I did this two ways.
The first way used a combination of pure processing power and logic. You know the even numbers - 62, 64, 66, and 68 will never be prime. 65 will always be divisible by 5. That leaves 61, 63, 67, and 69. 63 and 69 are divisible by 9 and 3 respectively (just plug and chug). That leaves you with 2 primes - 61 and 67 - the sum of which is 128.
The second way - applying divisibility rules - seems more elegant and more efficient in the long run:
For 2: All even numbers. This eliminates 62, 64, 66, and 68.
For 3: Add up all the digits of the number. If the sum is divisible by 3, the number is divisible by 3. This eliminates both 63 and 69.
For 4: If the last 2 digits of the number are divisible by 4, the entire number is divisible by 4. This does not apply to any of our remaining numbers - 61, 65, and 67.
For 5: All numbers ending in 5 or 0. This eliminates 65, leaving only 61 and 67 in the running for primes.
For 6: If the number is divisible by BOTH 2 & 3 (see first two rules), then the number is also divisible by 6. This does not apply to 61 and 67 since neither is divisible by 2 or 3.
For 7: No concise rule that I know of. I recommend long-division or mental math. Does anyone know a quick way to check for divisibility by7? Regardless, you can run through the multiples of 7 and realize none fall on 61 or 67 so this divisor doesn't check out.
For 8: If the last 3 digits are divisible by 8, then the whole number is divisible by 8. I didn't stick to this rule when using the divisibility rules approach. You know 2 is a prime factor of 8 meaning any number that has 8 as a multiple must be even. Since neither 61 nor 67 is even, 8 cannot be a divisor.
For 9: Add up all the digits and see if the resulting sum is divisible by 9. This does not work when applied to 61 or 67 and is redundant for 63 (we knocked 63 out in the "For 3" rule).
Therefore the only 2 primes in the set are 61 and 67.
Does anyone know a good way to check for divisibility by 7? Thanks.