Question 282537: If x, y and z are three consecutive odd integers such that 10< x < y < z<20 and if y and z
are prime numbers, what is the value of x + y?
If you can please guide me step by step what to do to solve this problem
I would really appreciated.
Found 2 solutions by jim_thompson5910, richwmiller:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! First list the primes from 10 to 20: 11, 13, 17, 19
The value of 'y' can't be 11 since there are no integers between 10 and 11. In other words, there are no integer solutions to 10 < x < 11
Similarly, 'y' can't be 19 since there are no integer solutions to 19 < z < 20.
So the value of 'y' must be either 13 or 17. If 'y' is 13, then 'z' must be 15 (since all three are consecutive odd integers). However, 'z' is supposed to be prime and 15 is NOT prime. So 'y' can't be equal to 13 either.
By process of elimination, 'y' must be equal to 17. So z=y+2=17+2=19 (since all three are consecutive odd integers) and 'x' is two integers down from 'y' which means that x=y-2=17-2=15
So x+y=15+17=32
Answer by richwmiller(17219)  (Show Source):
You can put this solution on YOUR website! Did this one a few days ago.
Odd integers between 10 and 20 are 11,13,15,17,19
so possible groups are
11,13,15
13,15,17
15,17,19
15 is not prime so can't be second nor third
that leaves only 15,17 19
so 15+17=32
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