If P1 + P2 + P3 = 59, where P1, P2 and P3 are prime numbers and P1 < P2 < P3, then how many values can P3 take?
If P1 + P2 + P3 = 59, where P1, P2 and P3 are prime numbers and P1 < P2 < P3, then how many values can P3 take?
P1 + P2 + P3 = 59
P1 and P2 → minimize
Then P3 → minimum i.e. 47
P1 | P2 | P3 |
5 | 7 | 47 |
5 | 11 | 43 |
3 | 13 | 43 |
5 | 13 | 41 |
5 | 17 | 37 |
3 | 19 | 37 |
5 | 23 | 31 |
7 | 23 | 29 |
So, only possible value of P3 can take → 29, 31, 37, 41, 43, and 47
Total 6 values
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