Question

The given answer was 67 i want the explanation for answer

Answer 1

$<marquee>☺☺Hi there☺☺</marquee>$

Lets suppose the three numbers be a , b and c . where a <b<c

Now , if the sum of three prime numbers is to be 100 ( even ) , then one of the numbers must be 2 . as

odd+odd + even = even

even + even + even = even ( not possible as only even prime number is 2 )

now difference between other 2 is 36 hence we get a is 2 .

a + b + c = 100

b + c = 98

c -36 + c = 98

c = 67 ; b = 31 ; a = 2
So the largest no. among three numbers is 67.


Thanks ♥♥

Answer 2

Okay you consider the first option 73 if you subtract 36 from it, you will get 42 which is not at all prime.Then if you take 91 it is a multiple of 13.13×7=91.So 91 is not prime.57 is not prime because it's a multiple of 3.Then you have the only possible option is 67.Now consider 67 it is a prime number.Now if you subtract 36 from it you will get 31 which too is prime.So the three prime numbers whose sum is 100 is 31+67+2.