2016 AMC 8 Problems/Problem 15

goat

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 $\boxed{\textbf{(C)}\ 32}$ .

~CHECKMATE2021

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 $2\cdot16=\boxed{\textbf{(C)}~32}.$

~ CHECKMATE2021

Solution 3 (Lifting the exponent)

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 $2^5=\boxed{\textbf{(C)} 32}$ .

~CHECKMATE2021

Solution 4 (Brute Force)

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 $\boxed{\textbf{(C)}~32}.$

-themathgood

Note: This is not the fastest way, and is not recommended.

I completely agree even though I did it./reader

Video Solution by OmegaLearn

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

~ pi_is_3.14

Video Solution

https://youtu.be/mZCOgH2kVuE

~savannahsolver

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/fWEwuLKZ7jY

~Education, the Study of Everything

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

Page

Toolbox

Search