exponent rules: $$2^{48} = (2^6)^8 = 64^8$$ * Now, substitute this back into the original expression: $$(2^{48} - 1) = (64^8 - 1^8)$$ * Here, $x = 64$, $y = 1$, and $n = 8$ (which is even). * According to the divisibility rule, $(64^8 - 1^8)$ is divisible by: 1. $(x - y) = 64 - 1 = 63$ 2. $(x + y) = 64 + 1 = 65$ * Both 63 and 65 fall exactly between 60 and 70. ### Exam Strategy & Shortcut When dealing with $2^n - 1$ where you need factors in a specific range, immediately look for a power of 2 near that range. $2^6 = 64$ is right in the middle of 60 and 70. Reformat $2^{48}$ as $(2^6)^8$. Applying the basic $a^2 - b^2 = (a-b)(a+b)$ logic recursively instantly gives you $64-1=63$ and $64+1=65$. ### Common Pitfall Students often try to prime factorize $2^{48}-1$ directly using complex binomial expansions or get stuck dividing by small primes like 3 or 5. Always manipulate the base to match the target range first. ### Final Answer Therefore, the correct answer is **63 and 65**.