Example 1. Given that 1000002000001 has a prime factor greater than 9000. Find it.
Example 2. Let A = {(a, b, c) : a, b, c are prime numbers satisfying a < b < c and a + b + c = 30}. Find the number of elements in A.
Of course, you can find the solution just below, but it is highly recommended that, you first try to solve it on your own.
Just remember the words of Paul Halmos, who says
the only way to learn mathematics is to do mathematics.
Solution 1. Seeing the given number 1000002000001, it’s difficult to proceed further. But a closer look might help us. We can write
1000002000001
= 1000000000000 + 2000000 + 1
= 1012 + 2 × 106 + 1
= (106)2 + 2 × 106 + 1
= (106 + 1)2
= [(102)3 + 13]2
= [(102 + 1){(102)2 − 102 + 1}]2
= (102 + 1)2[(102)2 − 102 + 1]2
= (100 + 1)2[10000 − 100 + 1]2
= 1012 × 99012.
Therefore, the prime sought is 9901.
Solution 2. In essence the question asks in how many ways 30 can be expressed as a sum of three distinct primes.
Now, 30 is an even number and we know that the sum of three even numbers is odd. Not all three of these primes can be odd. Following table shall help us.
| a | b | c | a + b + c |
| Even | Even | Odd | Odd |
| Even | Even | Even | Even |
| Even | Odd | Even | Odd |
| Even | Odd | Odd | Even |
| Odd | Even | Even | Odd |
| Odd | Even | Odd | Even |
| Odd | Odd | Even | Even |
| Odd | Odd | Odd | Odd |
Since 2 is the only even prime and a < b < c, all of a, b and c cannot be even.
Now consider a = 2 and hence b + c = 28. Since b < c, the only possible values b can take are 3, 5, 7, 11 and 13. From these assumed values of b, c = 28 – b is also a prime only when b = 5 and 11.
So there are only two members in A and they are (2, 5, 23) and (2, 11, 17).