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?
Option 1 -
6
Option 2 -
10
Option 3 -
8
Option 4 -
4
Asked by Shiksha User
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1 Answer
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Correct Option - 1
Detailed Solution: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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