Problem

What is the largest power of $2$ that is a divisor of $13^4 - 11^4$ ?

$\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128$

Solution 1

First, we use difference of squares on $13^4 - 11^4 = (13^2)^2 - (11^2)^2$ to get $13^4 - 11^4 = (13^2 + 11^2)(13^2 - 11^2)$ . Using difference of squares again and simplifying, we get $(169 + 121)(13+11)(13-11) = 290 \cdot 24 \cdot 2 = (2\cdot 8 \cdot 2) \cdot (3 \cdot 145)$ . Realizing that we don't need the right-hand side because it doesn't contain any factor of 2, we see that the greatest power of $2$ that is a divisor $13^4 - 11^4$ is $\boxed{\textbf{(C)}\ 32}$ .

Solution 2 (a variant of Solution 1)

Just like in the above solution, we use the difference-of-squares factorization, but only once to get $13^4-11^4=(13^2-11^2)(13^2+11^2).$ We can then compute that this is equal to $48\cdot290.$ Note that $290=2\cdot145$ (we don't need to factorize any further as $145$ is already odd) thus the largest power of $2$ that divides $290$ is only $2^1=2,$ while $48=2^4\cdot3,$ so the largest power of $2$ that divides $48$ is $2^4=16.$ Hence, the largest power of $2$ that is a divisor of $13^4-11^4$ is $2\cdot16=\boxed{\textbf{(C}~32}.$

Aops-g5-gethsemanea2 (talk) 05:16, 4 January 2023 (EST)

Solution 3 (Lifting the exponent)

Let $k=13^4-11^4.$ We wish to find the largest power of $2$ that divides $n$ .

Denote $v_p(k)$ as the largest exponent of $p$ in the prime factorization of $n$ . In this problem, we have $p=2$ .

Lifting the Exponent on $n$ ,

$v_2(13^4-11^4)=v_2(13-11)+v_2(2)+v_2(13+11)$ $=v_2(2)+v_2(2)+v_2(24)$ $=1+1+3=5.$

Therefore, exponent of the largest power of $2$ that divids $13^4-11^4$ is $5,$ so the largest power of $2$ that divides this number is $2^5=\boxed{\textbf{(C)} 32}$ .

-Benedict T (countmath1)

Video Solution by OmegaLearn

https://youtu.be/HISL2-N5NVg?t=3705

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Video Solution

https://youtu.be/mZCOgH2kVuE

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