Math Aunty

Sandwich party, cake style!

Anyways:

As soon as I read this question, I had a theory that the answer wasn’t looking for a definitive number because all exponential functions go to infinity… so how could I answer this question using words! They would bring the same amount of sandwiches! So I decided to test my theory using the first 10 powers of 2 and the first 10 powers of 4… and this amazing prime number calculator http://www.math.com/students/calculators/source/prime-number.htm

I know that all powers of 4 are elements in the set of powers of 2, in other words

A = {a: a = 4^n, n = N}

B = {b: b = 2^n, n = N}

A is a subset of B.

So I started to look at the numbers around the elements in B because I know they’re also in A. I will bold the ones both in A and B.

(3, 4, 5) 3 and 5 are both prime numbers.

(7, 8, 9) 9 is not a prime number.

(15, 16, 17) 15 is not a prime number.

(31, 32, 33) 33 is not a prime number.

(63, 64, 65) neither 63 nor 65 is prime.

(127, 128, 129) 129 is not prime.

(255, 256, 257) 255 is not prime.

(511, 512, 513) Neither are prime.

(1023, 1024, 1025) Neither are prime

 

So it appears that after the first sandwich, there aren’t any sandwiches that are prime! So to further test my theory, I use the fact that A is a subset of B and I know to prove my conjecture (That they both brought 1 sandwich) I need to prove (or explain well enough to be a pretty good explanation):

(2^n) – 1 (n can not equal 1) is not prime and (2^n) + 1 (n can not equal 1) is not prime.

 

I know that there are NO primes that end with 5 or 0 (except 5, but since neither 2^n or 4^n can equal 6, that example is irrelevant) because all numbers ending in 5 or 0 are divisible by 2 and 5. All powers of 2 end in 2, 4, 6, or 8. All powers of 4 end in 4 or 6.

All the powers of 2 where n is a multiple of 4 end in 6! Because (2^n)-1 will be some number ending in 5, I know that all powers of 2 that are a multiple of 4 can not make a prime sandwich.

By the same logic above, I can be assured that all numbers 2^n that end in 4 also can’t make a prime sandwich because (2^n)+1 will be a number that ends in 5. Since all numbers that end in 4 are multiples of 2 (except n = 1 because that’s our famous (3, 4, 5) I can conclude that there are no number 2^n where n is a multiple of 2 (except n = 2) that makes a prime sandwich. All of 2^n that is a multiple of 2 is also a number in 4^n so except n = 1, there is no other sandwich ((4^n) – 1, 4^n, (4^n)+1) that Emek can bring except (3, 4, 5)

Also, I have realized that if a is any integer that is divisible by 3, 2^a + 1 is also an integer that is divisible by 3. Since 2^n are all even, they end in either 2, 4, 6, 8. I have already eliminated numbers that end in 6 and 4. So numbers that end in 2 and 8 are left. It appears to be that all (2^n) + 1 if (2^n) ends in 4 or 6, is a number that ends in 3 or 9 which is also a divisible by 3. I’m almost done with my proof, I just need to know how to prove that ((2^(2n+1)) + 1)/3 is an integer (because all I have left are all sandwiches whose meat is 2^n where n is an odd integer.

((2^((2n)+1)) + 1)/3 = a (where a is any integer)

2^((2n) + 1) + 1 = 3a (where a is any integer which is a multiple of 3)

DONE!!!

So based on my research, I know that Emek can only bring 1 sandwich (3, 4, 5) and Sandy can only bring 1 sandwich (3, 4, 5) which are both equally delicious so neither can bring a better sandwich.

🙂

Going through this tedious process made me realize that (if what I was doing was considered number theory) I kinda like number theory!