Which three prime numbers make 121 if 1 is 10 more than the other?

Answer 1

Because only two prime numbers are multiplied together to make 121, I assume you mean which three prime numbers can be added together to make 121. I will also assume that all three prime numbers must be different.

Since one number is 10 more than another, we have the equation p + (p + 10) + q = 2p + 10 + q = 121, which can be changed to 2p + q = 111. Since 2p will be an even number, q must be an odd number in order to sum to 121 which is an odd number. The smallest odd prime is 3. If we substitute 3 for q, we have 2p + 3 = 111, so 2p = 108, and p would be 54. Thus, the correct value for p must be less than 54. So, let's look at primes less than 54 to find ones that will produce primes when 10 is added to them.

The prime numbers less than 54 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, and 53, but it cannot be the even prime 2. Our possible pairs are {3, 13}, {7, 17}, {13, 23}, {19, 29}, {31, 41}, {37, 47}, and {43, 53}. Any number ending with 3, will have the value of 2p ending in a 6, which when subtracted from 111 will have a final digit of 5, which will not be prime, so we can eliminate {3, 13} and {13, 23}. Try the first of each remaining pair as p in the equation q = 111 - 2p to see if the resulting value for q is prime. For the value of p = 7, q is the Prime number 97. For the value of p = 19, q is the prime number 73. For the value of p = 37, q is the prime number 37, which we will exclude as a possibility because then there is duplication of the prime number. So, the two possibilities are p = 7 and p = 19.

Thus the possible answers are
121 = 7 + 17 + 97
121 = 19 + 29 + 73