hellospeedmind

a(n) approaches 0.

I just plugged in a couple of values and it led to a pattern. For example, plug in 0 to n, and using the fact that a(0)=0, you will get a(1)=1. Then plug in 1 to n, and you will get a(2)=1/2. This will be an infinite series, and for every consecutive term, the result will be divided by 2.

1, 1/2, 1/4, 1/8, ... , 1/2^(n)

- hellospeedmind

According to Wolfram Alpha, \(\frac{120301232_4}{8}\) is equal to \(\frac{50743}{4}\) in base 10, which means that the remainder is 6. Does that help?

https://www.wolframalpha.com/input/?i=%28120301232_4%29%2F8

Thank you for your solution.

Ok. Thank you

Yes. I got this problem from my friend. When I searched it up, I found a very similar AMC 10 problem. That's probably where he got it from.

https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_24

Factor out the expression.

\(2^{28}(2^{6}-1)\)

\(2^{28}(64-1)\)

\(2^{28}*63\)

Since all the factors of \(2^{28}\) are 2's, for now, the smallest prime factor is 2.

Next factor out 63 into \(3^{2}*7\). 

The prime factors in this expression are 2,3, and 7 now.

7, being the largest, is the greatest prime factor of this expression.

The first condition should be followed for numbers that are lower than or equal to 1.

The second condition should be used for numbers that are higher than 1.

Since f^-1(-3) follows the first condition, do 2-(-3)=5.

Similarly, since f^-1(0) also follows the first condition, do 2-0=2.

Since f^-1(3) follows the second condition, as 3 is bigger than 1, do 2*(3)-3^2 = 6-9 = -3

5 + 2 - 3 = 4

The answer is 4. 

Ok. Thank you for your answer

I don't think you counted the 11 case. 2511 would also work because 2+5+1+1=9

For AB I got 11, 02, 20, 29, 38, 47, 56, 65, 74, 83, and 92. I'm not sure if that's all of them.