AMC 8 2016 Problem 15

AMC8 2016 年第 15 题

整除与因数★★★☆☆

What is the largest power of 2 that is a divisor of

13^4 - 11^4

是

A.8
B.16
C.32
D.64
E.128

正确答案 · Correct Answer

解析 · Solution

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 32. 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

2\cdot16=32

~ CHECKMATE2021 Let $n=13^4-11^4.$ 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 $p=2$. By the Lifting the Exponent Lemma on $n$, $v_2(13^4-11^4)=v_2(13-11)+v_2(4)+v_2(13+11)-1$ $=v_2(2)+v_2(4)+v_2(24)-1$ $=1+2+3-1=5.$ 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 $2^5=32$. We can simply take 13 to the 4th power, which is 28561. We subtract that by 11 to the 4th power, which is 14641 (You can use Pascal's Triangle to find this). Finally, subtract the numbers to get 13920. To test the options, since we need the largest one, we can go from top down. Testing, we see that both D and E are decimals, and 32 works. So, our answer is 32. -themathgood

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