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HKD1710
If a, b, and c are three consecutive odd integers such that 10 < a < b < c < 20 and if b and c are prime numbers, what is the value of a + b?

(A) 24
(B) 28
(C) 30
(D) 32
(E) 36
 
Note:

1. a,b,c are consecutive odd integers.
2.b and c are prime numbers.

Range: 10<a<b<c<20.

All the odd withing the range: 11,13,15,17,19.

***we have to keep in mind that last 2 consecutive odd and prime.

Thus , a<b<c = 15<17<19.

a + b = 15 + 17 = 32.

The best answer is D.
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If a, b, and c are three consecutive odd integers such that 10 < a < b < c < 20 and if b and c are prime numbers, what is the value of a + b?

a, b, and c are three consecutive odd integers and b and c are prime numbers
two consecutive odd prime numbers between 10 and 20 are (11, 13) and (17,19)

if,b=11,a=9 (rejected because not in the given range)
so,b=17,c=19
and a=15
so,a+b=15+17=32

correct answer D
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Archit3110
only possible relation would be:

10<15<17<19<20

15+17= 32 IMO D



If a, b, and c are three consecutive odd integers such that 10 < a < b < c < 20 and if b and c are prime numbers, what is the value of a + b?

(A) 24
(B) 28
(C) 30
(D) 32
(E) 36
 


It is written nowhere that a is not prime, 24 IMO is also possible, badly posed question.

Posted from my mobile device
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Consecutive odd integers between 10 and 20 are 11, 13, 15, 17, 19.
B and C are prime numbers. Since A is at least 11, then B and C must be 17 and 19.

Hence A = 15, B = 17, C = 19.

A + B = 32

Answer: D
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