M26-17

Official Solution:

What is the largest prime number that can be obtained by adding three 2-digit numbers formed from the list {1, 3, 6, 7, 7, 7}?

A. 97
B. 151
C. 209
D. 211
E. 219


What is the largest possible sum of three 2-digit numbers that can be formed using the numbers {1, 3, 6, 7, 7, 7}? To find the maximum sum, we start by maximizing the first digit of each number. We try 76, 73, and 71, which gives us a sum of 220, but since it's even, it cannot be a prime number.

Next, we try the next largest sum by switching the digits in 76 to get 67, giving us the numbers 67, 73, and 71. We then need to check whether the sum of these three numbers, which is 211, is a prime number.

The most efficient way to prove that a large enough positive integer \(x\) is prime, is to prove that it cannot be divided by any primes up to \(\sqrt{x}\). Therefore, to prove that 211 is prime, we need to prove that it cannot be divided by any primes up to \(\sqrt{211}\), which is less than 15 (since \(15^2 = 225\)). It is easy to see that 211 is not divisible by 2, 3, or 5. Therefore, we need to check divisibility by 7, 11, and 13. To check whether 211 is divisible by any of the primes 7, 11, or 13, we find easily identifiable multiples of these primes close to 211. We find that 210 = 7*30 is a multiple of 7, so 211 is not divisible by 7. 220 = 11*20 is a multiple of 11, so 211 is not divisible by 11. 260 = 13*20 is a multiple of 13, so is 260 - 13*4 = 208, thus 211 is not divisible by 13 either. Therefore, 211 is not divisible by any primes up to \(\sqrt{211}\), so it must be a prime number.


Answer: D